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  • Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 5

Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 5

Answers (2)

Answer:

x=0 is the point of  local maxima and the value of local maxima is \frac{1}{2}

Hint:

Use first derivative test to find the value and point of local maxima and local minima.

Given:

f\left ( x \right )=\frac{1}{x^{2}+2}

Solution:

f\left ( x \right )=\frac{1}{x^{2}+2}

Differentiating  f\left ( x \right ) with respect to ‘x’ then,

\frac{d}{d x}\{f(x)\}=\frac{d\left(\frac{1}{x^{2}+2}\right)}{d x}=\frac{d\left(x^{2}+2\right)-1}{d x} \quad\left[\because \frac{d}{d x}\left(x^{n}+a\right)^{n}=n\left(x^{n}+a\right)^{n-1} \frac{d}{d x}\left(x^{n}+a\right)\right]

                      =-1\left(x^{2}+2\right)-1-1 \cdot \frac{d}{d x}\left(x^{2}+2\right)     

=-1\left(x^{2}+2\right)^{-2}\left\{\frac{d\left(x^{2}\right)}{d x}+\frac{d(2)}{d x}\right\} \ \ \ \ \ \ \ \ \left[\because \frac{d}{d x}\left(x^{n}+a\right)=\frac{d\left(x^{n}\right)}{d x}+\frac{d(a)}{d x}\right]

\begin{aligned} &=\frac{-1}{\left(x^{2}+2\right)^{2}}\left[2 x^{2-1}+0\right] \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}, \frac{d}{d x} \text { constant }=0\right. \\ &\therefore f^{i}(x)=\frac{-1}{\left(x^{2}+2\right)^{2}} 2 x=\frac{-2 x}{\left(x^{2}+2\right)^{2}} \end{aligned}

  By first derivative test, for local maxima or local minima, we have

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0 \quad \Rightarrow \quad \frac{x}{\left(x^{2}+2\right)^{2}}=0[\because-2 \neq 0] \\ &\Rightarrow x=0 \end{aligned}

 

                                                                     +                                     -

                                           -∞                                     0                                ∞

since {f}'\left ( x \right ) changes from +ve to -ve when  through x=0
So, x=0 is the point of local maxima.
The value of local maxima of {f}'\left ( x \right ) at is-x=0
f(0)=\frac{1}{0^{2}+2}=\frac{1}{2}

 

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Answer:

x=0 is the point of  local maxima and the value of local maxima is \frac{1}{2}

Hint:

Use first derivative test to find the value and point of local maxima and local minima.

Given:

f\left ( x \right )=\frac{1}{x^{2}+2}

Solution:

f\left ( x \right )=\frac{1}{x^{2}+2}

Differentiating  f\left ( x \right ) with respect to ‘x’ then,

\frac{d}{d x}\{f(x)\}=\frac{d\left(\frac{1}{x^{2}+2}\right)}{d x}=\frac{d\left(x^{2}+2\right)-1}{d x} \quad\left[\because \frac{d}{d x}\left(x^{n}+a\right)^{n}=n\left(x^{n}+a\right)^{n-1} \frac{d}{d x}\left(x^{n}+a\right)\right]

                      =-1\left(x^{2}+2\right)-1-1 \cdot \frac{d}{d x}\left(x^{2}+2\right)     

=-1\left(x^{2}+2\right)^{-2}\left\{\frac{d\left(x^{2}\right)}{d x}+\frac{d(2)}{d x}\right\}\left[\because \frac{d}{d x}\left(x^{n}+a\right)=\frac{d\left(x^{n}\right)}{d x}+\frac{d(a)}{d x}\right]

\begin{aligned} &=\frac{-1}{\left(x^{2}+2\right)^{2}}\left[2 x^{2-1}+0\right] \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}, \frac{d}{d x} \text { constant }=0\right. \\ &\therefore f^{i}(x)=\frac{-1}{\left(x^{2}+2\right)^{2}} 2 x=\frac{-2 x}{\left(x^{2}+2\right)^{2}} \end{aligned}

  By first derivative test, for local maxima or local minima, we have

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0 \quad \Rightarrow \quad \frac{x}{\left(x^{2}+2\right)^{2}}=0[\because-2 \neq 0] \\ &\Rightarrow x=0 \end{aligned}

 

                                                                     +                                     -

                                           -∞                                     0                                ∞

since {f}'\left ( x \right ) changes from +ve to -ve when  through x=0
So, x=0 is the point of local maxima.
The value of local maxima of {f}'\left ( x \right ) at is-x=0
f(0)=\frac{1}{0^{2}+2}=\frac{1}{2}

 

Posted by

Infoexpert

View full answer

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