Continuity And Differentiability Class 12th Mathematics Part I CBSE Solution

Class 12^th Mathematics Part I CBSE Solution

Exercise 5.1

1. Prove that the function f (x) = 5x - 3 is continuous at x = 0, at x = - 3 and at…
2. Examine the continuity of the function f (x) = 2x^2 - 1 at x = 3.…
3. f (x) = x - 5 Examine the following functions for continuity.
4. f (x) = 1/x-5 Examine the following functions for continuity.
5. f (x) = x^2 - 25/x+5 Examine the following functions for continuity.…
6. f (x) = | x - 5| Examine the following functions for continuity.
7. Prove that the function f (x) = xn is continuous at x = n, where n is a positive…
8. Is the function f defined by f (x) = x , x less than equal to 1 5 , x1…
9. f (x) = 2x+3 , x less than equal to 2 2x-3 , x2 Find all points of discontinuity…
10. Find all points of discontinuity of f, where f is defined by
11. f (x) = |x|/x , x not equal 0 0 , Find all points of discontinuity of f, where f…
12. f (x) = x/|x| , x0 -1 Find all points of discontinuity of f, where f is defined…
13. f (x) = x+1 , x geater than or equal to 0 x^2 + 1 , x0 Find all points of…
14. f (x) = ll x^3 - 3 , x less than equal to 2 x^2 + 1 , x2 Find all points of…
15. f (x) = x^10 - 1 , x less than equal to 1 x^2 , x1 Find all points of…
16. Is the function defined by f (x) = x+5 , x less than equal to 1 x-5 x1 a…
17. f (x) = 3 0 less than equal to x less than equal to 1 4 1x3 5 3 less than equal…
18. Discuss the continuity of the function f, where f is defined by
19. f (x) = cc - 2 & x less than equal to -1 2x& - 1 less than equal to x less than…
20. Find the relationship between a and b so that the function f defined by f (x) =…
21. For what value of λ is the function defined by f (x) = r lambda (x^2 - 2x) , x…
22. Show that the function defined by g(x) = x - [x] is discontinuous at all…
23. Is the function defined by f (x) = x^2 - sin x + 5 continuous at x = π?…
24. Discuss the continuity of the following functions: (a) f (x) = sin x + cos x…
25. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.…
26. Find all points of discontinuity of f, where f (x) = sinx/x , x0 x+1 , x geater…
27. Determine if f defined by f (x) = r x^2sin 1/x , x not equal 0 0 x = 0 is a…
28. Examine the continuity of f, where f is defined by f (x) = sinx-cosx, x not…
29. Find the values of k so that the function f is continuous at the indicated…
30. Find the values of k so that the function f is continuous at the indicated…
31. Find the values of k so that the function f is continuous at the indicated…
32. f (x) = kx+1 , x less than equal to 5 3x-5 , x5 x = 5 Find the values of k so…
33. Find the values of a and b such that the function defined by f (x) = c 5 x less…
34. Show that the function defined by f (x) = cos (x^2) is a continuous function.…
35. Show that the function defined by f (x) = | cos x| is a continuous function.…
36. Examine that sin | x| is a continuous function.
37. Find all the points of discontinuity of f defined by f (x) = | x| - | x + 1|.…

Exercise 5.2

1. sin (x^2 + 5) Differentiate the functions with respect to x.
2. cos (sin x) Differentiate the functions with respect to x.
3. sin (ax + b) Differentiate the functions with respect to x.
4. sec (tan (root x)) Differentiate the functions with respect to x.…
5. sin (ax+b)/cos (cx+d) Differentiate the functions with respect to x.…
6. cos x^3 . sin^2 (x^5) Differentiate the functions with respect to x.…
7. Differentiate the functions with respect to x. 2 root cot (x^2)
8. cos(√x) Differentiate the functions with respect to x.
9. Prove that the function f given by f (x) = | x - 1|, x ∈ R is not differentiable…
10. Prove that the greatest integer function defined by f (x) = [x], 0 x 3 is not…

Exercise 5.3

1. 2x + 3y = sin x Find dy/dx in the following:
2. 2x + 3y = sin y Find dy/dx in the following:
3. ax + by^2 = cos y Find dy/dx in the following:
4. xy + y^2 = tan x + y Find dy/dx in the following:
5. x^2 + xy + y^2 = 100 Find dy/dx in the following:
6. x^3 + x^2 y + xy^2 + y^3 = 81 Find dy/dx in the following:
7. sin^2 y + cos xy = π Find dy/dx in the following:
8. sin^2 x + cos^2 y = 1 Find dy/dx in the following:
9. y = sin^-1 (2x/1+x^2) Find dy/dx in the following:
10. y = tan^-1 (3x-x^3/1-3x^2) ,- 1/root 3 x 1/root 3 Find dy/dx in the following:…
11. y = cos^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
12. y = sin^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
13. y = cos^-1 (2x/1+x^2) ,-1x1 Find dy/dx in the following:
14. y = sin^-1 (2x root 1-x^2) , - 1/root 2 x 1/root 2 Find dy/dx in the following:…
15. y = sec^-1 (1/2x^2 + 1) , 0x 1/root 2 Find dy/dx in the following:…

Exercise 5.4

1. e^x/sinx Differentiate the following w.r.t. x:
2. e^sin^-1x Differentiate the following w.r.t. x:
3. e^x^3 Differentiate the following w.r.t. x:
4. sin (tan-1 e-x) Differentiate the following w.r.t. x:
5. log (cos ex) Differentiate the following w.r.t. x:
6. e^x + e^x^2 + l l +e^x^5 Differentiate the following w.r.t. x:
7. root e^root x , x0 Differentiate the following w.r.t. x:
8. log (log x), x 1 Differentiate the following w.r.t. x:
9. cosx/logx , x0 Differentiate the following w.r.t. x:
10. cos (log x + ex), x 0 Differentiate the following w.r.t. x:

Exercise 5.5

1. cos x . cos 2x . cos 3x Differentiate the functions given in w.r.t. x.…
2. root (x-1) (x-2)/(x-3) (x-4) (x-5) Differentiate the functions given in w.r.t.…
3. (log x)cos x Differentiate the functions given in w.r.t. x.
4. xx - 2sin x Differentiate the functions given in w.r.t. x.
5. (x + 3)^2 . (x + 4)^3 . (x + 5)^4 Differentiate the functions given in w.r.t. x.…
6. (x + 1/x)^x + x^(1 + 1/x) Differentiate the functions given in w.r.t. x.…
7. (log x)x + xlog x Differentiate the functions given in w.r.t. x.
8. (sin x)x + sin-1 √x Differentiate the functions given in w.r.t. x.…
9. xsin x + (sin x)cos x Differentiate the functions given in w.r.t. x.…
10. x^xcosx^x + x^2 + 1/x^2 - 1 Differentiate the functions given in w.r.t. x.…
11. (xcosx)^x + (xsinx)^1/x Differentiate the functions given in w.r.t. x.…
12. xy + yx = 1 Find dy/dx of the functions.
13. yx = xy Find dy/dx of the functions.
14. (cos x)y = (cos y)x Find dy/dx of the functions.
15. xy = e(x - y) Find dy/dx of the functions.
16. Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 +…
17. Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below: (i)…
18. If u, v and w are functions of x, then show that in two ways - first by…

Exercise 5.6

1. x = 2at^2 , y = at^4 If x and y are connected parametrically by the equations…
2. x = a cos θ, y = b cos θ If x and y are connected parametrically by the…
3. x = sin t, y = cos 2t If x and y are connected parametrically by the equations…
4. x = 4t, y = 4/t If x and y are connected parametrically by the equations given…
5. x = cos θ - cos 2θ, y = sin θ - sin 2θ If x and y are connected parametrically…
6. x = a (θ - sin θ), y = a (1 + cos θ) If x and y are connected parametrically by…
7. x = sin^3t/root cos2t , y = cos^3t/root cos2t If x and y are connected…
8. x = a (cost+logtan t/2) y = asin If x and y are connected parametrically by the…
9. x = a sec θ, y = b tan θ If x and y are connected parametrically by the…
10. x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ) If x and y are connected…
11. If x = root a^sin^-1t , y = root a^cos^-1t , show that dy/dx = - y/x If x and y…

Exercise 5.7

1. x^2 + 3x + 2 Find the second order derivatives of the function
2. x^20 Find the second order derivatives of the function
3. x . cos x Find the second order derivatives of the function
4. log x Find the second order derivatives of the function
5. x^3 log x Find the second order derivatives of the function
6. ex sin 5x Find the second order derivatives of the function
7. e6x cos 3x Find the second order derivatives of the function
8. tan-1 x Find the second order derivatives of the function
9. log (log x) Find the second order derivatives of the function
10. sin (log x) Find the second order derivatives of the function
11. If y = 5 cos x - 3 sin x, prove that d^2y/dx^2 + y = 0
12. If y = cos-1 x, Find d^2 y/dx^2 in terms of y alone.
13. If y = 3 cos (log x) + 4 sin (log x), show that x^2 y2 + xy1 + y = 0…
14. If y = Aemx + Benx, show that d^2y/dx^2 - (m+n) dy/dx + mny = 0
15. If y = 500e7x + 600e-7x, show that d^2y/dx^2 = 49y .
16. If ey (x + 1) = 1, show that d^2y/dx^2 = (dy/dx)^2
17. If y = (tan-1 x)^2 , show that (x^2 + 1)^2 y2 + 2x (x^2 + 1) y1 = 2…

