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

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

Answers (1)

Answer:  3

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

Given: f(x)=x^{2}-2 x+4 \text { on }[1,5]

Solution:

f(x)=x^{2}-2 x+4

f(x)  is a polynomial function.

It is continuous in [1, 5] 

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

(Which is defined in [1, 5])

\therefore f(x) is differentiable in [1, 5]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[1,5] \\ &f^{\prime}(c)=\frac{f(5)-f(1)}{5-1} \\ &2 c-2=\frac{\left((5)^{2}-2(5)+4\right)-\left(2(1)^{2}-2(1)+4\right)}{4} \end{aligned}

\begin{aligned} &2 c-2=\frac{(25-10+4)-(1-2+4)}{4} \\ &2 c-2=\frac{19-3}{4} \end{aligned}

\begin{aligned} &2 c-2=\frac{16}{4} \\ &2 c-2=4 \end{aligned}

\begin{aligned} &2 c=4+2 \\ &2 c=6 \\ &c=\frac{6}{2} \\ &c=3 \end{aligned}

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