Match List - I with List - II.
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List - I |
List - II |
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(A) The minimum value of $f(x)=8 x^2-4 x+7$ is |
(I) 48 |
|
(B) The maximum value of $f(x)=x+\frac{1}{x}, x<0$ is |
(II) 13 |
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(C) The maximum slope of the cure $y=-2 x^3+6 x^2+7 x+26$ is |
(III) -2 |
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(D) The minimum value of $f(x)=x^2+\frac{128}{x}$ is |
(IV) $\frac{13}{2}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
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