# RD Sharma Class 12 Ex 15.1 Solutions Chapter 15 Mean Value Theorems

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## RD Sharma Class 12 Ex 15.1 Solutions Chapter 15 Mean Value Theorems

### (i) f(x) = 3 + (x â€“ 2)2/3 on [1, 3]

Solution:

Rolleâ€™s theorem states that if a function f is continuous on the closed interval [a, b] and

differentiable on the open interval (a, b) such that f(a) = f(b), then fâ€²(x) = 0 for some x with a â‰¤ x â‰¤ b.

We have, f'(x) =

â‡’ f'(x) =

â‡’  f'(x) =

â‡’  f'(x) =

To analyse differentiability at x = 2:

â‡’  f'(2) =

= 2 / 0 â‰  0

Since f'(2) â‰  0, f(x) is not differentiable in the interval [1, 3].

Hence, the Rolleâ€™s Theorem is not applicable for the given function f(x) in interval [1, 3].

### (ii) f(x) = [x] for -1 â‰¤ x â‰¤ 1, where [x] denotes the greatest integer not exceeding x

Solution:

Rolleâ€™s theorem requires a function to be continuous on the closed interval [a, b].

The continuity of f(x) needs to be analyzed at x = 1:

(RHL at x = 1) = limxâ‡¢(1+h)[x]

= limhâ‡¢0[1 + h]

= 1              â€¦.. (1)

(LHL at x = 1) = limxâ‡¢(1-h)[x]

= limhâ‡¢0[1 â€“ h]

= 0            â€¦â€¦ (2)

Since f(x) is not continuous at x = 1 or the interval [-1, 1].

Hence, the Rolleâ€™s Theorem is not applicable for the given function f(x) in interval [-1, 1].

### (iii) f(x) = sin(1/x) for -1 â‰¤ x â‰¤ 1

Solution:

Rolleâ€™s theorem requires a function to be continuous on the closed interval [a, b].

The continuity of f(x) needs to be analyzed at x = 0:

(RHL at x = 0) = limxâ‡¢(0+h)sin(1/x)

= limhâ‡¢0sin(1/h)

= k                  â€¦..(1)

(LHL at x = 0) = limxâ‡¢(0-h)sin(1/x)

= limhâ‡¢0sin(1/0-h)

= limhâ‡¢0sin(-1/h)

= -limhâ‡¢0sin(1/h)

= âˆ’k               â€¦.(2)

Since f(x) is not continuous at x = 1 or the interval [-1,1].

Hence, the Rolleâ€™s Theorem is not applicable for the given function f(x) in interval [-1, 1].

### (iv) f(x) = 2x2 â€“ 5x + 3 on [1, 3]

Solution:

Clearly f(x) being a polynomial function shall be continuous on the interval.

We need to verify whether f(a) = f(b) for the applicability of theorem.

Here, f(1) = 2(1)2 â€“ 5(1) + 3

â‡’ f(1) = 2 â€“ 5 + 3

â‡’ f(1) = 0                             â€¦â€¦ (1)

Now, f(3) = 2(3)2 â€“ 5(3) + 3

â‡’ f(3) = 2(9) â€“ 15 + 3

â‡’ f(3) = 18 â€“ 12

â‡’ f(3) = 6                              â€¦â€¦ (2)

From eq(1) and (2), we can say that, f(1) â‰  f(3).

We observe that f(a) â‰  f(b)

Hence, the Rolleâ€™s Theorem is not applicable for the given function f(x) in interval [1, 3].

### (v) f (x) = x2/3 on [-1, 1]

Solution:

Rolleâ€™s theorem states that if a function f is continuous on the closed interval [a, b] and

differentiable on the open interval (a, b) such that f(a) = f(b), then fâ€²(x) = 0 for some x with a â‰¤ x â‰¤ b.

f'(x) =

â‡’  f'(x) =

Clearly f'(x) is undefined at x = 0.

Hence, the Rolleâ€™s Theorem is not applicable for the given function f(x) in interval [-1, 1].

### (vi)

Solution:

Rolleâ€™s theorem requires a function to be continuous on the closed interval [a, b].

