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  • Verify the Rolle’s theorem for each of the functions f(x) = x (x - 1)2 in [0, 1].

Verify the Rolle’s theorem for each of the functions f(x) = x (x - 1)^2 in [0, 1].

Answers (1)

Solution

Given: f(x) = x (x - 1)^2

\\ \Rightarrow f(x)=x\left(x^{2}+1-2 x\right) \\ -f(x)=x^{3}+x-2 x^{2} \text { in }[0,1]

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:
On expanding f(x) = x(x - 1)\textsuperscript{2}, we get f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2\\ }

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x \in R

\Rightarrow f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2} \ is \ continuous\ at \ x \in [0,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2\\ }

Since, f(x) is a polynomial and every polynomial function is differentiable for all x \in R

\Rightarrow f(x)\ is \ differentiable \ at \ [0,1]

Hence, condition 2 is satisfied.

Condition 3:

\\f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2} \\f(0) = 0 \\f(1) = (1)\textsuperscript{3} + (1) - 2(1)\textsuperscript{2} = 1 + 1 - 2 = 0

Hence, f(0) = f(1)

Hence, condition 3 is also satisfied.

Now, let us show that c \in (0,1)  such that f’(c) = 0

f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2}

On differentiating above with respect to x, we get

f'(x) = 3x\textsuperscript{2} + 1 - 4x

Put x = c in above equation, we get

f'(c) = 3c\textsuperscript{2} + 1 - 4c

\because all the three conditions of Rolle’s theorem are satisfied

f’(c) = 0

3c\textsuperscript{2} + 1 - 4c = 0

On factorising, we get

\\ \Rightarrow 3c\textsuperscript{2} - 3c - c + 1 = 0 \\ \Rightarrow 3c(c - 1) - 1(c - 1) = 0 \\ \Rightarrow (3c - 1) (c - 1) = 0 \\ \Rightarrow (3c - 1) = 0$ or $(c - 1) = 0

\begin{aligned} &\Rightarrow c=\frac{1}{3} \text { or } c=1\\ &\text { So, value of }\\ &c=\frac{1}{3} \in(0,1) \end{aligned}

Thus, Rolle’s theorem is verified.

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