Question

Let $f\left(x\right)=cosx(sinx+\sqrt{si{n}^{2}x+si{n}^2\theta }),\theta$ is a given const, then max of f(x)is

• A. $\sqrt{1+cs{c}^2\theta }$
• B. $\sqrt{1+si{n}^2\theta }$
• C. $|cos\theta |$
• D. $|sin\theta |$

Answer 1

The correct option is B

$\sqrt{1+si{n}^2\theta }$

$f\left(x\right)=cosx(sinx+\sqrt{si{n}^{2}x+si{n}^2\theta })f\left(x\right)=\frac{1}{secx}(sinx+\sqrt{si{n}^{2}x+si{n}^2\theta })\Rightarrow secxf\left(x\right)=sinx+\sqrt{si{n}^2+si{n}^2\theta }\Rightarrow \left(secxf\right(x)-sinx{)}^2=si{n}^{2}x+si{n}^2\theta =se{c}^{2}x{f}^2(x)+si{n}^{2}x-2sinxsecxf(x)=si{n}^{2}x+si{n}^2\theta =f^2(x\left)\right[1+ta{n}^{2}x]-2f(x)tanx=si{n}^2\theta =f^2(x)ta{n}^{2}x-2f(x)tanx+f^2(x)-si{n}^2\theta =0\Rightarrow 4f^2(x)\ge 4f^2(x\left)\right(f^2\left(x\right)-si{n}^2\theta )=f^2(x)\le 1+si{n}^2\theta \Rightarrow |f(x)|\ge \sqrt{1+si{n}^2\theta }$