KCET · Maths · Limits
The value of \( C \) in Mean value theorem for the function \( f(x)=x^{2} \) in \( [2,4] \) is
- A \( 03 \)
- B \( 02 \)
- C \( 04 \)
- D \( 77 / 2 \)
Answer & Solution
Correct Answer
(B) \( 02 \)
Step-by-step Solution
Detailed explanation
We know that, value of \( C \) in mean value theorem for the function \( f(x) \) in \( [a, b] \) is given by
\( C=\frac{f^{\prime}(b)-f^{\prime}(a)}{b-a} \) Here \( f(x)=x^{2} \) and \( [a, b] \) is \( [2,4] \). Then, \( C=\frac{2(4)-2(2)}{4-2} \) \( =\frac{8-4}{2}=2 \)
\( C=\frac{f^{\prime}(b)-f^{\prime}(a)}{b-a} \) Here \( f(x)=x^{2} \) and \( [a, b] \) is \( [2,4] \). Then, \( C=\frac{2(4)-2(2)}{4-2} \) \( =\frac{8-4}{2}=2 \)
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