Exercise 5.8

1. Verify Rolle’s theorem for the function f (x) = x^2 + 2x - 8, x ∈ [- 4, 2].…
2. Examine if Rolle’s theorem is applicable to any of the following functions. Can…
3. If f : [- 5, 5] → R is a differentiable function and if f′(x) does not vanish…
4. Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3 in the interval [a, b], where…
5. Verify Mean Value Theorem, if f (x) = x^3 - 5x^2 - 3x in the interval [a, b],…
6. Examine the applicability of Mean Value Theorem for all three functions given in…

Miscellaneous Exercise

1. Differentiate w.r.t. x the function (3x^2 - 9x + 5)^9
2. sin^3 x + cos^6 x Differentiate w.r.t. x the function
3. (5x)3cos 2x Differentiate w.r.t. x the function
4. sin^-1 (x root x) , 0 less than equal to x less than equal to 1 Differentiate…
5. cos^-1/root 2x+7 ,-2x2 Differentiate w.r.t. x the function
6. cot^-1 (root 1+sinx + root 1-sinx/root 1+sinx - root 1-sinx) , 0x pi /2…
7. (log x)log x, x 1 Differentiate w.r.t. x the function
8. cos (a cos x + b sin x), for some constant a and b. Differentiate w.r.t. x the…
9. (sinx-cosx)^(sinx-cosx) , pi /4 x 3 pi /4 Differentiate w.r.t. x the function…
10. xx + xa + ax + aa, for some fixed a 0 and x 0 Differentiate w.r.t. x the…
11. x^x^2 - 3 + (x-3)^x^2 , for x 3 Differentiate w.r.t. x the function…
12. Find dy/dx, if y = 12 (1 - cos t), x = 10 (t - sin t), - pi /2 t pi /2…
13. Find dy/dx, if y = sin^-1x+sin^-1root 1-x^2 , 0 x 1
14. If x root 1+y+y root 1+x = 0 for , - 1 x 1, prove that dy/dx = - 1/(1+x)^2…
15. If (x - a)^2 + (y - b)^2 = c^2 , for some c 0, prove that [1 + (dy/dx)^2]^3/2/…
16. If cos y = x cos (a + y), with cos a 1, prove that dy/dx = cos^2 (a+y)/sina…
17. If x = a (cos t + t sin t) and y = a (sin t - t cos t), find d^2 y/dx^2 .…
18. If f (x) = |x|^3 , show that f″(x) exists for all real x and find it.…
19. Using mathematical induction prove that d/dx (x^n) = nx^n-1 for all positive…
20. Using the fact that sin (A + B) = sin A cos B + cos A sin B and the…
21. Does there exist a function which is continuous everywhere but not…
22. If | ccc f (x) & g (x) & h (x) 1 a | , prove that dy/dx = | ccc 1& (x) &…
23. If, y = e^acos^-1x ,-1 less than equal to x less than equal to 1 show that…

---

Exercise 5.1

Question 1.

Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Answer:

The given function is f(x) = 5x -3

At x = 0, f(0) = 5 × 0 – 3 = -3

Thus, 

Therefore, f is continuous at x = 0

At x = -3, f(-3) = 5 × (-3) – 3 = -18

Thus, 

Therefore, f is continuous at x = -3

At x = 5, f(5) = 5 × 5 – 3 = 22

 = 5 × 5 – 3 = 22

Thus, 

Therefore, f is continuous at x = 5

Question 2.

Examine the continuity of the function f (x) = 2x² – 1 at x = 3.

Answer:

The given function is f(x) = 2x² – 1

At x = 3, f(x) = f(3) = 2 × 3² – 1 = 17

Left hand limit (LHL):

Right hand limit(RHL):

As, LHL= RHL = f(3)
Therefore, f is continuous at x = 3

Question 3.

Examine the following functions for continuity.

f (x) = x – 5

Answer:

a) The given function is f(x) = x – 5

We know that f is defined at every real number k and its value at k is k – 5.

We can see that  = k – 5 = f(k)

Thus, 

Therefore, f is continuous at every real number and thus, it is continuous function.

Question 4.

Examine the following functions for continuity.

Answer:

The given function is 

For any real number k ≠ 5, we get,

Also, 

Thus, 

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

Question 5.

Examine the following functions for continuity.

Answer:

The given function is 

For any real number k ≠ 5, we get,

Also, 

Thus, 

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

Question 6.

Examine the following functions for continuity.

f (x) = | x – 5|

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.ee, k < 5, or k = 5 or k >5

Now, Case I: k<5

Then, f(k) = 5 – k

 = 5 – k = f(k)

Thus, 

Hence, f is continuous at all real number less than 5.

Case II: k = 5

Then, f(k) = f(5) = 5 – 5 = 0

 = 5 – 5 = 0

 = 5 – 5 = 0

Hence, f is continuous at x = 5.

Case III: k > 5

Then, f(k) = k – 5

 = k – 5 = f(k)

Thus, 

Hence, f is continuous at all real number greater than 5.

Therefore, f is a continuous function.

Question 7.

Prove that the function f (x) = xⁿ is continuous at x = n, where n is a positive integer.

Answer:

It is given that function f (x) = xⁿ

We can see that f is defined at all positive integers, n and the value of f at n is nⁿ.

 = nⁿ

Thus, 

Therefore, f is continuous at x =n, where n is a positive integer.

Question 8.

Is the function f defined by



Continuous at x = 0? At x = 1? At x = 2?

Answer:

It is given that 

Case I: x = 0

We can see that f is defined at 0 and its value at 0 is 0.

LHL = RHL = f(0)
Hence, f is continuous at x = 0.

Case II: x = 1

We can see that f is defined at 1 and its value at 1 is 1.

For x < 1
f(x) = x

Hence, LHL:

 = 1

For x > 1
f(x) = 5

therefore, RHL

 = 5

Hence, f is not continuous at x = 1.

Case III: x = 2

As, 

We can see that f is defined at 2 and its value at 2 is 5
LHL:

here f(2 - h) = 5, as h → 0 ⇒ 2 - h → 2

RHL:

LHL = RHL = f(2)

here f(2 + h) = 5, as h → 0 ⇒ 2 + h → 2
Hence, f is continuous at x = 2.

Question 9.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k >2

Now, Case I: k < 2

Then, f(k) = 2k + 3

 = 2k + 3= f(k)

Thus, 

Hence, f is continuous at all real number less than 2.

Case II: k = 2

 = 2×2 + 3 = 7

 = 2×2 - 3 = 1

Hence, f is not continuous at x = 2.

Case III: k > 2

Then, f(k) = 2k - 3

 = 2k – 3 = f(k)

Thus, 

Hence, f is continuous at all real number greater than 2.

Therefore, x = 2 is the only point of discontinuity of f.

Question 10.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 5 cases i.e., k < -3, k = -3, -3 < k < 3, k = 3 or k > 3

Now, Case I: k < -3

Then, f(k) = -k + 3

 = -k + 3= f(k)

Thus, 

Hence, f is continuous at all real number x < -3.

Case II: k = -3

f(-3) = -(-3) + 3 = 6

 =-(-3) + 3 = 6

 = -2×(-3) = 6

Hence, f is continuous at x = -3.

Case III: -3 < k < 3

Then, f(k) = -2k

 = -2k = f(k)

Thus, 

Hence, f is continuous in (-3,3).

Case IV: k = 3

 = -2×(3) = -6

 = 6 × 3 + 2 = 20

Hence, f is not continuous at x = 3.

Case V: k > 3

Then, f(k) = 6k + 2

 = 6k + 2= f(k)

Thus, 

Hence, f is continuous at all real number x < 3.

Therefore, x = 3 is the only point of discontinuity of f.

Question 11.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

We know that if x > 0

⇒ |x| = -x and

x > 0

⇒ |x| = x

So, we can rewrite the given function as:

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 0, or k = 0 or k >0.

Now, Case I: k < 0

Then, f(k) = -1

 = -1= f(k)

Thus, 

Hence, f is continuous at all real number less than 0.

Case II: k = 0

 = -1

 = 1

Hence, f is not continuous at x = 0.

Case III: k > 0

Then, f(k) = 1

 = 1 = f(k)

Thus, 

Hence, f is continuous at all real number greater than 1.

Therefore, x = 0 is the only point of discontinuity of f.

Question 12.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

We know that if x < 0

⇒ |x| = -x

So, we can rewrite the given function as:

⇒ f(x) = -1 for all x ϵ R

Let k be the point on a real line.

Then, f(k) = -1

 = -1= f(k)

Thus, 

Therefore, the given function is a continuous function.

Question 13.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k >1

Now, Case I: k < 1

Then, f(k) = k² + 1

 = k² + 1= f(k)

Thus, 

Hence, f is continuous at all real number less than 1.

Case II: k = 1

Then, f(k) = f(1) = 1 + 1 = 2

 = 1² + 1 = 2

 = 1 + 1 = 2

Hence, f is continuous at x = 1.

Case III: k > 1

Then, f(k) = k + 1

 = k + 1 = f(k)

Thus, 

Hence, f is continuous at all real number greater than 1.

