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need solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 4

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Answer:  Not applicable

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

Given: f(x)=\frac{1}{4 x-1}, 1 \leq x \leq 4


f(x)=\frac{1}{4 x-1} \text { on }[1,4]

Here, 4 x-1>0

f'(x) has the unique values for all x except \frac{1}{4}

So, f (x) is continuous in [1, 4]

f(x)=\frac{1}{4 x-1}


\begin{aligned} &f(x)=(-1)(4 x-1)^{-2}(4) \\ &f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}} \end{aligned}

f (x) is differentiable in (1, 4)

So, there exists a point c \in (1, 4)

\begin{aligned} &f^{\prime}(c)=\frac{f(4)-f(1)}{4-1} \\ &f^{\prime}(c)=\frac{f(4)-f(1)}{3} \end{aligned}

\frac{-4}{(4 c-1)^{2}}=\frac{\frac{1}{4(4)-1}-\frac{1}{4(1)-1}}{3}

\frac{-4}{(4 c-1)^{2}}=\frac{\frac{1}{15}-\frac{1}{3}}{3}

-3(4)=(4 c-1)^{2}\left(\frac{1}{15}-\frac{1}{3}\right)

\begin{aligned} &-12=(4 c-1)^{2}\left(\frac{3-15}{45}\right) \\ &-12=(4 c-1)^{2}\left(\frac{-12}{45}\right) \end{aligned}

-12 \times \frac{45}{-12}=(4 c-1)^{2}

\begin{aligned} &(4 c-1)^{2}=45 \\ &4 c-1=\sqrt{45} \\ &4 c-1=\pm 3 \sqrt{5} \end{aligned}

\begin{aligned} &c=\frac{\pm 3 \sqrt{5}+1}{4} \\ &c=\frac{3 \sqrt{5}+1}{4} \approx 1.92 \in(1,4) \end{aligned}

Thus, Lagrange’s Theorem is verified.

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