Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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

Answers (1)

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}

 

Posted by

Infoexpert

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question