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  • Need solution for R.D. Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.3 Question 1 Sub Question 8.

Need solution for R.D. Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.3 Question 1 Sub Question 8.

Answers (1)

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Answer:

Point of local maxima value is 4 and it’s local maximum value is 16. Also point of local minima is -4 and it’s local minimum value is  -16

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If f^{\prime}\left(c_{1}\right)>0 then c_1 is point of local minima.

If f^{\prime}\left(c_{2}\right)>0 then c_2 is point of local maxima .

where c_1 &c_2 are critical points.

Put c_1 and c_2 in f(x) to get minimum value & maximum value.

Given:

f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5

Explanation:

We have,

\begin{aligned} &f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5 \\ &f^{\prime}(x)=\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=-\frac{x}{\sqrt{32-x^{2}}}-\left(\frac{2 x \sqrt{32-x^{2}}+\frac{x^{3}}{\sqrt{32-x^{2}}}}{\left(32-x^{2}\right)}\right) \ldots u \sin g \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \end{aligned}

Now for local minima and local maxima

\begin{aligned} &f^{\prime}(x)=0 \\ &\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}}=0 \\ &32-x^{2}=x^{2} \\ &x^{2}=16 \\ &x=\pm 4 \end{aligned} 

At x=4,

\begin{aligned} f^{\prime \prime}(4) &=\frac{-4}{\sqrt{32-4^{2}}}-\left[\frac{\left(8\left(32-4^{2}\right)+4^{3}\right)}{\left(32-4^{2}\right)\left(\sqrt{32-4^{2}}\right.}\right] \\ &=-1-\frac{192}{64} \\ &=-3<0 \end{aligned}

So, x=4 is a point of local maxima.

At x=4,

\begin{aligned} f(4) &=4 \sqrt{32-(4)^{2}} \\ &=4 \sqrt{32-16} \\ &=16 \end{aligned}

At x=-4,

\begin{aligned} f^{\prime \prime}(-4) &=-\frac{4(-4)}{\sqrt{32-(-4)^{2}}}-\left[\frac{8\left(32-4^{2}\right)-(-4)^{3}}{\left(32-4^{2}\right) \sqrt{32-4^{2}}}\right] \\ &=0+2=3>0 \end{aligned}

So, x=-4 is the point of local minimum.

At x=-4

\begin{aligned} f(x) &=(-4) \sqrt{32-(-4)^{2}} \\ &=-4 \sqrt{32-16}=-16 \end{aligned}

Thus, point of local maxima is at x=4 is and its max. value is 16 and & point of local minima is at x=-4 is and its minimum value is -16.

 

 

 

 

 

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