Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Verify the Rolle’s theorem for each of the functions f(x) = log (x2 + 2) - log3 in [- 1, 1].

Verify the Rolle’s theorem for each of the functions
f(x) = log (x^2 + 2) - log3 in [- 1, 1].

Answers (1)

Given: f(x) = log (x^2 + 2) - log3

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:
f(x) = log (x^2 + 2) - log3

Since, f(x) is a logarithmic function and logarithmic function is continuous for all values of x.

\Rightarrow f(x) = log (x\textsuperscript{2} + 2) - log3 is continuous at x \in [-1,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = log (x\textsuperscript{2} + 2) - log3

\begin{aligned} &\text { Chain Rule }\\ &f(x)=g(h(x))\\ &f^{\prime}(x)=g^{\prime}(h(x)) h^{\prime}(x) \end{aligned}

\begin{aligned} &\Rightarrow f(x)=\log \left(\frac{x^{2}+2}{3}\right)\left[\because \log m-\log n=\log \frac{m}{n}\right]\\ &\text { On differentiating above with respect to } x \text { , we get }\\ &f^{\prime}(x)=\frac{1}{\frac{x^{2}+2}{3}} \times \frac{d}{d x}\left(\frac{x^{2}+2}{3}\right)\left[\because \frac{d}{d x} \log x=\frac{1}{x}\right]\\ &\Rightarrow f^{\prime}(x)=\frac{3}{x^{2}+2} \times \frac{d}{d x}\left(\frac{x^{2}}{3}+\frac{2}{3}\right)\\ &\Rightarrow f^{\prime}(x)=\frac{3}{x^{2}+2} \times\left(\frac{2 x}{3}+0\right)\\ &\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{3}{\mathrm{x}^{2}+2} \times \frac{2 \mathrm{x}}{3} \end{aligned}

\\\Rightarrow f^{\prime}(x)=\frac{2 x}{x^{2}+2} \\\Rightarrow f(x) is differentiable at [-1,1]
Hence, condition 2 is satisfied.
Condition 3:
\\f(x)=\log \left(\frac{x^{2}+2}{3}\right) \\f(-1)=\log \left(\frac{(-1)^{2}+2}{3}\right)=\log \frac{1+2}{3}=\log 1=0 \\f(1)=\log \left(\frac{(1)^{2}+2}{3}\right)=\log \frac{1+2}{3}=\log 1=0 \\\therefore f(-1)=f(1)
Hence, condition 3 is also satisfied.

Now, let us show that c \in(-1,1) such that f(c)=0
\\ f(x)=\log \left(\frac{x^{2}+2}{3}\right)$ \\$\Rightarrow f^{\prime}(x)=\frac{2 x}{x^{2}+2}
Put x=c in above equation, we get
\\\Rightarrow f^{\prime}(c)=\frac{2 c}{c^{2}+2}=0\Rightarrow c=0\epsilon (-1,1)$ \\$\because,$ all the three conditions of Rolle's theorem are satisfied $f(c)=0

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question