Question 14.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k > 2

Now, Case I: k < 2

Then, f(k) = k³ - 3

 = k³ - 3= f(k)

Thus, 

Hence, f is continuous at all real number less than 2.

Case II: k = 2

Then, f(k) = f(2) = 2³ - 3 = 5

 = 2³ - 3 = 5

 = 2² + 1 = 5

Hence, f is continuous at x = 2.

Case III: k > 2

Then, f(k) = 2² + 1 = 5

 = 2² + 1 = 5 = f(k)

Thus, 

Hence, f is continuous at all real number greater than 2.

Question 15.

Find all points of discontinuity of f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

Case I: k < 1

Then, f(k) = k¹⁰ - 1

 = k¹⁰ - 1= f(k)

Thus, 

Hence, f is continuous at all real number less than 1.

Case II: k = 1

Then, f(k) = f(1) = 1¹⁰ - 1 = 0

 = 1¹⁰ - 1 = 0

 = 1² = 1

Hence, f is not continuous at x = 1.

Case III: k > 1

Then, f(k) = 1² = 1

 = 1² = 1 = f(k)

Thus, 

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

Question 16.

Is the function defined by



a continuous function?

Answer:

The given function is 

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

Case I: k < 1

Then, f(k) = k + 5

 = k + 5 = f(k)

Thus, 

Hence, f is continuous at all real number less than 1.

Case II: k = 1

Then, f(k) = f(1) = 1 + 5 = 6

 = 1 + 5 = 6

 = 1 - 5 = -4

Hence, f is not continuous at x = 1.

Case III: k > 1

Then, f(k) = k -5

 = k - 5

Thus, 

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

Question 17.

Discuss the continuity of the function f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the interval [0,10].

Let k be the point in the interval [0,10].

Then, we have 5 cases i.e., 0≤ k < 1, k = 1, 1 < k < 3, k = 3 or 3 < k ≤ 10.

Now, Case I: 0≤ k < 1

Then, f(k) = 3

 = 3= f(k)

Thus, 

Hence, f is continuous in the interval [0,10).

Case II: k = 1

f(1) = 3

 = 3

 = 4

Hence, f is not continuous at x = 1.

Case III: 1 < k < 3

Then, f(k) = 4

 = 4 = f(k)

Thus, 

Hence, f is continuous in (1, 3).

Case IV: k = 3

 = 4

 = 5

Hence, f is not continuous at x = 3.

Case V: 3 < k ≤ 10

Then, f(k) = 5

 = 5 = f(k)

Thus, 

Hence, f is continuous at all points of the interval (3, 10].

Therefore, x = 1 and 3 are the points of discontinuity of f.

Question 18.

Discuss the continuity of the function f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1 or k < 1.

Now, Case I: k < 0

Then, f(k) = 2k

 = 2k= f(k)

Thus, 

Hence, f is continuous at all points x, s.t. x < 0.

Case II: k = 0

f(0) = 0

 = 2 × 0 = 0

 = 0

Hence, f is continuous at x = 0.

Case III: 0 < k < 1

Then, f(k) = 0

 = 0 = f(k)

Thus, 

Hence, f is continuous in (0, 1).

Case IV: k = 1

Then f(k) = f(1) = 0

 = 0

 = 4 × 1 = 4

Hence, f is not continuous at x = 1.

Case V: k < 1

Then, f(k) = 4k

 = 4k = f(k)

Thus, 

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, x = 1 is the only point of discontinuity of f.

Question 19.

Discuss the continuity of the function f, where f is defined by

Answer:

The given function is 

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < -1, k = -1, -1 < k < 1, k = 1 or k > 1.

Now, Case I: k < 0

Then, f(k) = -2

 = -2= f(k)

Thus, 

Hence, f is continuous at all points x, s.t. x < -1.

Case II: k = -1

f(k) = f(=1) = -2

 = -2

 = 2 × (-1) = -2

Hence, f is continuous at x = -1.

Case III: -1 < k < 1

Then, f(k) = 2k

 = 2k = f(k)

Thus, 

Hence, f is continuous in (-1, 1).

Case IV: k = 1

Then f(k) = f(1) = 2 × 1 = 2

 = 2 × 1 = 2

 = 2

Hence, f is continuous at x = 1.

Case V: k > 1

Then, f(k) = 2

 = 2 = f(k)

Thus, 

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, f is continuous at all points of the real line.

Question 20.

Find the relationship between a and b so that the function f defined by

 is continuous at x = 3.

Answer:

It is given function is 

It is given that f is continuous at x = 3, then, we get,

 ………………….(1)

And

 = 3a + 1

 = 3b + 1

f(3) = 3a + 1

Thus, from (1), we get,

3a + 1 = 3b + 3 = 3a + 1

⇒ 3a +1 = 3b + 1

⇒ 3a = 3b + 2

⇒ a = b + 

Therefore, the required the relation is a = b + .

Question 21.

For what value of λ is the function defined by

 Continuous at x = 0? What about continuity at x = 1?

Answer:

It is given that 

It is given that f is continuous at x = 0, then, we get,

And

 = 0

 = 1

Thus, there is no value of  for which f is continuous at x = 0

f(1) = 4x + 1 = 4 × 1 + 1 = 5

= 4 × 1 + 1 = 5

Then, 

Hence, for any values of, f is continuous at x = 1

Question 22.

Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Answer:

It is given that g(x) = x – [x]

We know that g is defined at all integral points.

Let k be ant integer.

Then,

g(k) = k – [-k] = k + k = 2k

And

Therefore, g is discontinuous at all integral points.

Question 23.

Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?

Answer:

It is given that f (x) = x² – sin x + 5

We know that f is defined at x = π

So, at x = π,

f(x) = f(π) = π² -sin π + 5 = π² – 0 + 5 = π² + 5

Now, 

Let put x = π + h

If 

Thus, 

Therefore, the function f is continuous at x = π.

Question 24.

Discuss the continuity of the following functions:

(a) f (x) = sin x + cos x

(b) f (x) = sin x – cos x

(c) f (x) = sin x . cos x

Answer:

We known that g and k are two continuous functions, then,

g + k, g – k and g.k are also continuous.

First we have to prove that g(x) = sinx and k(x) = cosx are continuous functions.

Now, let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if 

g(h) = sinh

= sinhcos0 + coshsin0

= sinh + 0

= sinh

Thus, 

Therefore, g is a continuous function…………(1)

Now, let k(x) = cosx

We know that k(x) = cosx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if 

Now k(h) = cosh

= coshcos0 - sinhsin0

= cosh - 0

= cosh

Thus, 

Therefore, k is a continuous function……………….(2)

So, from (1) and (2), we get,

(a) f(x) = g(x) + k(x) = sinx + cosx is a continuous function.

(b) f(x) = g(x) - k(x) = sinx - cosx is a continuous function.

(c) f(x) = g(x) × k(x) = sinx × cosx is a continuous function.

Question 25.

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Answer:

We know that if g and h are two continuous functions, then,

(i) 

(ii)

(iii)

So, first we have to prove that g(x) = sinx and h(x) = cosx are continuous functions.

Let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = k + h

So, if 

g(k) = sink

= sinkcos0 + cosksin0

= sink + 0

= sink

Thus, 

Therefore, g is a continuous function…………(1)

Let h(x) = cosx

We know that h(x) = cosx is defined for every real number.

Let k be a real number. Now, put x = k + h

So, if 

h(k) = sink

= coskcos0 - sinksin0

= cosk - 0

= cosk

Thus, 

Therefore, g is a continuous function…………(2)

So, from (1) and (2), we get,

Thus, cosecant is continuous except at x = np, (n ϵ Z)

Thus, secant is continuous except at x = , (n ϵ Z)

Thus, cotangent is continuous except at x = np, (n ϵ Z)

Question 26.

Find all points of discontinuity of f, where

Answer:

It is given that 

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k < 0,

Then f(k) = 

Thus, f is continuous at all points x that is x < 0.

Case II: k > 0,

Then f(k) = c + 1

Thus, f is continuous at all points x that is x > 0.

Case III: k = 0

Then f(k) = f(0) = 0 + 1 = 1

 = 1

 = 1

Hence, f is continuous at x = 0.

Therefore, f is continuous at all points of the real line.

Question 27.

Determine if f defined by

 is a continuous function?

Answer:

It is given that 

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) = 

Thus, f is continuous at all points x that is x ≠ 0.

Case II: k = 0

Then f(k) = f(0) = 0

We know that -1 ≤  ≤ 1, x ≠ 0

⇒ x² ≤  ≤ 0

⇒ 

⇒ 

Similarly, 

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

Question 28.

Examine the continuity of f, where f is defined by

Answer:

It is given that 

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) = sink - cosk

Thus, f is continuous at all points x that is x ≠ 0.

Case II: k = 0

Then f(k) = f(0) = 0

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

Question 29.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Answer:

It is given that 

Also, it is given that function f is continuous at x =,

So, if f is defined at x =  and if the value of the f at x =  equals the limit of f at x = .

We can see that f is defined at x =  and f = 3

Now, let put x = 

Then, 

⇒ 

⇒ 

Therefore, the value of k is 6.

Question 30.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Answer:

It is given that 

Also, it is given that function f is continuous at x = 2,

So, if f is defined at x = 2 and if the value of the f at x = 2 equals the limit of f at x = 2.

