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  • Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 13

Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 13

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Answer

x=\frac{1}{4}  is the point of local minima and the values of local minima at x=\frac{1}{4} \ \ \ is \ \ \ \ \frac{-1}{512}

Hint

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

Given:

 f(x)=x^{3}(2 x-1)^{3}   

Solution: -   

 f(x)=x^{3}(2 x-1)^{3}   

Differentiating f(x) w \cdot r . t^{\prime} x^{\prime} then

\begin{aligned} f^{\prime}(x) &=\frac{d}{d x}\left\{x^{3}(2 x-1)^{3}\right\} \\ &=x^{3} \frac{d}{d x}(2 x-1)^{3}+(2 x-1)^{3} \frac{d}{d x} x^{3} \quad\left\{\because \frac{d}{d x}(y x)=y \frac{d}{d x} x+x \frac{d}{d x} y\right\} \\ &=x^{3} 3(2 x-1)^{3-1} \frac{d}{d x}(2 x-1)+(2 x-1)^{3} 3 x^{3-1} \quad\left\{\because \frac{d}{d x} x^{n}=n x^{n-1} \& \frac{d}{d x}(a x+b)^{n}=n(a x+b)^{n-1} \frac{d}{d x}(a x+b)\right\} \\ &=x^{3} 3(2 x-1)^{2}\left\{\frac{d}{d x}(2 x)-\frac{d}{d x}(1)^{2}\right\}+(2 x-1)^{3} 3 x^{2} \quad\left\{\because \frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right\} \end{aligned}

            =x^{3} 3(2 x-1)^{2}\left\{2 \frac{d}{d x}(x)-0\right\}+(2 x-1)^{3} 3 x^{2} \quad\left\{\because \frac{d}{d x}(a x)=a \frac{d}{d x}(x), \frac{d}{d x}(x)=1, \frac{d}{d x} \text { constant }=0\right\}

            \\ =3 x^{3}(2 x-1)^{2}(2 \cdot 1)+3 x^{2}(2 x-1)^{3} \\ =6 x^{3}(2 x-1)^{2}+3 x^{2}(2 x-1)^{3} \\ =3 x^{2}(2 x-1)^{2}(2 x+2 x-1) \\

f^{\prime}(x) =3 x^{2}(2 x-1)^{2}(4 x-1)

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

\Rightarrow f^{\prime}(x)=0

\Rightarrow 3 x^{2}(2 x-1)^{2}(4 x-1)=0

\Rightarrow x^{2}(2 x-1)^{2}(4 x-1)=0 \quad[\because 3 \neq 0]

\Rightarrow x^{2}=0 \quad$ or $(2 x-1)^{2}=0 \Rightarrow(4 x-1)=0

\Rightarrow x=0 \quad$ or $(2 x-1)=0 \Rightarrow 2 x=1$ or $\Rightarrow 4 x=1 \Rightarrow x=1 / 4

                                                            \Rightarrow x=1 / 2

                              \therefore x=0, \frac{1}{2}, \frac{1}{4}

                                              -                       -                              +                      + 

                                   -∞                    0                     1/4                  1/2                    ∞

 Since  f(x) changes from –ve to +ve when x increases through  \frac{1}{4}   so   x=\frac{1}{4}   is the point of local minima

The value of the local minima of f\left ( x \right ) at x=\frac{1}{4}

\begin{aligned} f\left(\frac{1}{4}\right) &=\left(\frac{1}{4}\right)^{3}\left(2 \cdot \frac{1}{4}-1\right)^{3} \\ &=\frac{1}{4^{3}}\left(\frac{1}{2}-1\right)^{3}=\frac{1}{64}\left(\frac{1-2}{2}\right)^{3}=\frac{1}{64}\left(\frac{-1}{2}\right)^{3} \\ &=\frac{1}{64} \cdot \frac{-1}{8}=\frac{-1}{512} \end{aligned}

 

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