Match List - I with List - II.

Choose the correct answer from the options given below:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

The correct answer is Option (2) - (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

$\text{(A)}\; f(x)=8x^2-4x+7$

$x=\frac{-(-4)}{2\cdot 8}=\frac{4}{16}=\frac{1}{4}$

$f\left(\frac{1}{4}\right)=8\cdot\frac{1}{16}-4\cdot\frac{1}{4}+7=\frac{1}{2}-1+7=\frac{13}{2}$

$\Rightarrow \text{matches (IV)}$

$\text{(B)}\; f(x)=x+\frac{1}{x},\; x<0$

$f'(x)=1-\frac{1}{x^2}=0 \Rightarrow x=-1$

$f(-1)=-1-1=-2$

$\Rightarrow \text{matches (III)}$

$\text{(C)}\; y=-2x^3+6x^2+7x+26$

$y'=-6x^2+12x+7$

$\text{Maximum of slope (quadratic)} = \frac{-b}{2a}=\frac{-12}{2(-6)}=1$

$y'(1)=-6+12+7=13$

$\Rightarrow \text{matches (II)}$

$\text{(D)}\; f(x)=x^2+\frac{128}{x}$

$f'(x)=2x-\frac{128}{x^2}=0 \Rightarrow 2x^3=128 \Rightarrow x=4$

$f(4)=16+32=48$

$\Rightarrow \text{matches (I)}$

A–IV,\; B–III,\; C–II,\; D–I