UPSC NDA I 2021 Mathematics Solved Paper

CONCEPT: Second Derivative Test: Let $f$ be a function defined on an interval $I$. - Calculate $f'(x)$ - Solve $f'(x)=0$ and find the roots of $f'(x)=0$. Suppose $x=c$ is the root of $f'(x)=0$. - Calculatef" $(x)$ and put $x=c$ to get the value of $f$ "(c). - If $f''(c) < 0$ then $x=c$ is a point of local maxima. - If $f$ " $(c) > 0$ then $x=c$ is a point of local minima. - If $f''(c)=0$ then we need to use the first derivative test. Note: Maxima and Minima on a closed Interval: Let $f(x)$ be a given function defined on $[a, b]$. Let the local minimum value of $f(x)$ be $m$ and let the local maximum value of $f(x)$ be $M$. Then Minimum value of $f(x)$ on $[a, b]$ is the smallest of $m, f(a)$ and $f(b)$ Maximum value of $f(x)$ on $[a, b]$ is the greatest of $M, f(a)$ and $f(b)$ CALCULATION: Let $f(x)=\sin 2 x \cdot \cos 2 x$ $\Rightarrow f'(x)=2 \cos^2 2x-2 \sin^2 2x$ $\Rightarrow f'(x)=2 \cdot (\cos^2 2x-\sin^2 2x)$ As we know that, $\cos 2x=\cos^2 x-\sin^2 x$ $\Rightarrow f'(x)=2 \cos 4x$ If $f'(x)=0$ then $2 \cos 4x=0 \Rightarrow x=\frac{\pi}{8}$ $\Rightarrow f'(x)=-8 \sin 4x$ $\Rightarrow f''(x)=-8 < 0$ So, $x=\frac{\pi}{8}$ is the point of maxima. So, the maximum value of $f(x)=\sin 2x \cdot \cos 2x$ is given by $f\left(\frac{\pi}{8}\right)=\sin\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{4}\right)=\frac{1}{2}$ Hence, correct option is 1 .