Question

For $\theta >\frac{\pi }{3}$, the value of $f\left(\theta \right)={\sec }^2\theta +{\cos }^2\theta$ always lies in the interval

• A. $(0,2)$
• B. $[0,1]$
• C. $(1,2)$
• D. $[2,\infty )$

Answer 1

The correct option is D $[2,\infty )$

To find the optimum, we take the derivative and equate to zero.

given, $f\left(\theta \right)=co{s}^2\theta +se{c}^2\theta$

taking derivative w.r.to $\theta$ and equating to zero we get,

$⟹2cos\theta (-sin\theta )+2sec\theta (sec\theta \times tan\theta )=0$

$⟹cos\theta (-sin\theta )+sec\theta (sec\theta \times \frac{sin\theta }{cos\theta })=0$

$⟹cos\theta (-sin\theta )+sec\theta (sec\theta \times sin\theta \times sec\theta )=0$

$⟹cos\theta (-sin\theta )+se{c}^3\theta \left(sin\theta \right)=0$

$⟹sin\theta (-cos\theta +se{c}^3\theta )=0$

$⟹sin\theta =0$ and $(-cos\theta +se{c}^3\theta )=0$

since, $\theta >\frac{\pi }{3}$

Therefore, $sin\theta =0⟹\theta =n\pi$

$-cos\theta +se{c}^3\theta =0$

$⟹-cos\theta +\frac{1}{co{s}^3\theta }=0$

$⟹\frac{1-co{s}^4\theta }{co{s}^3\theta }=0$

$⟹1-co{s}^4\theta =0$

$⟹co{s}^4\theta =1⟹cos\theta =1$

since, $\theta >\frac{\pi }{3}$

Therefore, $cos\theta =1⟹\theta =n\pi$

Substituting $\theta =\pi$ we get

$(cos180{)}^2+(sec180{)}^2=1+1=2$

Therefore, for $\theta =n\pi$ and $\theta >\frac{\pi }{3}$ we get

$f\left(\theta \right)=[2,\infty )$