We can see that f is defined at x = 2 and

f(2) = k(2)² = 4k

⇒ 

⇒ k × 2² = 3 = 4k

⇒ 4k = 3 = 4k

⇒ 4k = 3

⇒ k = 

Therefore, the required value of k is .

Question 31.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Answer:

It is given that 

Also, it is given that function f is continuous at x = k,

So, if f is defined at x = p and if the value of the f at x = k equals the limit of f at x = k.

We can see that f is defined at x = p and

f(π) = kπ + 1

⇒ 

⇒ kπ + 1 = cosπ = kπ + 1

⇒ kπ + 1 = -1 = kπ + 1

⇒ k = 

Therefore, the required value of k is.

Question 32.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Answer:

It is given that 

Also, it is given that function f is continuous at x = 5,

So, if f is defined at x = 5 and if the value of the f at x = 5 equals the limit of f at x = 5.

We can see that f is defined at x = 5 and

f(5) = kx + 1 = 5k + 1

⇒ 

⇒ 5k + 1 = 15 -5 = 5k + 1

⇒ 5k + 1 = 10

⇒ 5k = 9

⇒ k = 

Therefore, the required value of k is.

Question 33.

Find the values of a and b such that the function defined by

 is a continuous function.

Answer:

It is given function is 

We know that the given function f is defined at all points of the real line.

Thus, f is continuous at x = 2, we get,

⇒ 

⇒ 5 = 2a + b = 5

⇒ 2a + b = 5………………(1)

Thus, f is continuous at x = 10, we get,

⇒ 

⇒ 10a + b = 21 =21

⇒ 10a + b = 21………………(2)

On subtracting eq. (1) from eq. (2), we get,

8a = 16

⇒ a = 2

Thus, putting a = 2 in eq. (1), we get,

2 × 2 + b = 5

⇒ 4 + b = 5

⇒ b = 1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 resp.

Question 34.

Show that the function defined by f (x) = cos (x²) is a continuous function.

Answer:

It is given function is f(x) = cos (x²)

This function f is defined for every real number and f can be written as the composition of two function as,

f = goh, where, g(x) = cosx and h(x) = x²

First we have to prove that g(x) = cosx and h(x) = x² are continuous functions.

We know that g is defined for every real number.

Let k be a real number.

Then, g(k) =cos k

Now, put x = k + h

If 

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, g(x) = cosx is continuous function.

Now, h(x) = x²

So, h is defined for every real number.

Let c be a real number, then h(c) = c²

Therefore, h is a continuous function.

We know that for real valued functions g and h,

Such that (fog) is continuous at c.

Therefore, f(x) = (goh)(x) = cos(x²) is a continuous function.

Question 35.

Show that the function defined by f (x) = | cos x| is a continuous function.

Answer:

It is given that f(x) = |cosx|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = cosx

First we have to prove that g(x) = |x| and h(x) = cosx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And 

Thus, 

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus, 

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = cosx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If 

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(cosx) = |cosx| is a continuous function.

Question 36.

Examine that sin | x| is a continuous function.

Answer:

It is given that f(x) = sin|x|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = sinx

First we have to prove that g(x) = |x| and h(x) = sinx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And 

Thus, 

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus, 

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = sinx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If 

= sinkcos0 + cosksin0

= sink

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(sinx) = |sinx| is a continuous function.

Question 37.

Find all the points of discontinuity of f defined by f (x) = | x| – | x + 1|.

Answer:

It is given that f(x) = |x| - |x + 1|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = |x + 1|

Then, f = g - h

First we have to prove that g(x) = |x| and h(x) = |x + 1| are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And 

Thus, 

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus, 

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

g(x) = |x + 1| can be written as

Now, h is defined for all real number.

Let k be a real number.

Case I: If k < -1,

Then h(k) = -(k + 1)

And 

Thus, 

Therefore, h is continuous at all points x, i.e., x < -1

Case II: If k > -1,

Then h(k) = k + 1 and

Thus, 

Therefore, h is continuous at all points x, i.e., x > -1.

Case III: If k = -1,

Then, h(k) = h(-1) = -1 + 1 = 0

Therefore, g is continuous at x = -1

From the above 3 cases, we get that h is continuous at all points.

Hence, g and h are continuous function.

Therefore, f = g – h is also a continuous function.

---

Exercise 5.2

Question 1.

Differentiate the functions with respect to x.

sin (x² + 5)

Answer:

Given: sin(x² + 5)

Let y = sin(x² + 5)

= cos(x² + 5).(2x + 0)

= cos(x² + 5).(2x)

= 2x.cos(x² + 5)

Question 2.

Differentiate the functions with respect to x.

cos (sin x)

Answer:

Given: cos(sinx)

Let y = cos(sinx)

= -sin(sinx).cosx

= -cosx.sin(sinx)

Question 3.

Differentiate the functions with respect to x.

sin (ax + b)

Answer:

Given: sin(ax + b)

Let y = sin(ax + b)

= cos(ax + b).(a + 0)

= cos(ax + b). (a)

= a.cos(ax + b)

Question 4.

Differentiate the functions with respect to x.

Answer:

Given: sec (tan(√x))

Let y= sec (tan(√x))

Question 5.

Differentiate the functions with respect to x.

Answer:

Given:

Let 

We know that 

= a cos (ax + b) sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d)

Question 6.

Differentiate the functions with respect to x.

cos x³ . sin² (x⁵)

Answer:

Given: cos x³ . sin² (x⁵)

Let y = cos x³ . sin² (x⁵)

We know that, 

= cosx³.2sin(x⁵).cos(x⁵)(5x⁴) + sin²(x⁵).(-sinx³).(3x²)

= 10x⁴.cosx³.sin(x⁵).cos(x⁵)-(3x²).sin² (x⁵).(sinx³)

Question 7.

Differentiate the functions with respect to x. 

Answer:

Let

we know that,

Applying both the formula, we get,

Now,

Therefore,

[Using sin 2x = 2 sin x cos x]

Question 8.

Differentiate the functions with respect to x.

cos(√x)

Answer:

Given: cos√x

Let y = cos√x

Question 9.

Prove that the function f given by f (x) = | x – 1|, x ∈ R is not differentiable at x = 1.

Answer:

Given: f(x)=|x-1|, x ∈R

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left hand limit of function f at x=1

because, {h < 0 ⇒ |h|= -h}

= -1

Now, let we consider the right hand limit of function f at x=1

because, {h>0 ⇒ |h|= h}

= 1

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

Question 10.

Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

Answer:

Given: f(x) =[x], 0 < x <3

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left-hand limit of function f at x=1

because, {h<0=> |h|= -h}

Let we consider the right hand limit of function f at x=1

= 0

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

Let we consider the left hand limit of function f at x=2

=  =

Now, let we consider the right hand limit of function f at x=2

= 0

Because, left hand limit is not equal to right hand limit of function f at x=2, so f is not differentiable at x=2.

---

Exercise 5.3

Question 1.

Find dy/dx in the following:

2x + 3y = sin x

Answer:

It is given that 2x + 3y = sin x

Differentiating both sides w.r.t. x, we get,

Question 2.

Find dy/dx in the following:

2x + 3y = sin y

Answer:

It is given that 2x + 3y = sin y

Differentiating both sides w.r.t. x, we get,

Question 3.

Find dy/dx in the following:

ax + by² = cos y

Answer:

It is given that ax + by² = cos y

Differentiating both sides w.r.t. x, we get,

Question 4.

Find dy/dx in the following:

xy + y² = tan x + y

Answer:

It is given that xy + y² = tan x + y

Differentiating both sides w.r.t. x, we get,

Question 5.

Find dy/dx in the following:

x² + xy + y² = 100

Answer:

It is given that x² + xy + y² = 100

Differentiating both sides w.r.t. x, we get,

Question 6.

Find dy/dx in the following:

x³ + x²y + xy² + y³ = 81

Answer:

It is given that x³ + x²y + xy² + y³ = 81

Differentiating both sides w.r.t. x, we get,

Question 7.

Find dy/dx in the following:

sin² y + cos xy = π

Answer:

It is given that sin² y + cos xy = π

Differentiating both sides w.r.t. x, we get,

Question 8.

Find dy/dx in the following:

sin² x + cos² y = 1

Answer:

It is given that sin² x + cos² y = 1

Differentiating both sides w.r.t. x, we get,

Question 9.

Find dy/dx in the following:

Answer: Let x = tan A

then, 
A = tan^-1x

And

 
Question 10.

Find dy/dx in the following:

Answer:

It is given that:


Assumption: Let x = tan θ, putting it in y, we get,



we know by the formula that,

Putting this in y, we get,

y = tan^-1(tan3θ)

y = 3(tan^-1x)

Differentiating both sides, we get,

Question 11.

Find dy/dx in the following:

Answer:

It is given that,

y = 

On comparing both sides, we get,

Now, differentiating both sides, we get,

Question 12.

Find dy/dx in the following:

Answer:

It is given that y = 

Now, we can change the numerator and the denominator,
We know that we can write,


Therefore, by applying the formula: (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab, we get,

Dividing the numerator and denominator by cos (y/2), we get,

Now, we know that:

Now, differentiating both sides, we get,

Question 13.

Find dy/dx in the following:

Answer:

It is given that y = 

Differentiating both sides w.r.t. x, we get,

Question 14.

