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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise 17.3 Question 2 Sub Question 3 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 17 Maxima and Minima  Exercise 17.3 Question 2 Sub Question 3 Maths Textbook Solution.

Answers (1)

Answer:

x=1 is point of inflexion

x=-1 is the point of local minimum & x=-\frac{1}{5} is the point of local maximum

Hint:

Using chain rule of derivative

Given:

\mathrm{f}(\mathrm{x})=-(\mathrm{x}-1)^{3}(\mathrm{x}+1)^{2}

Explanation:

Differentiating f with respect to x

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[-(\mathrm{x}-1)^{3}(\mathrm{x}+1)^{2}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+1)^{2}+(\mathrm{x}+1)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1)^{3}\right] \end{aligned}

\begin{aligned} &=-\left[(x-1)^{3} \cdot 2(x+1)+3(x+1)^{2}(x-1)^{2}\right] \\ &=-(x-1)^{2}(x+1)[2 x-2+3 x+3] \\ &=-(x-1)^{2}(x+1)(5 x+1) \\ &=-\left(x^{2}-2 x+1\right)\left(5 x^{2}+x+5 x+1\right) \end{aligned}

\begin{aligned} &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}+6 \mathrm{x}^{3}+\mathrm{x}^{2}-10 \mathrm{x}^{3}-12 \mathrm{x}^{2}-2 \mathrm{x}+5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}-4 \mathrm{x}^{3}-6 \mathrm{x}^{2}+4 \mathrm{x}+1\right) \end{aligned}

Put \mathrm{f}^{\prime}(\mathrm{x})=0

\begin{aligned} &-1(x-1)^{2}(x+1)(5 x+1)=0 \\ &(x-1)^{2}(x+1)(5 x+1)=6 \\ &\Rightarrow(x-1)^{2}=0 \end{aligned}

\begin{aligned} &x-1=0 \\ &x=1 \\ &x+1=0 \quad, \quad 5 x+1=0 \\ &x=-1 \quad, \quad x=-\frac{1}{5} \end{aligned}

Thus x=1 and  x=-1and x=-\frac{1}{5} are the possible points of local minima and maxima

Differentiating f’(x) with respect to x

\begin{aligned} &f^{\prime \prime}(x)=\frac{d}{d x}\left[-\left(5 x^{4}-4 x^{3}-6 x^{2}+4 x+1\right)\right] \\ &=-\left(20 x^{3}-12 x^{2}-12 x+4\right) \\ &=-20 x^{3}+12 x^{2}+12 x-4 \end{aligned}

when x=1,

\begin{aligned} \mathrm{f}^{\prime \prime}(1) &=-20(1)^{3}+12(1)^{2}+12(1)-4 \\ &=-20+12+12-4=0 \end{aligned}

Thus this test is fail as x=1 is point of inflexion

when x=-1

\begin{aligned} \mathrm{f}^{\prime \prime}(-1) &=-20(-1)^{3}+12(-1)^{2}+12(-1)-4 \\ &=20+12-12-4=16>0 \end{aligned}

So , x = -1  is a point of local minimum

When x=-\frac{1}{5}

\begin{aligned} \mathrm{f}\left(-\frac{1}{5}\right)=&-20\left(-\frac{1}{5}\right)^{3}+12\left(-\frac{1}{5}\right)^{2}+12\left(-\frac{1}{5}\right)-4 \\ &=-336 / 125<0 \end{aligned}

So, x=-1/5 is the point of local maximum

 

Posted by

infoexpert21

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