The continuity of f(x) needs to be analyzed at x = 1:

(LHL at x = 1) = lim(x->1-h) (-4x + 5)

= lim(h->0) [-4(1 â€“ h) + 5]

= -4 + 5

= 1

(RHL at x = 1) =  lim(x->1+h) (2x â€“ 3)

= lim(h->0) [2(1 + h) â€“ 3]

= 2(0) â€“ 3

= -1

We observe that LHL â‰  RHL.

Clearly f(x) is not continuous at x = 1.

Hence, the Rolleâ€™s Theorem is not applicable for the f(x) function in interval [-1, 1].

### (i) f (x) = x2 â€“ 8x + 12 on [2, 6]

Solution:

Given function is f (x) = x2 â€“ 8x + 12 on [2, 6].

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

We only need to check whether f(a) = f(b) or not.

Thus, f(2) = 22 â€“ 8(2) + 12

â‡’ f(2) = 4 â€“ 16 + 12

â‡’ f(2) = 0

Now, f(6) = 62 â€“ 8(6) + 12

â‡’ f(6) = 36 â€“ 48 + 12

â‡’ f(6) = 0

Since f(2) = f(6), Rolleâ€™s theorem is applicable for function f(x) on [2, 6].

f(x) = x2 â€“ 8x + 12

f'(x) = 2x â€“ 8

Then f'(c) = 0

2c â€“ 8 = 0

c = 4 âˆˆ (2, 6)

Hence, the Rolleâ€™s theorem is verified

### (ii) f(x) = x2 â€“ 4x + 3 on [1, 3]

Solution:

Given function is f (x) = x2 â€“ 4x + 3 on [1, 3]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

We only need to check whether f(a) = f(b) or not.

So f (1) = 12 â€“ 4(1) + 3

â‡’ f (1) = 1 â€“ 4 + 3

â‡’ f (1) = 0

Now, f (3) = 32 â€“ 4(3) + 3

â‡’ f (3) = 9 â€“ 12 + 3

â‡’ f (3) = 0

âˆ´ f (1) = f(3), Rolleâ€™s theorem applicable for function â€˜fâ€™ on [1, 3].

â‡’ fâ€™(x) = 2x â€“ 4

We have fâ€™(c) = 0, c Ïµ (1, 3), from the definition of Rolleâ€™s Theorem.

â‡’ fâ€™(c) = 0

â‡’ 2c â€“ 4 = 0

â‡’ 2c = 4

â‡’ c = 4/2

â‡’ c = 2 âˆˆ (1, 3)

Hence, Rolleâ€™s Theorem is verified.

### (iii) f (x) = (x â€“ 1) (x â€“ 2)2 on [1, 2]

Solution:

Given function is f (x) = (x â€“ 1) (x â€“ 2)2 on [1, 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.

We only need to check whether f(a) = f(b) or not.

So f (1) = (1 â€“ 1) (1 â€“ 2)2

â‡’ f (1) = 0(1)2

â‡’ f (1) = 0

Now, f (2) = (2 â€“ 1)(2 â€“ 2)2

â‡’ f (2) = 02

â‡’ f (2) = 0

Since f (1) = f (2), Rolleâ€™s Theorem applicable for function f(x) on [1, 2].

f(x) = (x â€“ 1)(x â€“ 2)2

f'(x) = (x â€“ 1) Ã— 2(x â€“ 2) + (x â€“ 2)2

f'(x) = (x â€“ 2) (3x â€“ 4)

Then f'(c) = 0

(c â€“ 2)(3c â€“ 4) = 0

c = 2

or

c = 4/3 âˆˆ (1, 2)

Hence, Rolleâ€™s Theorem is verified.

### (iv) f (x) = x (x â€“ 1)2 on [0, 1]

Solution:

Given function is f (x) = x(x â€“ 1)2 on [0, 1]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is, on R.

We only need to check whether f(a) = f(b) or not.

