Question

. lf $y=1+{\sec }^2\theta +{\cos }^2\theta ,\theta \ne 0$ then

• A. $y=0$
• B. $y\le 2$
• C. $y\ge 3$
• D. $y>2$

Answer 1

The correct option is C $y\ge 3$

$f\left(\theta \right)=y=1+{\sec }^2\theta +{\cos }^2\theta ,\theta \ne 0$
we have to find range of $y$,
$y=1+\frac{1}{{\cos }^2\theta }+{\cos }^2\theta$
as $\cos \theta$ is a continuous function,
Let $g\left(\theta \right)=\frac{1}{{\cos }^2\theta }+{\cos }^2\theta$
Put $x=\cos \theta$
$g\left(x\right)=\frac{1}{x^2}+x^2$
$\therefore$ for minima, diff. $g\left(x\right)$ with respect to x and equal to 0.
$g^{\prime }\left(x\right)=\frac{-2}{x^3}+2x=0$
$x^4=1$
${\cos }^2\theta =1$
$\Rightarrow g\left(x\right)=2$
$\therefore$ minimum value of $y=1+min\left(g\right(x\left)\right)=1+2=3$
$yϵ[3,\infty ]$