Find dy/dx in the following:

Answer:

It is given that y = 

Differentiating both sides w.r.t. x, we get,

Question 15.

Find dy/dx in the following:

Answer:

It is given that y = 

Differentiating w.r.t. x, we get,

---

Exercise 5.4

Question 1.

Differentiate the following w.r.t. x:

Answer:

Let y = 

By using the quotient rule, we get

Question 2.

Differentiate the following w.r.t. x:

Answer:

Let y = 

Now, by using the chain rule, we get,

Thus, 

Question 3.

Differentiate the following w.r.t. x:

Answer:

Let y = 

So, by using the chain rule, we get,

Question 4.

Differentiate the following w.r.t. x:

sin (tan^–1 e^–x)

Answer:

Let y = sin(tan^-1 e^x)

So, by using chain rule, we get

Question 5.

Differentiate the following w.r.t. x:

log (cos e^x)

Answer:

Let y = log(cose^x)

So, by using the chain rule, we get,

Question 6.

Differentiate the following w.r.t. x:

Answer:

Let 

Question 7.

Differentiate the following w.r.t. x:

Answer:

Let 

Then, 

Now, differentiating both sides we get,

Question 8.

Differentiate the following w.r.t. x:

log (log x), x > 1

Answer:

let y = log(logx)

So, by using chain rule, we get,

Question 9.

Differentiate the following w.r.t. x:

Answer:

Let 

So, by using the quotient rule, we get,

Question 10.

Differentiate the following w.r.t. x:

cos (log x + e^x), x > 0

Answer:

Let y = cos(logx + e^x)

So, by using chain rule, we get,

---

Exercise 5.5

Question 1.

Differentiate the functions given in w.r.t. x.

cos x . cos 2x . cos 3x

Answer:

Given: cos x . cos 2x . cos 3x

Let y= cos x . cos 2x . cos 3x

Taking log on both sides, we get

log y = log(cos x . cos 2x . cos 3x)

⇒log y = log(cos x) + log(cos 2x) + log(cos 3x)

Now, differentiate both sides with respect to x

Question 2.

Differentiate the functions given in w.r.t. x.

Answer:

Given: 

Let 

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Question 3.

Differentiate the functions given in w.r.t. x.

(log x)^{cos x}

Answer:

Given: (log x)^{cos x}

Let y=(log x)^{cos x}

Taking log on both sides, we get

log y = log(log x)^{cos x}

⇒log y = cos x.log(log x)

Now, differentiate both sides with respect to x

Question 4.

Differentiate the functions given in w.r.t. x.

x^x – 2^{sin x}

Answer:

Given: x^x – 2^{sin x}

Let y= x^x – 2^{sin x}

Let y = u - v

⇒ u = x^x and v = 2^{sin x}

For, u = x^x

Taking log on both sides, we get

log u = log x^x

⇒log u = x.log(x)

Now, differentiate both sides with respect to x

For, v = 2^{sin x}

Taking log on both sides, we get

log v = log 2^{sin x}

⇒log v = sin x. log (2)

Now, differentiate both sides with respect to x

Because, y = u - v

dy/dx = x^x(1 + logx) - 2^sinx.cosx.log2

Question 5.

Differentiate the functions given in w.r.t. x.

(x + 3)². (x + 4)³. (x + 5)⁴

Answer:

Given: (x + 3)². (x + 4)³. (x + 5)⁴

Let y= (x + 3)². (x + 4)³. (x + 5)⁴

Taking log on both sides, we get

log y = log((x + 3)². (x + 4)³. (x + 5)⁴)

⇒log y = log(x + 3)² + log(x + 4)³ + log(x + 5)⁴

⇒log y = 2.log(x + 3) + 3.log(x + 4) + 4.log(x + 5)⁴

Now, differentiate both sides with respect to x

= (x + 3)(x + 4)²(x + 5)³(9x² + 70x + 133)

Question 6.

Differentiate the functions given in w.r.t. x.

Answer:

Given: 

Let y= 

Also, Let y = u + v

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Because, y = u + v

Question 7.

Differentiate the functions given in w.r.t. x.

(log x)^x + x^{log x}

Answer:

Given: (log x)^x + x^{log x}

Let y= (log x)^x + x^{log x}

Let y = u + v

⇒ u = (log x)^x and v = x^{log x}

For, u =(log x)^x

Taking log on both sides, we get

log u =log (log x)^x

⇒log u = x.log (log(x))

Now, differentiate both sides with respect to x

For, v = x^{log x}

Taking log on both sides, we get

log v =log( x^{log x)}

⇒log v = log x. log x

Now, differentiate both sides with respect to x

Because, y = u + v

Question 8.

Differentiate the functions given in w.r.t. x.

(sin x)^x + sin^–1 √x

Answer:

Given: 

Let y= 

Let y = u + v

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Now, differentiate both sides with respect to x

Because, y = u + v

Question 9.

Differentiate the functions given in w.r.t. x.

x^{sin x} + (sin x)^{cos x}

Answer:

Given: x^{sin x} + (sin x)^{cos x}

Let y= x^{sin x} + (sin x)^{cos x}

Let y = u + v

⇒ u = x^{sin x} and v = (sin x)^{cos x}

For, u = x^{sin x}

Taking log on both sides, we get

log u =log (x^{sin x} )

⇒log u = sin x.log(x)

Now, differentiate both sides with respect to x

For, v = (sin x)^{cos x}

Taking log on both sides, we get

log v =log(sin x)^{cos x}

⇒log v = cos x. log (sin x)

Now, differentiate both sides with respect to x

Because, y = u + v

Question 10.

Differentiate the functions given in w.r.t. x.

Answer:

Given: 

Let y= 

Let y = u + v

Taking log on both sides, we get

⇒log u = x.cos x.log x

Now, differentiate both sides with respect to x

Taking log on both sides, we get

⇒ log v = log (x² + 1) – log (x² – 1)

Now, differentiate both sides with respect to x

Because, y = u + v

Question 11.

Differentiate the functions given in w.r.t. x.

Answer:

Given: 

Let y= 

Let y = u + v

Taking log on both sides, we get

⇒log u = x.log(x cos x)

⇒ log u = x(log x + log (cos x))

⇒ log u = x(log x) +x(log (cos x))

Now, differentiate both sides with respect to x

Taking log on both sides, we get

Now, differentiate both sides with respect to x

Because, y = u + v

Question 12.

Find dy/dx of the functions.

x^y + y^x = 1

Answer:

Given: x^y + y^x = 1

Let y= x^y + y^x = 1

Let u = x^y and v = y^x

Then, ⇒ u + v = 1

For, u = x^y

Taking log on both sides, we get

Log u =log x^y

⇒log u = y.log(x)

Now, differentiate both sides with respect to x

For, v = y^x

Taking log on both sides, we get

Log v =log y^x

⇒log v = x.log(y)

Now, differentiate both sides with respect to x

Question 13.

Find dy/dx of the functions.

y^x = x^y

Answer:

Given: y^x = x^y

Taking log on both sides, we get

log y^x =log x^y

⇒x log y = y log x

Now, differentiate both sides with respect to x

Question 14.

Find dy/dx of the functions.

(cos x)^y = (cos y)^x

Answer:

Given: (cos x)^y = (cos y)^x

Taking log on both sides, we get

log (cos x)^y =log (cos y)^x

⇒y log (cos x) = x log (cos y)

Now, differentiate both sides with respect to x

Question 15.

Find dy/dx of the functions.

xy = e^{(x – y)}

Answer:

Given: xy = e^{(x – y)}

Taking log on both sides, we get

log (xy) = log (e^{(x – y)})

⇒ log x + log y = (x - y) log e

⇒ log x + log y = (x - y) .1

⇒ log x + log y = (x - y)

Now, differentiate both sides with respect to x

Question 16.

Find the derivative of the function given by f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f ′(1).

Answer:

Given: f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸)

Taking log on both sides, we get

log f (x) =log (1 + x) + log (1 + x²) + log (1 + x⁴) + log (1 + x⁸)

Now, differentiate both sides with respect to x

Question 17.

Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three ways mentioned below:

(i) by using product rule

(ii) by expanding the product to obtain a single polynomial.

(iii) by logarithmic differentiation.

Do they all give the same answer?

Answer:

Given: (x² – 5x + 8) (x³ + 7x + 9)

Let y=(x² – 5x + 8) (x³ + 7x + 9)

(i) By applying product rule differentiate both sides with respect to x

(ii) by expanding the product to obtain a single polynomial

y = (x² – 5x + 8) (x³ + 7x + 9)

y = x⁵ + 7x³ + 9x² - 5x⁴ – 35x² - 45x + 8x³ + 56x + 72

y = x⁵ - 5x⁴ + 15x³ - 26x² + 11x + 72

Now, differentiate both sides with respect to x

(iii) by logarithmic differentiation

y = (x² – 5x + 8) (x³ + 7x + 9)

Taking log on both sides, we get

log y = log ((x² – 5x + 8) (x³ + 7x + 9))

log y = log (x² – 5x + 8) + log (x³ + 7x + 9)

Now, differentiate both sides with respect to x

From equation (i),(ii)and(iii), we can say that value of given function after differentiating by all the three methods is same.

Question 18.

If u, v and w are functions of x, then show that



in two ways – first by repeated application of product rule, second by logarithmic differentiation.