So f (0) = 0 (0 â€“ 1)2

â‡’ f (0) = 0

Now f (1) = 1 (1 â€“ 1)2

â‡’ f (1) = 02

â‡’ f (1) = 0

Since f (0) = f (1), Rolleâ€™s theorem applicable for function f(x) on [0, 1].

f (x) = x (x â€“ 1)2

f'(x) = (x â€“ 1)2+ x Ã— 2(x â€“ 1)

= (x â€“ 1)(x â€“ 1 + 2x)

= (x â€“ 1)(3x â€“ 1)

Then f'(c) = 0

(c â€“ 1)(3c â€“ 1) = 0

c = 1 or 1/3 âˆˆ (0, 1)

Hence, Rolleâ€™s Theorem is verified.

### (v) f (x) = (x2 â€“ 1) (x â€“ 2) on [-1, 2]

Solution:

Given function is f (x) = (x2 â€“ 1) (x â€“ 2) on [â€“ 1, 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.

We only need to check whether f(a) = f(b) or not.

So f(â€“1) = (( â€“ 1)2 â€“ 1)( â€“ 1 â€“ 2)

â‡’ f ( â€“ 1) = (1 â€“ 1)( â€“ 3)

â‡’ f ( â€“ 1) = (0)( â€“ 3)

â‡’ f ( â€“ 1) = 0

Now, f (2) = (22 â€“ 1)(2 â€“ 2)

â‡’ f (2) = (4 â€“ 1)(0)

â‡’ f (2) = 0

âˆ´ f (â€“ 1) = f (2), Rolleâ€™s theorem applicable for function f on[ â€“ 1, 2].

â‡’ fâ€™(x) = 3x2 â€“ 4x â€“ 1

We have fâ€™(c) = 0 c âˆˆ (-1, 2), from the definition of Rolleâ€™s Theorem.

Clearly, f'(c) = 0

3c2 â€“ 4c â€“ 1 = 0

c = (2 Â±âˆš7)/ 3

c = 1/3(2 + âˆš7) or 1/3(2 â€“ âˆš7) âˆˆ (-1, 2)

Hence, Rolleâ€™s Theorem is verified.

### (vi) f(x) = x(x â€“ 4)2 on [0, 4]

Solution:

Given function is f (x) = x(x â€“ 4)2 on [0, 4]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

We only need to check whether f(a) = f(b) or not.

So f (0) = 0(0 â€“ 4)2

â‡’ f (0) = 0

Now, f (4) = 4(4 â€“ 4)2

â‡’ f (4) = 4(0)2

â‡’ f (4) = 0

Since f (0) = f (4), Rolleâ€™s theorem applicable for function f(x) on [0, 4].

f(x) = x(x â€“ 4)2

f'(x) = x Ã— 2(x â€“ 4) + (x â€“ 4)2

= 2x2 â€“ 8x + x2 + 16 â€“ 8x

= 2x2 â€“ 16x + 16

Then

f'(c) = 2c2 â€“ 16c + 16

= (c â€“ 4)(3c â€“ 4)

c = 4 or 4/3 âˆˆ (0, 4)

Hence, Rolleâ€™s Theorem is verified.

### (vii) f (x) = x(x â€“ 2)2 on [0, 2]

Solution:

Given function is f (x) = x(x â€“ 2)2 on [0, 2].

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.

We only need to check whether f(a) = f(b) or not.

So f (0) = 0(0 â€“ 2)2

â‡’ f (0) = 0

Now, f (2) = 2(2 â€“ 2)2

â‡’ f (2) = 2(0)2

â‡’ f (2) = 0

f (0) = f(2), Rolleâ€™s theorem applicable for function f on [0,2].

c = 12/6 or 4/6

c = 2 or 2/3

So, c = 2/3 since c âˆˆ (0, 2)

Hence, Rolleâ€™s Theorem is verified.

### (viii) f (x) = x2 + 5x + 6 on [-3, -2]

Solution:

Given function is f (x) = x2 + 5x + 6 on [â€“ 3, â€“ 2].

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

We only need to check whether f(a) = f(b) or not.

