Q110 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

For \(m > 1, n > 1\), the value of \(c\) for which the Rolle's theorem is applicable for the function \(f(x)=x^{2 m-1}(a-x)^{2 n}\) in \((0, a)\) is

  • A
    \(\frac{2 a m-1}{m+2 n-1}\)
  • B
    \(\frac{a(m-n+1)}{2 m+2 n}\)
  • Correct answer
    C
    \(\frac{a(2 m-1)}{2 m+2 n-1}\)
  • D
    \(\frac{a(2 m+1)}{m+n-1}\)

Solution

Since it is given that Rolle's theorem is applicaple for the function \(f(x)=x^{2 m-1}(a-x)^{2 n}\) in \((0, a). \) \(\begin{aligned} & \text {So, } \\ & f^{\prime}(x)=(2 m-1) x^{2 m-2}(a-x)^{2 n}-2 n(a-x)^{2 n-1} x^{2 m-1} \\ & \text { at } x=c, m > 1, n > 1 \\ & f^{\prime}(c)=0 \\ & \Rightarrow \quad(2 m-1) c^{2 m-2}=2 n c^{2 m-1}(a-c)^{2 n-1} \\ & \Rightarrow \quad \frac{(2 m-1)}{c}=\frac{2 n}{a-c} \\ & \Rightarrow \quad c=\frac{a(2 m-1)}{2 n+2 m-1} \end{aligned}\) Hence, option (3) is correct.

Hence, the correct answer is option: 'C'.

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