UPSC NDA Math Model Paper 3

We have to calculate the maximum value of 16‌sin‌θ−12sin2θ
Let f(θ)=16‌sin‌θ−12sin2θ
Differentiating, both sides, we get
⇒f′(θ)=16‌cos‌θ−12×2 ×sin‌θ×cos‌θ =16‌cos‌θ−24×sin‌θ×cos‌θ
For maxima and minima, f′(θ)=0
⇒16‌cos‌θ−24×sin‌θ×cos‌θ=0
⇒8‌cos‌θ(2−3‌sin‌θ)=0
⇒8‌cos‌θ=0 and (2−3‌sin‌θ)=0
∴θ=90° and sin‌θ=2∕3 or θ=sin−1(2∕3)
Now we have to find second derivative,
⇒f"(θ)=−16‌sin‌θ−24 [cos2θ−sin2θ]
⇒f"(θ)=−16‌sin‌θ−24 [1−sin2θ−sin2θ] =−16‌sin‌θ−24[1−2sin2θ]
When θ=.90°⇒f(θ)|θ=90° =−16‌sin‌90°−24[1−2sin290°] =−16+24=8>0
So, function gives minimum value at θ=90°
When θ=sin−1(2∕3)⇒f"(θ)|θ=sin−1(23) =−16‌sin‌sin−1(2∕3)−24 [1−2sin2sin−1(2∕3)]
⇒f"(θ)=−16(2∕3)−24[1−2(2∕3)2] =−(32∕3)−(24∕9)<0
So, function gives maximum value at θ=sin−1(2∕3) or sin‌θ=2∕3
⇒f(θ)max=16(2∕3)−12(2∕3)2 =(32∕3)−(16∕3)=16∕3