So f(â€“3) = (â€“3)2 + 5(â€“3) + 6

â‡’ f(â€“3) = 9 â€“ 15 + 6

â‡’ f(â€“3) = 0

Now, f(â€“2) = (â€“2)2 + 5(â€“2) + 6

â‡’ f(â€“2) = 4 â€“ 10 + 6

â‡’ f(â€“2) = 0

Since f(â€“3) = f(â€“2), Rolleâ€™s theorem applicable for function f(x) on [â€“3, â€“2].

f(x) = x2 + 5x + 6

f'(x) = 2x + 5

Then

f(c) = 0

2c + 5 = 0

c = -5/2 âˆˆ (-3, -2)

Hence, Rolleâ€™s Theorem is verified.

### (i) f(x) = cos2(x â€“ Ï€/4) on [0, Ï€/2]

Solution:

Clearly the cosine function is continuous and differentiable everywhere.

Hence, the given function f(x) is continuous in [0, Ï€/2] and differentiable in [0, Ï€/2]

So,

f(0) = cos2(0 â€“ Ï€/4) = 0

f(Ï€/2) = cos2(Ï€/2 â€“ Ï€/4) = 0

Since f(0) = f(Ï€/2), Rolleâ€™s theorem applicable for function f(x) on [0, Ï€/2].

Now, f'(x) =

= -2sin(2x â€“ Ï€/2)

If f'(c) = 0

â‡’  -2sin(2c â€“ Ï€/2) = 0

â‡’ c = Ï€/4 âˆˆ [0, Ï€/2]

Clearly c belongs to the given interval.

Hence, Rolleâ€™s Theorem is verified.

### (ii) f(x) = sin 2x on [0, Ï€/2]

Solution:

Since sin 2x is everywhere continuous and differentiable.

So, sin2x is continuous on (0, Ï€/2) and differentiable on (0, Ï€/2).

f(0) = sin 0 = 0

f(Ï€/2) = sin Ï€/2 = 0

f(Ï€/2) = f(0) = 0

Thus, f(x) satisfies all the conditions of Rolleâ€™s theorem.

Now, we have to show that there exists c in (0, Ï€/2) such that f'(c) = 0.

We have, f'(x) = 0

â‡’ cos2x = 0

â‡’ x = Ï€/4

So, f'(c) = 2cos2c

2cos2c = 0

Thus, c = Ï€/4 in (0, Ï€/2).

â€‹Hence, Rolleâ€™s theorem is verified.

### (iii) f(x) = cos2x on [-Ï€/4, Ï€/4]

Solution:

Since cos 2x is everywhere continuous and differentiable.

So, cos2x is continuous on [-Ï€/4, Ï€/4] and differentiable on [-Ï€/4, Ï€/4].

cos(-Ï€/4) = cos2(-Ï€/4) = cos(-Ï€/2) = 0

cos(Ï€/4) = cos2(Ï€/4) = cos(Ï€/2) = 0

Also, cos(-Ï€/4) = cos(Ï€/4)

Thus, f(x) satisfies all the conditions of Rolleâ€™s theorem.

Now, we have to show that there exists c in (0, Ï€/2) such that f'(c) = 0.

We have f'(x) = 0

â‡’ sin2c = 0

â‡’ 2c = 0

So, c = 0 as c âˆˆ (-Ï€/4, Ï€/4)

Hence, Rolleâ€™s Theorem is verified for the given function f(x).

### (iv) ex sin x on [0, Ï€]

Solution:

The given function is a composition of exponential and trigonometric functions. Hence it is differential and continuous everywhere.

So, the given function f(x) is continuous on [0, Ï€] and differentiable on (0, Ï€).

f(0) = esin 0 = 0

f(Ï€) = eÏ€ sin Ï€ = 0

f(0) = f(Ï€)

Now, we have to show that there exists c in (0, Ï€) such that f'(c) = 0.

So f'(x) = ex (sin x + cos x)

â‡’ ex (sin x + cos x) = 0

â‡’ sin x + cos x = 0

Dividing both sides by cos x, we have

tan x = -1

â‡’  x = Ï€ â€“ Ï€/4 = 3Ï€/4

Since c = 3Ï€/4 in (0, Ï€) such that f'(c) = 0.

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (v) f(x) = ex cos x on [âˆ’Ï€/2, Ï€/2]

Solution:

The given function is a composition of exponential and trigonometric functions. Hence it is differential and continuous everywhere.

