Let $f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}$. Then $\mathrm{f}\left(\frac{7 \pi}{12}\right) \mathrm{f}''\left(\frac{7 \pi}{12}\right)$ is equal to

Differentiation — JEE Main · 2023 · 08 Apr 2023 · Shift 1 Question with Solution

Let

f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x}, x \in [0, π] - {\frac{π}{4}}

. Then

f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12})

is equal to

Answer: (B)

Rewrite

f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x} = \frac{2 s i n ( x + \frac{π}{4} ) - 2}{2 s i n ( x - \frac{π}{4} )} = \frac{s i n ( x + \frac{π}{4} ) - 1}{s i n ( x - \frac{π}{4} )}

. Using the identity

sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2}

sin (x + \frac{π}{4}) - sin \frac{π}{2} = 2 cos (\frac{x}{2} + \frac{3 π}{8}) sin (\frac{x}{2} - \frac{π}{8})

. After simplification,

f (x) = - tan (\frac{x}{2} - \frac{π}{8})

. Then

f^{'} (x) = - \frac{1}{2} sec^{2} (\frac{x}{2} - \frac{π}{8})

and

f^{''} (x) = - \frac{1}{2} sec^{2} (\frac{x}{2} - \frac{π}{8}) tan (\frac{x}{2} - \frac{π}{8})

. At

x = \frac{7 π}{12}

\frac{x}{2} - \frac{π}{8} = \frac{7 π}{24} - \frac{3 π}{24} = \frac{π}{6}

. Thus

f (\frac{7 π}{12}) = - tan \frac{π}{6} = - \frac{1}{3}

and

f^{''} (\frac{7 π}{12}) = - \frac{1}{2} \cdot \frac{4}{3} \cdot \frac{1}{3} = - \frac{2}{3 3}

. Therefore

f (\frac{7 π}{12}) \cdot f^{''} (\frac{7 π}{12}) = (- \frac{1}{3}) \cdot (- \frac{2}{3 3}) = \frac{2}{9}

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option