Answer:

To prove:

Let y=u.v.w=u.(v.w)

(a) by applying product rule differentiate both sides with respect to x

(b) Taking log on both sides, we get

as, y=u.v.w

log y = log (u.v.w)

log y = log u + log v + log w

Now, differentiate both sides with respect to x

---

Exercise 5.6

Question 1.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = 2at², y = at⁴

Answer:

It is given that

x = 2at², y = at⁴

So, now

= 2a.2t

= 4at ………… (1)

And

= a.4.t³

= 4at³………… (2)

Therefore, form equation (1) and (2). we get

Hence, the value of  is t²

Question 2.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = a cos θ, y = b cos θ

Answer:

It is given that

x = a cos θ, y = b cos θ

Then, we have

= a(-sinƟ)

= -asinƟ………… (1)

= b(-sinƟ)

= -bsinƟ …… (2)

From equation (1) and (2), we get

Hence, the value of  is 

Question 3.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = sin t, y = cos 2t

Answer:

It is given that

x = sin t, y = cos 2t

Then, we have

= cost ………… (1)

= -2sin2t………… (2)

So, equation (1) and (2), we get

 , Since sin2t = 2sintcost

= -4sint

Hence, the value of  is -4sint

Question 4.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = 4t, y = 4/t

Answer:

It is given that

x = 4t, y = 

Then, we have

= 4 ……… (1)

 ………… (2)

Therefore, from equation (1) and (2), we get

Hence, the value of  is 

Question 5.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = cos θ – cos 2θ, y = sin θ – sin 2θ

Answer:

It is given that

x = cos θ – cos 2θ, y = sin θ – sin 2θ

Then, we have

= -sinθ – (-2sin2θ)

= 2sin2θ - sinθ …………… (1)

= cosθ – 2cos2θ

= -bsinƟ ………… (2)

From equation (1) and (2), we get,

Hence, the value of  is 

Question 6.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = a (θ – sin θ), y = a (1 + cos θ)

Answer:

It is given that

x = a (θ – sin θ), y = a (1 + cos θ)

Then, we have

= a(1-cosθ) …………………… (1)

= a[0 + (-sinθ)]

= -asinƟ …………………… (2)

From equation (1) and (2), we get

Hence, the value of  is 

Question 7.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

Answer:

It is given that

Then, we have

 …………………… (1)

 …………………… (2)

Therefore, from equation (1) and (2), we get

 , [since, cos2t= (2cos²t-1) and also cos2t= (1-2sin²t)]

 , [Since, cos3t = 4cos³t-3cost, sin3t = 3sint – 4sin³t]

= -cot3t

Hence, the value of  is -cot3t

Question 8.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

Answer:

It is given that

Then, we have

 …………………… (1)

= acost…………………… (2)

From equation (1) and (2), we get

= tant

Hence, the value of  is tant

Question 9.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = a sec θ, y = b tan θ

Answer:

It is given that

x = a sec θ, y = b tan θ

Then, we have

= asecθtanθ…………………… (1)

= bsec²θ…………………… (2)

From equation (1) and (2), we get,

Hence, the value of  is 

Question 10.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)

Answer:

It is given that

x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)

Then, we have

= a[-sinθ + θcosθ + sinθ]

= aθcosθ…………………… (1)

= a[cosθ + θsinθ - cosθ]

= aθsinθ…………………… (2)

From (1) and (2) we get,

= tanθ

Hence, the value of  is tanθ

Question 11.

If x and y are connected parametrically by the equations given in without eliminating the parameter, Find dy/dx.

If, show that 

Answer:

It is given that

Now,

Similarly, 

Let us consider,

Taking Log on both sides, we get

Therefore, 

 ……………………………..(1)

Now, Consider

Taking Log on both sides, we get

Therefore, 

 ……………..(2)

So, from equation (1) and (2), we get

Therefore, L.H.S. = R.H.S.

Hence Proved

---

Exercise 5.7

Question 1.

Find the second order derivatives of the function

x² + 3x + 2

Answer:

Let us take y= x² + 3x + 2

Now,

= 2x + 3

Therefore,

= 2 + 0

= 2

Question 2.

Find the second order derivatives of the function

x²⁰

Answer:

Let us take y= x²⁰

Now,

= 20x¹⁹

Therefore,

= 20 × 19 × x¹⁸

= 380 x¹⁸

Question 3.

Find the second order derivatives of the function

x . cos x

Answer:

: Let us take y= x . cos x

Now,

= cosx.1 + x(-sinx)

= cosx -xsinx

Therefore,

= -sinx –(sinx + xcosx)

= - (xcosx + 2sinx)

Question 4.

Find the second order derivatives of the function

log x

Answer:

Let us take y = logx

Now,

Therefore,

Question 5.

Find the second order derivatives of the function

x³ log x

Answer:

Let us take y = x³ log x

Now,

= logx.3x² + x³.1/x

= logx .3x² + x²

= x²(1 + 3logx)

Therefore,

= 2x + 6xlogx + 3x

= 5x + 6xlogx

= x(5 + 6logx)

Question 6.

Find the second order derivatives of the function

e^x sin 5x

Answer:

Let us take y = e^x sin 5x

Now,

= e^xsin5x+e^xcos5x.5

= e^x (sin5x + 5cos5x)

= e^x (sin5x + 5cos5x) + e^x (5cos5x – 25sin5x)

= e^x (10cos5x – 24sin5x)

= 2e^x (5cos5x – 12sin5x)

Question 7.

Find the second order derivatives of the function

e^6x cos 3x

Answer:

Let us take y = e^6x cos 3x

Now,

= 6e^6xcos3x - 3e^6xsin3x

= 36e^6xcos3x – 18e^6xsin3x -3[sin3x.e^6x.6 + e^6x.cos3x.3]

= 36e^6xcos3x – 18e^6xsin3x – 18e^6xsin3x -9e^6xcos3x

= 27e^6xcos3x -36e^6xsin3x

= 9e^6x(3cos3x – 4sin3x)

Question 8.

Find the second order derivatives of the function

tan^–1 x

Answer:

Let us take y = tan^–1 x Now,

Question 9.

Find the second order derivatives of the function

log (log x)

Answer:

Let us take y = log (log x)

Now,

= (xlogx)^-1

Question 10.

Find the second order derivatives of the function

sin (log x)

Answer:

Let us take y = sin (log x)

Now,

Then,

Question 11.

If y = 5 cos x – 3 sin x, prove that 

Answer:

It is given that y = 5 cos x – 3 sin x

Now, on differentiating we get,

= 5(-sinx) – 3(cosx)

= -(5sinx + cosx)

Then,

= - [5cosx +3(-sinx)]

= -[5cosx-3sinx]

= -y

Therefore,

Hence Proved.

Question 12.

If y = cos^–1 x, Find d²y/dx² in terms of y alone.

Answer:

It is given that y = cos^–1 x

Now,

Therefore,

 ………………………(1)

Now it is given that y = cos^–1 x

⇒ x= cosy

Now putting the value of x in equation (1), we get

Question 13.

If y = 3 cos (log x) + 4 sin (log x), show that x² y₂ + xy₁ + y = 0

Answer:

It is given that y = 3 cos (log x) + 4 sin (log x)

Now, on differentiating we get,

Again differentiating we get,

Therefore,

x² y₂ + xy₁ + y

= -sin(logx) – 7cos(logx) + 4cos(logx) – 3sin(logx) + 3cos(logx) + 4sin(logx)

= 0

So, x² y₂ + xy₁ + y = 0

Hence Proved

Question 14.

If y = Ae^mx + Be^nx, show that 

Answer:

According to given equation, we have,

y = Ae^mx + Be^nx

Then,

= Ame^mx + Bne^nx

Now, on again differentiating we get,

= Am²e^mx + Bn²e^nx

= Am²e^mx + Bn²e^nx – (m+n)( Ame^mx + Bne^nx) + mn(Ae^mx + Be^nx)

= Am²e^mx + Bn²e^nx- Am²e^mx - Bmne^nx -Amne^mx - Bn²e^nx + Amne^mx + Bmne^nx

= 0

Hence Proved

Question 15.

If y = 500e^7x + 600e^–7x, show that .

Answer:

According to given equation, we have,

y = 500e^7x + 600e^–7x

= 3500e^7x - 4200e^-7x

Now, on again differentiating we get,

= 7×3500.e^7x + 7×4200.e^-7x

= 49×500e^7x + 49×600e^-7x

= 49(500e^7x + 600e^-7x)

= 49y

Hence Proved

Question 16.

If e^y (x + 1) = 1, show that= 

Answer:

It is given that

e^y (x + 1) = 1

Now, taking logarithm on both the sides we get,

On differentiating both sides we get,

Again, on differentiating we get,

Hence Proved

Question 17.

If y = (tan^–1 x)², show that (x² + 1)² y₂ + 2x (x² + 1) y₁ = 2

Answer:

: It is given that

y = (tan^–1 x)²

On differentiating we get,

Again differentiating, we get,

So, (1+x²)²y₂ + 2x(1+x²)y₁ = 2

where, 

Hence Proved

---

Exercise 5.8

Question 1.

Verify Rolle’s theorem for the function f (x) = x² + 2x – 8, x ∈ [– 4, 2].

