Question

$3\sin x+4\cos x=y^2-2y+6$.If $x,y$ are its solutions then which of the following is true

• A. $y=1$
• B. $x=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$
• C. $xy=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$
• D. $\frac{x}{y}=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $y=1$
B $xy=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$
C $x=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$
D $\frac{x}{y}=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$

$3\sin x+4\cos x=y^2-2y+6$
$3\sin x+4\cos x={(y-1)}^2+5$
Now, maximum value of LHS i.e. $3\sin x+4\cos x$ is $\sqrt{9+16}=5$
which is possible only if $y-1=0\therefore y=1$
$\therefore 3\sin x+4\cos x=5$
Putting $3=r\cos \theta ,4=r\sin \theta \Rightarrow r=5$
$\Rightarrow 5\sin (x+\theta )=5$
$\Rightarrow x+\theta =\frac{\pi }{2}$
$\Rightarrow x=\frac{\pi }{2}-\theta$
$\Rightarrow x=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$
$\Rightarrow xy=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$ $(\because y=1)$
$\Rightarrow \frac{x}{y}=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{4}{3}\right)$