Question

If $f\left(x\right)={9.4}^x-{24.2}^x$ and $g\left(x\right)=\cos 2x-4{\sin }^{2}x-5{\cos }^{2}x$, then the value of $x$ and $y$ for which $f\left(x\right)+12=g\left(y\right)$ are

• A. $x=3,y=2n\pi :n\in I$
• B. $x={\log }_2\frac{4}{3};y=n\pi ;n\in I$
• C. $x={\log }_2\frac{4}{3},y=\frac{2n\pi }{3};n\in I$
• D. $x=2+{\log }_{4}9,y=n\pi ;n\in I$

Answer 1

The correct option is B $x={\log }_2\frac{4}{3};y=n\pi ;n\in I$

$f\left(x\right)={9.4}^x-{24.2}^x$

$f\left(x\right)=({3.2}^x-4{)}^2-16$

$\Rightarrow f\left(x\right)\ge -16$

$\Rightarrow f\left(x\right)+12\ge -4$

Now, $g\left(x\right)=(2{\cos }^{2}x-1)-4(1-{\cos }^{2}x)-5\left({\cos }^{2}x\right)$

$g\left(x\right)={\cos }^{2}x-5$

So, $g\left(y\right)\le -4$

So, there is only one possibility for $f\left(x\right)+12=g\left(y\right)$

$\Rightarrow$ both are equal to $-4$

$\Rightarrow {3.2}^x-4=0$ and ${\cos }^{2}y=1$

$\Rightarrow x={\log }_2\frac{4}{3}$ and $y=n\pi$