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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 4.

Need solution for R.D. Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.3  question 1 sub question 4.

Answers (1)

Answer:

Point of local maxima is 2 and it’s local maximum value is \frac{1}{2}

Hint:

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

If f''(c_1) >0 then c_1 is point of local minima.

If f"(c_2) <0then 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)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0

Explanation:

We have,

\begin{gathered} f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0 \\ \therefore f^{\prime}(x)=-\frac{2}{x^{2}}+\frac{4}{x^{3}} \\ f^{\prime \prime}(x)=\frac{4}{x^{2}}-\frac{12}{x^{4}} \end{gathered}

 

 For maxima and minima, f’(x)=0

\begin{array}{r} -\frac{2}{x^{2}}+\frac{4}{x^{3}}=0 \\ -\frac{2(x-2)}{x^{3}}=0 \\ x=2 \end{array}

  At x=2,

\begin{aligned} f^{\prime \prime}(2) &=\frac{4}{(2)^{3}}-\frac{12}{(2)^{2}} \\ &=\frac{4}{8}-\frac{12}{16} \\ &=\frac{1}{2}-\frac{3}{4} \\ &=-\frac{1}{4}<0 \end{aligned} 

So, x=2 is point of local maxima.

So, local max. value at x= 2 is 

\begin{aligned} f(2) &=\frac{2}{2}-\frac{2}{(2)^{2}} \\ &=1-\frac{2}{4}=\frac{1}{2} \end{aligned}

Hence, point of local maxima is 2 & its max. value is \frac{1}{2}

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