For m1, n1, the value of c for which the Rolle's theorem is applicable for the function f(x)=x2 m-1(a-x)2 n in (0, a) is

Question

For m>1, n>1, the value of c for which the Rolle's theorem is applicable for the function f(x)=x2 m-1(a-x)2 n in (0, a) is (A) (2am - 1/ m + 2n - 1) (B) (a(m - n + 1)/ 2m + 2n) (C) (a( 2m - 1)/ 2m + 2n - 1 ) (D) (a( 2m + 1)/m + n - 1 )

Tardigrade Admin · Accepted Answer

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) .

So,

f^{'} (x) = (2 m - 1) x^{2 m - 2} (a - x)^{2 n} - 2 n (a - x)^{2 n - 1} x^{2 m - 1}

at

x = c, m > 1, n > 1

f^{'} (c) = 0

\Rightarrow (2 m - 1) c^{2 m - 2} = 2 n c^{2 m - 1} (a - c)^{2 n - 1}

\Rightarrow \frac{( 2 m - 1 )}{c} = \frac{2 n}{a - c}

\Rightarrow c = \frac{a ( 2 m - 1 )}{2 n + 2 m - 1}

Q. 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

$\frac{2 am - 1}{m + 2 n - 1}$

$\frac{a ( m - n + 1 )}{2 m + 2 n}$

$\frac{a ( 2 m - 1 )}{2 m + 2 n - 1}$

$\frac{a ( 2 m + 1 )}{m + n - 1}$

Q. For m>1,n>1, the value of c for which the Rolle's theorem is applicable for the function f(x)=x2m−1(a−x)2n in (0,a) is

m+2n−12am−1​

2m+2na(m−n+1)​

2m+2n−1a(2m−1)​

m+n−1a(2m+1)​

Since it is given that Rolle's theorem is applicaple for the function f(x)=x2m−1(a−x)2n in (0,a). So, f′(x)=(2m−1)x2m−2(a−x)2n−2n(a−x)2n−1x2m−1 at x=c,m>1,n>1 f′(c)=0 ⇒(2m−1)c2m−2=2nc2m−1(a−c)2n−1 ⇒c(2m−1)​=a−c2n​ ⇒c=2n+2m−1a(2m−1)​

Q. 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

$\frac{2 am - 1}{m + 2 n - 1}$

$\frac{a ( m - n + 1 )}{2 m + 2 n}$

$\frac{a ( 2 m - 1 )}{2 m + 2 n - 1}$

$\frac{a ( 2 m + 1 )}{m + n - 1}$