So, the given function f(x) is continuous on [-Ï€/2, Ï€/2] and differentiable on (-Ï€/2, Ï€/2).

f(âˆ’Ï€/2) = eÏ€/2 cos(âˆ’Ï€/2) = 0

f(Ï€/2) = eÏ€/2 cos(Ï€/2) = 0

f(âˆ’Ï€/2) = f(Ï€/2)

Now, we have to show that there exists c in (-Ï€/2, Ï€/2) such that f'(c) = 0.

f'(x) = ex (cos x â€“ sin x)

â‡’ ex (cos x â€“ sin x) = 0

â‡’ sin x â€“ cos x = 0

Dividing both sides by cos x, we have

tan x = 1

â‡’ x = Ï€/4

Since c = Ï€/4 in (-Ï€/2, Ï€/2) such that fâ€™(c) = 0.

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (vi) f (x) = cos 2x on [0, Ï€]

Solution:

Since cos 2x is everywhere continuous and differentiable.

So, cos2x is continuous on [0, Ï€] and differentiable on (0, Ï€).

f(0) = cos(0) = 1

f(Ï€) = cos(2Ï€) = 1

f(0) = f(Ï€)

Thus, f(x) satisfies all the conditions of Rolleâ€™s theorem.

There must be a c belongs to (0, Ï€) such that fâ€™(c) = 0.

â‡’ sin 2c = 0

So, 2c = 0 or Ï€

c = 0 or Ï€/2

But, c = Ï€/2 as c âˆˆ (0, Ï€)

Hence, Rolleâ€™s Theorem is verified for the given function f(x).

### (vii) f(x) = sinx/ex on 0 â‰¤ x â‰¤ Ï€

Solution:

The given function is a composition of exponential and trigonometric functions.

Hence it is differential and continuous everywhere.

So, sinx/ex is continuous on [0, Ï€] and differentiable on (0, Ï€).

f(0) = sin(0)/e0 = 0

f(Ï€) = sin(Ï€)/eÏ€ = 1

f(0) = f(Ï€)

Now, we have to show that there exists c in (0, Ï€/2) such that f'(c) = 0.

f'(x) = ex (cos x â€“ sin x)

â‡’ ex (cos x â€“ sin x) = 0

â‡’ sin x â€“ cos x = 0

â‡’ tan x = 1

â‡’  x = Ï€/4

Since c = Ï€/4 in (-Ï€/4, Ï€/4) such that fâ€™(c) = 0.

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (viii) f(x) = sin 3x on [0, Ï€]

Solution:

Given function is f (x) = sin3x on [0, Ï€]

Since sine function is continuous and differentiable on R.

We need to check whether f(a) = f(b) or not.

So f (0) = sin 3(0)

â‡’ f (0) = sin0

â‡’ f (0) = 0

Now, f (Ï€) = sin 3Ï€

â‡’ f (Ï€) = sin(3 Ï€)

â‡’ f (Ï€) = 0

We have f (0) = f (Ï€), so there exist a c belongs to (0, Ï€) such that fâ€™(c) = 0.

â‡’ 3cos(3x) = 0

â‡’ cos 3x = 0

â‡’ 3x = Ï€ / 2

â‡’ x = Ï€ / 6 âˆˆ [0, Ï€]

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (ix) f(x) =  on [âˆ’1, 1]

Solution:

We know that the exponential function is continuous and differentiable everywhere.

So, is continuous on [-1, 1] and differentiable on (-1, 1).

f(-1) = e1-1 = 1

f(1) = e1-1 = 1

Also, f(-1) = f(1) = 1

Thus, Rolleâ€™s theorem is applicable. So there must exist c âˆˆ  [âˆ’1, 1] such that f'(c) = 0.

f'(x) =

â‡’ -2x e^{1 â€“ x^2}  = 0

â‡’ x = 0 âˆˆ  [âˆ’1, 1]

â€‹Hence, Rolleâ€™s theorem is verified.

### (x) f (x) = log (x2 + 2) â€“ log 3 on [-1, 1]

Solution:

Clearly the logarithmic function is continuous and differentiable in its own domain.

We need to verify whether f(a) = f(b) or not.

