CUET Preparation Today
If the function $f(x)=\frac{cx+d}{(x-1)(x-4)}$ has a turning point at the point (2, –1) then: |
c = 2, d = 0 c = 1, d = 0 c = 1, d = -1 c = 1, d = 1 |
c = 1, d = 0 |
$f'(x)=\frac{c(x-1)(x-4)-(cx+d)(2x-5)}{(x-1)^2(x-4)^2}$ So, $f'(2)=0⇒\frac{-2c+(2c+d)}{-4}=\frac{d}{4}⇒d=0$ Also, $-1=f(2)=\frac{2c+d}{-2}=-c⇒c=1$ |
