# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.8

### Question 1. Verify Rolleâ€™s theorem for the function f(x) = x^{2}+ 2x â€“ 8, x âˆˆ [â€“ 4, 2].

**Solution:**

Now f(x) = xÂ² + 2x â€“ 8 is a polynomial

So, f(x) is continuous in the interval [-4,2] and differentiable in the interval (- 4,2)

f(-4) = (-4)Â² + 2(-4) â€“ 8 = 16 â€“ 8 â€“ 8 = 0

f(2) = 2Â² + 4 â€“ 8 = 8 â€“ 8 = 0

f(-4) = f(2)

As Conditions of Rolleâ€™s theorem are

satisfied.Then there exists some c in (-4, 2) such that fâ€²(c) = 0

f'(x) = 2x + 2

fâ€™ (c) = 2c + 2 = 0

c = â€“ 1,

and -1 âˆˆ [-4,2]

Hence,

fâ€™ (c) = 0 at c = â€“ 1.

### Question 2. Examine if Rolleâ€™s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolleâ€™s theorem from these example?

### (i) f(x) = [x] for x âˆˆ [5, 9]

**Solution:**

In the interval [5, 9],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = 6,7,8

Hence, Rolleâ€™s theorem is

NOTapplicable

### (ii) f(x) = [x] for x âˆˆ [â€“ 2, 2]

**Solution:**

In the interval [â€“ 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Rolleâ€™s theorem is

NOTapplicable

### (iii) f(x) = x^{2}â€“ 1 for x âˆˆ [1, 2]

**Solution:**

Now f(x) = xÂ² – 1 is a polynomial

So, f(x) is continuous in the interval [1, 2] and differentiable in the interval (1,2)

f(1) = (1)Â² â€“ 1 = 0

f(2) = 2Â² â€“ 1 = 3

f(-4) â‰ f(2)

As Conditions of Rolleâ€™s theorem are

NOTsatisfied.Hence, Rolleâ€™s theorem is

NOTapplicable

### Question 3. If f : [â€“ 5, 5] â†’ R is a differentiable function and if fâ€²(x) does not vanish anywhere, then prove that f(â€“ 5) â‰ f(5).

**Solution:**

For Rolleâ€™s theorem

f is continuous in [a, b] ………(1)

f is derivable in [a, b] ………(2)

f (a) = f (b) ………(3)

then fâ€™ (c)=0, c âˆˆ (a, b)

So as, f is continuous and derivable

but f ‘(c) â‰ 0

It concludes, f(a) â‰ f(b)

f(-5) â‰ f(5)

### Question 4. Verify Mean Value Theorem, if f(x) = x^{2}â€“ 4x â€“ 3 in the interval [a, b], where a = 1 and b = 4.

**Solution:**

Now f(x) = xÂ² – 4x -3 is a polynomial

So, f(x) is continuous in the interval [1,4] and differentiable in the interval (1,4)

f(1) = (1)Â² – 4(1) â€“ 3 = -6

f(4) = 4Â² – 4(4) â€“ 3 = -3

fâ€²(c) = 2c – 4

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,4) such that

fâ€²(c) =

=

= 1

2c – 4 = 1

c = 5/2and c = 5/2 âˆˆ (1,4)

### Question 5. Verify Mean Value Theorem, if f(x) = x^{3}â€“ 5x^{2}â€“ 3x in the interval [a, b], where a = 1 and b = 3. Find all c âˆˆ (1, 3) for which fâ€²(c) = 0.

**Solution:**

Now f(x) = x

^{3}â€“ 5x^{2}â€“ 3x is a polynomialSo, f(x) is continuous in the interval [1,3] and differentiable in the interval (1,3)

f(1) = (1)

^{3}â€“ 5(1)^{2}â€“ 3(1) = -7f(3) = 3

^{3}â€“ 5(3)^{2}â€“ 3(3) = -27fâ€²(c) = 3c

^{2}– 5(2c) – 3fâ€²(c) = 3c

^{2}– 10c – 3As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,3) such that

fâ€²(c) =

=

=

=

3c

^{2}– 10c – 3 = -103c

^{2}– 10c + 7 = 03c2 – 3c – 7c + 7 = 0

3c (c-1) – 7(c -1) = 0

(3c -7) (c-1) = 0

c = 7/3 or c = 1

As, 1 âˆ‰ (1,3)

So,

c = 7/3 âˆˆ (1,3)According to the Rolle’s Theorem

As, f(3) â‰ f(1), Then there does not exist some c âˆˆ (1,3) such that fâ€²(c) = 0

### Question 6. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

### (i) f(x) = [x] for x âˆˆ [5, 9]

**Solution:**

In the interval [5, 9],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = 6,7,8

Hence, Mean value theorem is

NOTapplicable

### (ii) f(x) = [x] for x âˆˆ [â€“ 2, 2]

**Solution:**

In the interval [â€“ 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Mean value theorem is

NOTapplicable

### (iii) f(x) = x^{2}â€“ 1 for x âˆˆ [1, 2]

**Solution:**

Now f(x) = xÂ² – 1 is a polynomial

So, f(x) is continuous in the interval [1,2] and differentiable in the interval (1,2)

f(1) = (1)Â² – 1 = 0

f(2) = 2Â² -1 = 3

fâ€²(c) = 2c

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,2) such that

fâ€²(c) =

=

=

= 3

2c = 3

c = 3/2and

c = 3/2 âˆˆ (1,4)

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