So f (â€“ 1) = log((â€“ 1)2 + 2) â€“ log 3

â‡’ f (â€“ 1) = log (1 + 2) â€“ log 3

â‡’ f (â€“ 1) = log 3 â€“ log 3

â‡’ f ( â€“ 1) = 0

Now, f (1) = log (12 + 2) â€“ log 3

â‡’ f (1) = log (1 + 2) â€“ log 3

â‡’ f (1) = log 3 â€“ log 3

â‡’ f (1) = 0

We have got f (â€“ 1) = f (1).

So, there exists a c such that c Ïµ (â€“ 1, 1) such that fâ€™(c) = 0.

Now, f'(x) =

f'(c) = 0

â‡’ c = 0 Ïµ (â€“ 1, 1)

â€‹Hence, Rolleâ€™s theorem is verified.

### (xi) f(x) = sin x + cos x on [0, Ï€/2]

Solution:

Since sin and cos functions are continuous and differentiable everywhere,

f(x) is continuous on [0, Ï€/2] and differentiable on (0, Ï€/2).

So, f(0) = sin 0 + cos 0 = 1

f(Ï€/2) = sin Ï€/2 + cos Ï€/2 = 1

Also, f(0) = f(Ï€/2) = 1.

Hence, Rolleâ€™s theorem is applicable. So there must exist c âˆˆ  [âˆ’1, 1] such that f'(c) = 0.

Now, f'(x) = cos x â€“ sin x

â‡’ cos x â€“ sin x = 0

â‡’ tan x = 1

â‡’ x = Ï€/4

Thus, c = Ï€/4 on (0, Ï€/2).

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (xii) f (x) = 2 sin x + sin 2x on [0, Ï€]

Solution:

Since sin function continuous and differentiable over R, f(x) is continuous on [0, Ï€] and differentiable on (0, Ï€).

We need to verify whether f(a) = f(b) or not.

So f (0) = 2sin(0) + sin2(0)

â‡’ f (0) = 2(0) + 0

â‡’ f (0) = 0

Now, f (Ï€) = 2sin(Ï€) + sin2(Ï€)

â‡’ f (Ï€) = 2(0) + 0

â‡’ f (Ï€) = 0

We have f (0) = f (Ï€), so there exist a c belongs to (0, Ï€) such that fâ€™(c) = 0.

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

Now, f'(c) = 0

â‡’ 2cos x + cos 2x = 0

â‡’ (2cos c â€“ 1) (cos c + 1) = 0

â‡’ tan c = 1

â‡’ c = Ï€/3 in (0, Ï€)

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (xiii) f(x) =  on [-1, 0]

Solution:

Since sin function is continuous and differentiable over R, f(x) is continuous on [-1, 0] and differentiable on (-1, 0).

f(-1) = = 0

f(0) = 0 â€“ sin 0 = 0

Also f(-1) = f(0) = 0

Hence Rolleâ€™s theorem is applicable so there exist a c belongs to (0, Ï€) such that fâ€™(c) = 0.

f'(x) =

Now, f'(c) = 0

â‡’

â‡’

â‡’ c =

Thus, c Ïµ (-1, 0).

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (xiv) f(x) =  on [0, Ï€/6]

Solution:

Since sin function is continuous and differentiable everywhere, f(x) is continuous on [0, Ï€/6] and differentiable on (0, Ï€/6).

f(0) = 0 â€“ 0 = 0

f(Ï€/6) = 1 â€“ 1 = 0

Also, f(0) = f(Ï€/6) = 0

Hence Rolleâ€™s theorem is applicable so there exist a c belongs to (0, Ï€/6) such that fâ€™(c) = 0.

We have, f'(x) = 6/Ï€ â€“ 8sinxcosx

f'(x) = 6/Ï€ â€“ 4sin2x

f'(c) = 0

6/Ï€ â€“ 4sin2c = 0

4sin2c = 6/Ï€

sin2c = 3/2Ï€

2c = sin-1(21/44)

c = 1/2 sin-1(21/44)

c âˆˆ (-1/2, 1/2)

c âˆˆ (0, 11/21)

c âˆˆ (0, Ï€/6)

â€‹Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (xv) f(x) = 4sin x on [0, Ï€]

Solution:

Since sin and exponential functions are continuous and differentiable everywhere, f(x) is continuous on [0, Ï€] and differentiable on (0, Ï€).

f(0) = 4sin 0 = 1

f(Ï€) = 4sin Ï€ = 1

Also, f(0) = f(Ï€) = 1.

