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Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima  Exercise 17.3 Question 5 Maths Textbook Solution.

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x = 0 is local minima and its minimum value is 2 and x =-4 is  local maxima and its maximum value is -6




Differentiating f with respect to x

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(4(x+2)^{-1}+x\right) \\ &=-4(x+2)^{-2}+1 \\ &=-\frac{4}{(x+2)^{2}}+1 \end{aligned}

\begin{aligned} &=\frac{-4+x^{2}+4 x+4}{(x+2)^{2}} \\ &=\frac{x^{2}+4 x}{(x+2)^{2}} \\ &=\frac{x(x+4)}{(x+2)^{2}} \end{aligned}

Putf\left ( x \right )=0

\begin{aligned} &\frac{\mathrm{x}(\mathrm{x}+4)}{(\mathrm{x}+2)^{2}}=0 \\ &\mathrm{x}(\mathrm{x}+4)=0 \\ &\text { i.e } \mathrm{x}=0, \text { or } \\ &\mathrm{x}=-4 \end{aligned}

Thus x = 0 and x = -4 are the pointsof local maxima and minima

Differentiating f’ with respect to x

\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}}\right)

=\frac{(\mathrm{x}+2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+4 \mathrm{x}\right)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+2)^{2}}{(\mathrm{x}+2)^{4}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}}

=\frac{(x+2)^{2}(2 x+4)-\left(x^{2}+4 x\right) \cdot 2(x+2)}{(x+2)^{4}}

=\frac{2(x+2)\left[(x+2)^{2}-\left(x^{2}+4 x\right)\right]}{(x+2)^{4}}

=\frac{2(x+2)\left(x^{2}+4 x+4-x^{2}-4 x\right)}{(x+2)^{4}}

\begin{aligned} &=\frac{8(x+2)}{(x+2)^{4}} \\ &=\frac{8}{(x+2)^{3}} \end{aligned}

Put,x=0 in {f}''\left ( x \right )

\begin{aligned} \mathrm{f}^{\prime \prime}(0) &=\frac{8}{(0+2)^{3}} \\ &=\frac{8}{8} \\ &=1>0 \end{aligned}

S,x=0 is local minima. Hence,its minimum value will be f(0)=2

Andx= -4in {f}''(x)

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

So x=-4  is local maxima. Hence, its maximum value will be f(-4)=-6

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