Question

Consider the real valued function $f\left(x\right)=4\sin x+{\cos }^{2}x.$ Then which of the following statements is (are) FALSE?

• A. Range of $f$ is $[-4,4]$
• B. $f\left(\frac{\pi }{2}\right)-f\left(0\right)=4$
• C. Graph of $y=f\left(x\right)$ intersects $x$-axis at only one point in the interval $[0,\pi ]$
• D. $f(-x)=f\left(x\right)\Rightarrow x=n\pi ,$ where $n\in Z$

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(\frac{\pi }{2}\right)-f\left(0\right)=4$
C Graph of $y=f\left(x\right)$ intersects $x$-axis at only one point in the interval $[0,\pi ]$

$f\left(x\right)=4\sin x+{\cos }^{2}x=4\sin x+1-{\sin }^{2}x\Rightarrow f\left(x\right)=5-({\sin }^{2}x-4\sin x+4)\Rightarrow f\left(x\right)=5-(\sin x-2{)}^2$

We know that $-1\le \sin x\le 1$
$\Rightarrow -3\le \sin x-2\le -1\Rightarrow 1\le (\sin x-2{)}^2\le 9\Rightarrow -4\le 5-(\sin x-2{)}^2\le 4$
Hence, range of $f$ is $[-4,4]$

$f\left(\frac{\pi }{2}\right)-f\left(0\right)=4-1=3$

If $x\in [0,\pi ],$ then $f\left(x\right)>0$
Hence, graph of $y=f\left(x\right)$ can not intersect $x$-axis in the interval $[0,\pi ].$

$f(-x)=f\left(x\right)\Rightarrow -4\sin x+{\cos }^{2}x=4\sin x+{\cos }^{2}x\Rightarrow 8\sin x=0$
$\Rightarrow x=n\pi$, where $n\in Z$