Hence Rolleâ€™s theorem is applicable so there exist a c belongs to (0, Ï€) such that fâ€™(c) = 0.

We have, f'(x) = 4sin x cos x log 4

So, fâ€™(c) = 0

4sin c cos c log 4 = 0

4sin c cos c = 0

cos c = 0

c = Ï€/2 âˆˆ (0, Ï€)

Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (xvi) f (x) = x2 â€“ 5x + 4 on [1, 4]

Solution:

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

We need to verify whether f(a) = f(b) or not.

So f (1) = 12 â€“ 5(1) + 4

â‡’ f (1) = 1 â€“ 5 + 4

â‡’ f (1) = 0

Now, f (4) = 42 â€“ 5(4) + 4

â‡’ f (4) = 16 â€“ 20 + 4

â‡’ f (4) = 0

We have f (1) = f (4). So, there exists a c âˆˆ  (1, 4) such that fâ€™(c) = 0.

f'(x) = 2x â€“ 5

Now, f'(c) = 2c â€“ 5 = 0

Thus, c = 5 / 2 âˆˆ (1, 4).

Hence, Rolleâ€™s theorem is verified for the given function f(x).

### (xvii) f(x) = sin4 x + cos4 x on (0, Ï€/2)

Solution:

Since sin and cos functions are continuous and differentiable everywhere, f(x) is continuous on [0, Ï€/2] and differentiable on (0, Ï€/2).

So,

f(0) = sin4(0) + cos4(0) = 1

f(Ï€/2) = sin4(Ï€/2) + cos4(Ï€/2) = 1

f(0) = f(Ï€/2)

Hence Rolleâ€™s theorem is applicable so there exist a c belongs to (0, Ï€/2) such that fâ€™(c) = 0.

â‡’ f'(x) = 4sin3 x cos x â€“ 4cos3 x sin x

so, f'(c) = 4sin3 x cos x â€“ 4cos3 x sin x  = 0

4sin3 c cos c â€“ 4cos3 c sin c  = 0

-sin4c = 0

4x = 0 or 4x = Ï€

x = 0, or Ï€/4 âˆˆ (0, Ï€/2)

Hence, Rolleâ€™s theorem is verified.

### (xviii) f(x) = sin x â€“ sin 2x on [0, Ï€]

Solution:

Given function is f (x) = sin x â€“ sin2x on [0, Ï€]

We know that sine function is continuous and differentiable over R.

Now we have to check the values of the function â€˜fâ€™ at the extremes.

â‡’ f (0) = sin (0) â€“ sin 2(0)

â‡’ f (0) = 0 â€“ sin (0)

â‡’ f (0) = 0

â‡’ f (Ï€) = sin(Ï€) â€“ sin2(Ï€)

â‡’ f (Ï€) = 0 â€“ sin(2Ï€)

â‡’ f (Ï€) = 0

We have f (0) = f (Ï€). So, there exists a c âˆˆ (0, Ï€) such that fâ€™(c) = 0.

f'(x) = cos x â€“ 2cos 2x

so, f'(c) = 0

f'(c) = cos c â€“ 2cos 2c = 0

cos c â€“ (2cos2c + 1) = 0

2cos2 c â€“ cosc â€“ 1 = 0

(cos c â€“ 1)(2cos c + 1) = 0

cos c = 1, -1/2

c = cos(Ï€/2), cos(2Ï€/3)

So, c = Ï€/2, 2Ï€/3 âˆˆ (0, Ï€)

Hence, Rolleâ€™s theorem is verified.

### Question 4. Using Rolleâ€™s Theorem, find points on the curve y = 16 â€“ x2, x âˆˆ [-1, 1], where the tangent is parallel to the x-axis.