Answer:

The given function is f (x) = x² + 2x – 8 and x ∈ [-4, 2].

By Rolle’s Theorem, for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then there exists some c in (a, b) such that f '(c) = 0.

As f(x) = x² + 2x – 8 is a polynomial function,

(a) f(x) is continuous in [-4, 2]

(b) f'(x) = 2x + 2

So, f(x) is differentiable in (-4, 2).

(c) f(a) = f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 =16 – 16 = 0

f(b) = f(2) = (2)² + 2(2) – 8 = 4 + 4 – 8 = 8 – 8 = 0

Hence, f(a) = f(b).

∴ There is a point c ∈ (-4, 2) where f'(c) = 0.

f(x) = x² +2x – 8

f'(x) = 2x + 2

f'(c) = 0

⇒ f'(c) = 2c + 2 = 0

⇒2c = -2

⇒ c = -2 /2

⇒ c = -1 where c = -1 ∈ (-4, 2)

Hence, Rolle’s Theorem is verified.

Question 2.

Examine if Rolle’s theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s theorem from these examples?

(i) f (x) = [x] for x ∈ [5, 9]

(ii) f (x) = [x] for x ∈ [– 2, 2]

(iii) f (x) = x 2 – 1 for x ∈ [1, 2]

Answer:

By Rolle’s Theorem, for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then there exists some c in (a, b) such that f '(c) = 0.

If a function does not satisfy any of the above conditions, then Rolle’s Theorem is not applicable.

(i) f (x) = [x] for x ∈ [5, 9]

As the given function is a greatest integer function,

(a) f(x) is not continuous in [5, 9]

(b) Let y be an integer such that y ∈ (5, 9)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x = y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in [5, 9]

(c) f(a)= f(5) = [5] = 5

f(b) = f(9) = [9] = 9

f(a) ≠ f(b)

Here, f(x) does not satisfy the conditions of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = [x] for x∈[5, 9].

(ii) f (x) = [x] for x ∈ [– 2, 2]

As the given function is a greatest integer function,

(a) f(x) is not continuous in [-2, 2]

(b) Let y be an integer such that y ∈ (-2, 2)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x = y are not equal, f(x) is not differentiable at x = y.

So, f(x) is not differentiable in (-2, 2)

(c) f(a)= f(-2) = [-2] = -2

f(b) = f(2) = [2] = 2

f(a) ≠ f(b)

Here, f(x) does not satisfy the conditions of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = [x] for x∈[-2, 2].

(iii) f (x) = x² – 1 for x ∈ [1, 2]

As the given function is a polynomial function,

(a) f(x) is continuous in [1, 2]

(b) f'(x) = 2x

So, f(x) is differentiable in [1, 2]

(c) f(a) = f(1) = 1² – 1 = 1 – 1 = 0

f(b) = f(2) = 2² - 1 = 4 – 1 = 3

f(a) ≠ f(b)

Here, f(x) does not satisfy a condition of Rolle’s Theorem.

Rolle’s Theorem is not applicable for f(x) = x² – 1 for x∈[1, 2].

Question 3.

If f : [– 5, 5] → R is a differentiable function and if f′(x) does not vanish anywhere, then prove that f (– 5) ≠ f (5).

Answer:

Given: f : [-5, 5] → R is a differentiable function.

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that 

We know that a differentiable function is a continuous function.

So,

(a) f is continuous on [-5, 5]

(b) f is differentiable on (-5, 5)

∴ By Mean Value Theorem, there exists c ∈ (-5, 5) such that

⇒ 

⇒ 10 f'(c) = f(5) – f(-5)

It is given that f'(x) does not vanish anywhere.

∴ f'(c) ≠ 0

10 f'(c) ≠ 0

f(5) – f(-5) ≠0

f(5) ≠ f(-5)

Hence proved.

By Mean Value Theorem, it is proved that f(5) ≠ f(-5).

Question 4.

Verify Mean Value Theorem, if f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b =4.

Answer:

Given: f(x) = x² – 4x – 3 in the interval [1, 4]

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that 

As f(x) is a polynomial function,

(a) f(x) is continuous in [1, 4]

(b) f'(x) = 2x – 4

So, f(x) is differentiable in (1, 4).

∴ 

f(4) = 4² – 4(4) – 3 = 16 – 16 – 3 = -3

f(1) = 1² – 4(1) – 3 = 1 – 4 – 3 = -6

∴ There is a point c ∈ (1, 4) such that f'(c) = 1

⇒ f'(c) = 1

⇒ 2c – 4 = 1

⇒ 2c = 1+4 =5

⇒ c = 5/2 where c ∈ (1,4)

The Mean Value Theorem is verified for the given f(x).

Question 5.

Verify Mean Value Theorem, if f (x) = x³ – 5x² – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f′(c) = 0.

Answer:

Given: f(x) = x³ – 5x² – 3x in the interval [1, 3]

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that 

As f(x) is a polynomial function,

(a) f(x) is continuous in [1, 3]

(b) f'(x) = 3x² - 10x - 3

So, f(x) is differentiable in (1, 3).

∴ 

f(3) = 3³ – 5(3)² – 3(3) = 27 – 45 - 9 = -27

f(1) = 1³ – 5(1)² – 3(1)= 1 – 5 - 3 = -7

⇒ 

∴ There is a point c ∈ (1, 4) such that f'(c) = -10

⇒ f'(c) = -10

⇒ 3c² – 10c - 3 = -10

⇒ 3c² – 10c +7 =0

⇒ 3c² - 3c – 7c + 7 = 0

⇒ 3c(c – 1) – 7(c – 1) = 0

⇒(c – 1)(3c – 7) = 0

⇒ c = 1, 7/3 where c = 7/3 ∈ (1, 3)

The Mean Value Theorem is verified for the given f(x) and c = 7/3∈(1, 3) is the only point for which f'(c) = 0.

Question 6.

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Answer:

Mean Value Theorem states that for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that 

If a function does not satisfy any of the above conditions, then Mean Value Theorem is not applicable.

(i) f (x) = [x] for x ∈ [5, 9]

As the given function is a greatest integer function,

(a)f(x) is not continuous in [5, 9]

(b) Let y be an integer such that y ∈ (5, 9)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x=y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in [5, 9].

Here, f(x) does not satisfy the conditions of Mean Value Theorem.

Mean Value Theorem is not applicable for f(x) = [x] for x∈[5, 9].

(ii) f (x) = [x] for x ∈ [– 2, 2]

As the given function is a greatest integer function,

(a)f(x) is not continuous in [-2, 2]

(b) ) Let y be an integer such that y ∈ (-2, 2)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x=y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in (-2, 2)

Here, f(x) does not satisfy the conditions of Mean Value Theorem.

Mean Value Theorem is not applicable for f(x) = [x] for x∈[-2, 2].

(iii) f (x) = x² – 1 for x ∈ [1, 2]

As the given function is a polynomial function,

(a) f(x) is continuous in [1, 2]

(b) f'(x) = 2x

So, f(x) is differentiable in [1, 2].

Here, f(x) satisfies the conditions of Mean Value Theorem.

So, Mean Value Theorem is applicable for f(x).

∴ 

f(2) = 2² – 1 = 4 -1 = 3

f(1) = 1² – 1 = 1 – 1 = 0

⇒ 

∴ There is a point c ∈ (1, 2) such that f'(c) = 3

⇒ f'(c) = 3

⇒ 2c = 3

⇒ c = 3/2 where c ∈ (1, 2)

Mean Value Theorem is applicable for f(x) = x² – 1 for x∈[1, 2].

---

Miscellaneous Exercise

Question 1.

Differentiate w.r.t. x the function

(3x² – 9x + 5)⁹

Answer:

Let y = (3x² – 9x + 5)⁹

If u = v(w(x))

Then using chain rule 

∴ Differentiating y w.r.t. x using chain rule

= 9(3x² – 9x + 5)⁸×(6x – 9)

=9(3x² – 9x + 5)⁸ × 3(2x – 3)

=27(3x² – 9x + 5)⁸ (2x – 3)

∴ 

Question 2.

Differentiate w.r.t. x the function

sin³ x + cos⁶ x

Answer:

Let y = sin³ x + cos⁶ x

Differentiating both sides with respect to x

∴  ∵ ]

= 3 sin² x × cos x + 6 cos⁵ x × ( – sin x)

= 3 sinx cosx (sinx– 2 cos⁴ x)

∴  = 3 sinx cosx (sinx– 2 cos⁴ x)

Question 3.

Differentiate w.r.t. x the function

(5x)^{3cos 2x}

Answer:

Let y = (5x)^{3cos 2x}

Then log y = log (5x)^{3cos 2x}

⇒ log y = 3cox 2x × log 5x

Differentiating both sides with respect to x, we get

 [∵ ]

⇒ 

⇒ 

⇒ 

⇒ 

∴ 

Question 4.

Differentiate w.r.t. x the function

Answer:

Let y = 

Differentiating both sides with respect to x, we get

Using chain rule we get

⇒ 

⇒ 

⇒ 

⇒ 

∴ 

Question 5.

Differentiate w.r.t. x the function

Answer:

Let y = 

Differentiating both sides with respect to x, we get

Using Quotient rule

∴ 

Question 6.

Differentiate w.r.t. x the function

Answer:

Let y = 

⇒ 

Substituting the value of  in y.

