Let f(x) = frac{sin x + cos x - sqrt{2}}{sin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$.Dividing numerator and denominator by $\sqrt{2}$,we get $f(x) = \frac{\frac{1}{\

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(B) Given $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$.Dividing numerator and denominator by $\sqrt{2}$,we get $f(x) = \frac{\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x - 1}{\frac{1}{\sqrt{2}}\sin x - \frac{1}{\sqrt{2}}\cos x} = \frac{\sin(x + \frac{\pi}{4}) - 1}{\sin(x - \frac{\pi}{4})}$.Using $\sin 	heta - 1 = -2 \sin^2(\frac{\pi}{4} - \frac{	heta}{2})$ and $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$,we simplify $f(x) = 	an(\frac{\pi}{8} - \frac{x}{2})$.Thus,$f(x) = -	an(\frac{x}{2} - \frac{\pi}{8})$.$f'(x) = -\frac{1}{2} \sec^2(\frac{x}{2} - \frac{\pi}{8})$.$f''(x) = -\frac{1}{2} \cdot 2 \sec(\frac{x}{2} - \frac{\pi}{8}) \cdot \sec(\frac{x}{2} - \frac{\pi}{8}) 	an(\frac{x}{2} - \frac{\pi}{8}) \cdot \frac{1}{2} = -\frac{1}{2} \sec^2(\frac{x}{2} - \frac{\pi}{8}) 	an(\frac{x}{2} - \frac{\pi}{8})$.At $x = \frac{7\pi}{12}$,$\frac{x}{2} - \frac{\pi}{8} = \frac{7\pi}{24} - \frac{3\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$.$f(\frac{7\pi}{12}) = -	an(\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$.$f''(\frac{7\pi}{12}) = -\frac{1}{2} \sec^2(\frac{\pi}{6}) 	an(\frac{\pi}{6}) = -\frac{1}{2} \cdot (\frac{2}{\sqrt{3}})^2 \cdot \frac{1}{\sqrt{3}} = -\frac{1}{2} \cdot \frac{4}{3} \cdot \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}}$.Therefore,$f(\frac{7\pi}{12}) f''(\frac{7\pi}{12}) = (-\frac{1}{\sqrt{3}}) \cdot (-\frac{2}{3\sqrt{3}}) = \frac{2}{3 \cdot 3} = \frac{2}{9}$.

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

Similar Questions

If $f(x)=10 \cos x+(13+2 x) \sin x$,then $f^{\prime \prime}(x)+f(x)$ is equal to

Find the second order derivative of the function $\log (\log x)$.

If $y=e^{\sqrt{x \sqrt{x} \sqrt{x} \ldots}}, x>1$,then $\frac{d^2 y}{d x^2}$ at $x=\log _e 3$ is

If $y_k$ is the $k$-th derivative of $y$ with respect to $x$,and $y = \cos(\sin x)$,then $y_1 \sin x + y_2 \cos x$ is equal to

If $x^3 + y^3 = 6$ and $\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}$, where $m, n \in R$, then find the value of $\frac{m}{n}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module