Solution:

Given: y = 16 â€“ x2, x Ïµ [â€“ 1, 1]

â‡’ y (â€“ 1) = 16 â€“ (â€“ 1)2

â‡’ y (â€“ 1) = 16 â€“ 1

â‡’ y (â€“ 1) = 15

Now, y (1) = 16 â€“ (1)2

â‡’ y (1) = 16 â€“ 1

â‡’ y (1) = 15

We have y (â€“ 1) = y (1). So, there exists a c Ïµ (â€“ 1, 1) such that fâ€™(c) = 0.

We know that for a curve g, the value of the slope of the tangent at a point r is given by gâ€™(r).

â‡’ yâ€™ = â€“2x

We have yâ€™(c) = 0

â‡’ â€“ 2c = 0

â‡’ c = 0 Ïµ (â€“ 1, 1)

Value of y at x = 1 is

â‡’ y = 16 â€“ 02

â‡’ y = 16

Hence, the point at which the curve y has a tangent parallel to x â€“ axis (since the slope of x â€“ axis is 0) is (0, 16).

### (i) y = x2 on [-2, 2]

Solution:

Given: f(x) = x2

f'(x) = 2x

Also, f(-2) = f(2) = 22 = (-2)2 = 4

So, there exists a c Ïµ (-2, 2) such that fâ€™(c) = 0.

â‡’ 2c = 0

â‡’ c = 0

Therefore the required point is (0, 0).

### (ii) y =  on [âˆ’1, 1]

Solution:

Since exponential function is continuous and differentiable everywhere.

Here f(1) = f(-1) = 1

Thus, all the conditions of Rolleâ€™s theorem are satisfied.

Consequently, there exists at least one point c in (-1, 1) for which f'(c) = 0.

â‡’ f'(c) = 0

â‡’ -2c e^{1 â€“ c^2} = 0

â‡’ c = 0

Hence, (0, e) is the required point.

### (iii) y = 12 (x + 1) (x âˆ’ 2) on [âˆ’1, 2]

Solution:

Since polynomial function is continuous and differentiable everywhere.

Here f(2) = f(-1) = 0.

Thus, all the conditions of Rolleâ€™s theorem are satisfied.

Consequently, there exists at least one point c in (-1, 2) for which f'(c) = 0.

But f'(c) = 0

â‡’ 24c â€“ 12 = 0

â‡’ c = 1/2

Hence, (1/2, -27) is the required point.

### Question 6. If f: [5, 5] to R is a differentiable function and if f'(x) does not vanish anywhere, prove that f(-5) â‰  f(5).

Solution:

Here f(x) is differentiable on (5,5).

By Mean Value Theorem,

â‡’ 10 f'(c) = f(5) â€“ f(-5)

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

â‡’ f'(c) â‰  0

â‡’ 10 f'(c) â‰  0

â‡’  f(5) â€“ f(-5) â‰  0

â‡’  f(5) â‰  f(-5)

Hence Proved

### Can you say something about the converse of Rolleâ€™s Theorem from these functions?

Solution:

Rolleâ€™s theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then fâ€²(x) = 0 for some x with a â‰¤ x â‰¤ b.

(i) f(x) = [x] for x âˆˆ [ 5 , 9 ]

f(x) is not continuous at x = 5 and x = 9. Thus, f (x) is not continuous on [5, 9].

Also , f (5) = [5] = 5 and f (9) = [9] = 9

âˆ´ f (5) â‰  f (9)

The left hand limit of  f at x = n is âˆž and the right hand limit of f at x = n is 0.

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable on (5, 9).

Hence, Rolleâ€™s theorem is not applicable on function f(x) for x âˆˆ [5 , 9].

(ii)  f (x) = [x] for x âˆˆ  [-2 , 2]

f(x) is not continuous at x = âˆ’2 and x = 2. Thus, f (x) is not continuous on [âˆ’2, 2].

Also, f (-2) = [-2] = -2 and f (2) = [2] = 2

âˆ´ f (-2) â‰  f (2)

The left hand limit of f at x = n is âˆž and the right hand limit of f at x = n is âˆž.

Since the left and the right hand limits of f at x = n are not equal, thus, f is not differentiable on (âˆ’2, 2).

Hence, Rolleâ€™s theorem is not applicable on function f(x) = [x] for x âˆˆ [ -2, 2].

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