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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 40 of 50
Marks: +1, -0
The absolute maximum value of the function f(x)=f(x)= 4x−12x24x-\frac{1}{2}x^2 in the interval [−2,92]\left[-2,\frac{9}{2}\right] is
Solution:     👈: Video Solution
Given, f(x)=4x−12x2\text{Given, } f(x)=4x-\frac{1}{2}x^2
∴f′(x)=4−12(2x)=4−x\therefore f'(x)=4-\frac{1}{2}(2x)=4-x
put f′(x)=0\text{put } f'(x)=0
⇒4−x=0\Rightarrow 4-x=0
⇒x=4\Rightarrow x=4
Then, we evaluate the ff at critical point x=4x=4 and at the end points of the interval [−2,92]\left[-2,\frac{9}{2}\right].
f(4)=16−12(16)=16−8=8f(4)=16-\frac{1}{2}(16)=16-8=8
f(−2)=−8−12(4)f(-2)=-8-\frac{1}{2}(4)
=−8−2=−10=-8-2=-10
f(92)=4(92)−12(92)2f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2
=18−818=7.875=18-\frac{81}{8}=7.875
Thus, the absolute maximum value of ff on [−2,92]\left[-2,\frac{9}{2}\right] is 8 occurring at x=4x=4.
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