Match List – I with List – II.
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LIST I |
LIST II |
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A. The maximum value of $f(x)=4 x^3-48 x+100$ in the interval [1, 3] is: |
I. 0 |
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B. The minimum value of $f(x)=x^3-18 x^2+96 x$ in the interval [0, 9] is: |
II. 1 |
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C. The minimum value of $f(x)=3 x^2-6 x$ in the interval [0, 2] is: |
III. 64 |
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D. If x is real, the minimum value of $f(x)=x^2-8 x+17$ is |
IV. -3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-IV, D-II
$\text{A: } f(x)=4x^3-48x+100$
$f'(x)=12x^2-48=12(x^2-4)=0 \Rightarrow x=\pm2$
$\text{In }[1,3],\ x=2$
$f(1)=56,\ f(2)=36,\ f(3)=64 \Rightarrow \text{maximum}=64$
$\text{B: } f(x)=x^3-18x^2+96x$
$f'(x)=3x^2-36x+96=3(x^2-12x+32)=0 \Rightarrow x=4,8$
$f(0)=0,\ f(4)=160,\ f(8)=128,\ f(9)=162 \Rightarrow \text{minimum}=0$
$\text{C: } f(x)=3x^2-6x$
$f'(x)=6x-6=0 \Rightarrow x=1$
$f(0)=0,\ f(1)=-3,\ f(2)=0 \Rightarrow \text{minimum}=-3$
$\text{D: } f(x)=x^2-8x+17$
$\text{Minimum at } x=\frac{8}{2}=4$
$f(4)=1$
$\text{A-III,\ B-I,\ C-IV,\ D-II}$