Question

If $P=4\sin x+{\cos }^{2}x$, then which of the following is/are true ?

• A. Maximum value of $P$ is $5$
• B. Minimum value of $P$ is $-4$
• C. Maximum value of $P$ occurs when $\sin x=0$
• D. Minimum value of $P$ occurs when $\sin x=-1$

Answer 1

Answer: (B), (D)

Minimum value of $P$ occurs when $\sin x=-1$

$P=4\sin x+{\cos }^{2}x\Rightarrow P=4\sin x+(1-{\sin }^{2}x)$
Assuming $\sin x=t\Rightarrow t\in [-1,1]$
$P=4t+1-t^2\Rightarrow f\left(t\right)=-(t-2{)}^2+5,t\in [-1,1]$
$x$ coordinate of vertex is $2$ and $2\notin [-1,1]$
$f\left(1\right)=4$
$f(-1)=-4$
$\Rightarrow$ The mimimum value of $P$ is $-4$ at $\sin x=-1$
$\Rightarrow$ The maximum value of $P$ is $4$ at $\sin x=1$

Alternate Solution:
$P=4\sin x+{\cos }^{2}x\Rightarrow P=4\sin x+(1-{\sin }^{2}x)\Rightarrow P=-({\sin }^{2}x-4\sin x+4)+5$
$\Rightarrow P=5-(\sin x-2{)}^2$

$-1\le \sin x\le 1\Rightarrow -3\le \sin x-2\le -1\Rightarrow 1\le (\sin x-2{)}^2\le 9\Rightarrow -9\le -(\sin x-2{)}^2\le -1\Rightarrow -4\le 5-(\sin x-2{)}^2\le 4\Rightarrow -4\le P\le 4$

$P=4\Rightarrow 5-(\sin x-2{)}^2=4\Rightarrow \sin x-2=\pm 1\Rightarrow \sin x=1(\because \sin x\ne 3)$

$P=-4\Rightarrow 5-(\sin x-2{)}^2=-4\Rightarrow \sin x-2=\pm 3\Rightarrow \sin x=-1(\because \sin x\ne 5)$

The mimimum value of $P$ is $-4$ when $\sin x=-1.$

The maximum value of $P$ is $4$ when $\sin x=1.$