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  • provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (xii)

provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (xii)

Answers (1)

Answer:   \frac{-8+4 \sqrt{13}}{3}

Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: f(x)=x(x+4)^{2} \text { on }[0,4]

Solution:

\begin{aligned} &f(x)=x(x+4)^{2} \\ &f(x)=x\left(x^{2}+16+8 x\right) \\ &f(x)=x^{3}+8 x^{2}+16 x \end{aligned}

f(x)  is a polynomial function.

It is continuous in [0, 4] 

f^{\prime}(x)=3 x^{2}+16 x+16

(Which is defined in [0, 4])

\therefore f(x) is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,4] \\ &f^{\prime}(c)=\frac{f(4)-f(0)}{4-0} \end{aligned}

\begin{aligned} &3 c^{2}+16 c+16=\frac{(4)^{3}+8(4)^{2}+16(4)-(0)^{3}+8(0)^{2}+16(0)}{4} \\ &3 c^{2}+16 c+16=\frac{64+128+64-0}{4} \end{aligned}

\begin{aligned} &3 c^{2}+16 c+16=\frac{256}{4} \\ &3 c^{2}+16 c+16=64 \\ &3 c^{2}+16 c+16-64=0 \\ &3 c^{2}+16 c-48=0 \end{aligned}

\begin{aligned} &c=\frac{-16 \pm \sqrt{(16)^{2}-4(3)(-48)}}{6} \\ &c=\frac{-16 \pm \sqrt{256+576}}{6} \end{aligned}

c=\frac{-16 \pm \sqrt{832}}{6}

c=\frac{-8 \pm 4 \sqrt{13}}{3}

But in [0, 4], only \frac{-8 \pm 4 \sqrt{13}}{3}  exists.

 

 

Posted by

infoexpert26

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