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  • Explain solution RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 12 maths

Explain solution RD Sharma class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 12 maths

Answers (1)

Answer:

 (b)

Hint:

 Integration

Given:

 y = x4 - 2x3 + x2 + 3 , x - axis and minima of y

Explanation:

Finding the minima of y

\begin{aligned} &y = x^{4} - 2x^{3} + x^{2} + 3 \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=4x^{3}-6x^{2}+2x \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=12x^{2}-12x+2 \\ \end{aligned}

Now,

\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &4x^{3}-6x^{2}+2x=0\\ &2x(2x^{2}-3x+1)=0\\ &x=0 \qquad 2x^{2}-3x+1=0\\ & \qquad \qquad (2x-1)(x-1)=0\\ &x=0,\: 1,\: \frac{1}{2} \end{aligned}

At x = 0

\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=2> 0 \end{aligned}

At x = 1

\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=12-12+2 =2> 0 \end{aligned}

At

\begin{aligned} &x=\frac{1}{2} \end{aligned}

\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=-1 < 0 \end{aligned}

So,f(x) has minimum value at

\begin{aligned} &x=\frac{1}{2} \end{aligned}

The minimum value is y = 1

                    \begin{aligned} &1=x^{4}-2x^{3}+x^{2}+3\\ &x^{4}-2x^{3}+x^{2}+2=0\\ &x=1 \end{aligned}

\begin{aligned} &\int_{0}^{1} (x^{4}-2x^{3}+x^{2}+3)dx\\ &=\left [ \frac{x^{5}}{5}-2\left ( \frac{x^{4}}{x} \right )+\frac{x^{3}}{3}+3x \right ]_{0}^{1}\\ &=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=0\\ &=\frac{91}{30} \end{aligned}

Posted by

Gurleen Kaur

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