Question

Let $f\left(x\right)=\frac{1}{(\cos x-3{)}^2+(\sin x+4{)}^2}$ and $g\left(x\right)=\sqrt{3}\sin x+\cos x$. Then which of the following is/are true ?

• A. Maximum value of $f\left(x\right)$ is $\frac{1}{36}$
• B. Minimum value of $f\left(x\right)$ is $\frac{1}{36}$
• C. Maximum value of $g\left(x\right)$ occurs at $x=\frac{\pi }{3}$
• D. Sum of the minimum values of $f\left(x\right)$ and $g\left(x\right)$ is $-\frac{71}{36}$

Answer 1

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

The correct options are
B Minimum value of $f\left(x\right)$ is $\frac{1}{36}$
C Maximum value of $g\left(x\right)$ occurs at $x=\frac{\pi }{3}$
D Sum of the minimum values of $f\left(x\right)$ and $g\left(x\right)$ is $-\frac{71}{36}$

$f\left(x\right)=\frac{1}{(\cos x-3{)}^2+(\sin x+4{)}^2}$
$=\frac{1}{{\cos }^{2}x-6\cos x+9+{\sin }^{2}x+8\sin x+16}$
$=\frac{1}{8\sin x-6\cos x+26}$

Extreme values of $8\sin x-6\cos x+26$ are $26\pm \sqrt{(8{)}^2+(-6{)}^2}=26\pm 10=36,16$

$\therefore f_{max}=\frac{1}{16}$ and $f_{min}=\frac{1}{36}$

$g\left(x\right)=\sqrt{3}\sin x+\cos x=2(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x)$
$g\left(x\right)=2\sin (x+\frac{\pi }{6})$

$\therefore g_{max}$ occurs at $x=\frac{\pi }{3}$ and $g_{min}=-2$

Sum of minimum values of $f\left(x\right)$ and $g\left(x\right)$ is $\frac{1}{36}+(-2)=-\frac{71}{36}$