∴ y = 

⇒ y = x/2

Differentiating both sides with respect to x, we get

∴ 

Question 7.

Differentiate w.r.t. x the function

(log x)^{log x}, x > 1

Answer:

Let y = 

Taking logarithm on both sides

⇒ log y = log (log x)^{log x} = log x × log (log x)

Differentiating both sides with respect to x, we get

⇒ 

⇒ 

⇒ 

∴ 

Question 8.

Differentiate w.r.t. x the function

cos (a cos x + b sin x), for some constant a and b.

Answer:

Let y = cos (a cos x + b sin x)

a and b are some constants

y = cos (a cos x + b sin x)

Differentiating both sides with respect to x, we get

Using chain rule

⇒ 

⇒ 

⇒ 

∴ 

Question 9.

Differentiate w.r.t. x the function

Answer:

Let y = 

Taking logarithm both sides, we get

log y = log [(sin x – cos x)^{(sin x – cos x)}]

⇒ log y = (sin x – cos x) × log (sin x – cos x)

Differentiating both sides with respect to x, we get

⇒ 

⇒

⇒ 

∴ 

Question 10.

Differentiate w.r.t. x the function

x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0

Answer:

Let y = x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0

And let x^x = u, x^a = v, a^x = w and a^a = s

Then y = u + v + w + s

∴  ……..(I)

Now,

u = x^x

Taking logarithm both sides, we get

log u = log x^x

⇒ log u = x log x

Differentiating both sides w.r.t. x

⇒ 

⇒  …….(II)

v = x^a

Differentiating both sides with respect to x

⇒  ………..(III)

w = a^x

Taking logarithm both sides

log w = log a^x

log w = x log a

Differentiating both sides with respect to x

⇒ 

⇒ …………….(IV)

s = a^a

Differentiating both sides with respect to x

 … ……….(V)

Putting (II), (III), (IV) and (V) in (I)

∴ 

Question 11.

Differentiate w.r.t. x the function

, for x > 3

Answer:

Let y = 

And let  = u &  = v

∴ y = u + v

Differentiating both sides w.r.t. x we get

 ………..(I)

Now,

Taking logarithm both sides

⇒ 

Differentiating w.r.t. x, we get

⇒ 

⇒  ……………(II)

Also,

Taking logarithm both sides

⇒ 

Differentiating both sides w.r.t. x

⇒ 

⇒ 

⇒ ……………………….(II)

Substituting (II) and (III) in (I)

∴ 

Question 12.

Find dy/dx, if y = 12 (1 – cos t), x = 10 (t – sin t), 

Answer:

To find  we need to find out and 

So, 

Given, y = 12 (1 – cos t) and x = 10 (t – sin t)

x = 10 (t – sin t)

Differentiating with respect to t.

⇒ 

y = 12 (1 – cos t)

Differentiating with respect to t.

⇒ 

∴ 

∴ 

Question 13.

Find dy/dx, if , 0 < x < 1

Answer:

Given,

Differentiating with respect to x

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

∴ 

Question 14.

If  for , – 1 < x < 1, prove that 

Answer:

Given, 

Now, squaring both sides, we get

⇒ 

⇒ 

⇒ x² + x²y = y² + y²x

⇒ x² – y² = xy² – x²y

⇒ (x + y)(x – y) = xy (y – x)

⇒ x + y = –xy

⇒ y + xy = –x

⇒ y (1 + x) = –x

⇒ 

Differentiating both sides with respect to x, we get

Using Quotient Rule

∴ 

Hence, Proved

Question 15.

If (x – a)² + (y – b)² = c², for some c > 0, prove that  is a constant independent of a and b.

Answer:

Given, (x – a)² + (y – b)² = c²

Differentiating with respect to x, we get

⇒ 

⇒ 

∴ 

Differentiating again with respect to x

Using Quotient Rule

⇒ 

⇒ 

Substituting the value of dy/dx in the above equation

⇒ 

⇒ 

∴ 

∴ , which is independent of a and b

Hence, Proved

Question 16.

If cos y = x cos (a + y), with cos a ≠ ± 1, prove that 

Answer:

Given, cos y = x cos (a + y)

Differentiating both sides with respect to x

⇒ 

⇒ 

⇒ …………..(I)

Since, cos y = x cos (a + y) ⇒ x = cos y/cos (a + y)

Substituting the value of x in (I)

⇒ 

⇒ 

⇒ 

⇒ 

Hence, proved

Question 17.

If x = a (cos t + t sin t) and y = a (sin t – t cos t), find d²y/dx².

Answer:

Given, x = a (cos t + t sin t) and y = (sin t – t cos t)

To find  we need to find out and 

So,  and 

x = a (cos t + t sin t)

Differentiating with respect to t.

⇒ 

⇒ 

y = a (sin t – t cos t)

Differentiating with respect to t.

⇒ 

∴ 

Differentiating dy/dx with respect to t

And 

⇒ 

∴ 

Question 18.

If f (x) = |x|³, show that f″(x) exists for all real x and find it.

Answer:

When, x ≥ 0,

f(x) = |x|³ = x³

So, f’(x) = 3x²

And f’’(x) = d(f’(x))/dx = 6x

∴ f’’(x) = 6x

When x < 0,

f(x) = |x|³ = ( – x)³ = – x³

f’(x) = – 3x²

f’’(x) = – 6x

∴ 

Question 19.

Using mathematical induction prove that  for all positive integers n.

Answer:

To prove : P(n) :  = nx^{n – 1} for all positive integers n

For n = 1,

LHS =  = 1

RHS = 1 × x^{1 – 1} = 1

So, LHS = RHS

∴ P(1) is true.

∴ P(n) is true for n = 1

Let P(k) be true for some positive integer k.

i.e. P(k) = 

Now, to prove that P(k + 1) is also true

RHS = (k + 1)x^{(k + 1) – 1}

LHS = 

∴ LHS = RHS

Thus, P(k + 1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.

Hence, proved.

Question 20.

Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.

Answer:

sin (A + B) = sin A cos B + cos A sin B

Differentiating with respect to x, we get

⇒

⇒

⇒ 

∴ cos (A + B) = cos A cos B – sin A sin B

Question 21.

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Answer:

Considering the function

f(x) = |x| + |x + 1|

The above function f is continuous everywhere, but is not differentiable at x = 0 and x = – 1

Now, checking continuity

CASE I: At x < – 1

f(x) = – 2x – 1

f(x) is a polynomial

⇒ f(x) is continuous [∵ Every polynomial function is continuous]

CASE II: x > 0

f(x) = 2x + 1

f(x) is a polynomial

⇒ f(x) is continuous [∵ Every polynomial function is continuous]

CASE III: At – 1 < x < 0

f(x) = 1

f(x) is constant

⇒ f(x) is continuous

CASE IV: At x = – 1

A function will be continuous at x = – 1

If LHL = RHL = f( – 1)

i.e. 

LHL = 

Putting x = – 1

LHL = – 2 × ( – 1) – 1 = 2 – 1 = 1

RHL = 

f(x) = – 2x – 1

f( – 1) = – 2 × ( – 1) – 1 = 2 – 1 = 1

so, LHL = RHL = f( – 1)

⇒ f is continuous.

CASE V: At x = 0

A function will be continuous at x = 0

If LHL = RHL = f(0)

i.e. 

LHL =  = 1

RHL = 

Putting x = 0

RHL = 2 × 0 + 1 = 1

f(x) = 2x + 1

f(0) = 2 × 0 + 1 = 0 + 1 = 1

so, LHL = RHL = f(0)

⇒ f is continuous.

Thus f(x) = |x| + |x + 1| is continuous for all values of x.

Checking differentiability

CASE I: At x < – 1

f(x) = – 2x – 1

f’(x) = – 2

f(x) is polynomial.

⇒ f(x) is differentiable

CASE II: At x > 0

f(x) = 2x + 1

f’(x) = 2

f(x) is polynomial.

⇒ f(x) is differentiable

CASE III: At – 1 < x < 0

f(x) = 1

f(x) is constant.

⇒ f(x) is differentiable

CASE IV: At x = – 1

f is differentiable at x = – 1 if

LHD = RHD = f’( – 1)

i.e. 

LHD = 

LHD = 

RHD = 

Since, LHD ≠ RHD

∴ f is not differentiable at x = – 1

CASE V: At x = 0

f is differentiable at x = 0 if

LHD = RHD = f’(0)

i.e. 

LHD = 

RHD = 

Since, LHD ≠ RHD

∴ f is not differentiable at x = 0

So, f is not differentiable at exactly two point x = 0 and x = 1, but continuous at all points.

Question 22.

If , prove that 

Answer:

Let y = 

Differentiation of determinant  is given by

∴

Since, a, b, c and l, m, n are constants so, their differentiation is zero.

Also in a determinant if all the elements of row or column turns to be zero then the value of determinant is zero.

∴ 

∴ 

Hence, proved.

Question 23.

If,  show that 

Answer:

Given, 

Taking logarithm both sides, we get

⇒ log y = a cos ^{– 1}x log e

⇒ log y = a cos ^{– 1}x [log e = 1]

Differentiating both sides with respect to x

⇒ 

Squaring both sides

⇒ 

Differentiating both sides

⇒ 

⇒ 

⇒ 

∴ 

Hence, proved