Question

If $f\left(x\right)={\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x-1;x\in R$, then the equation $f\left(x\right)=0$, has

• A. no solution
• B. one solution
• C. two solutions
• D. more than two solutions

Answer 1

Answer: (B)

one solution

$f\left(x\right)={\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x-1$

Substituting $\sin \theta =\frac{3}{5}$ and $\cos \theta =\frac{4}{5}$, we get

$f\left(x\right)={\sin \theta }^x+{\cos \theta }^x-1$

Now for equation

$f\left(x\right)=0$

$\Rightarrow {\sin \theta }^x+{\cos \theta }^x-1=0$

$\Rightarrow {\sin \theta }^x+{\cos \theta }^x=1={\sin \theta }^2+{\cos \theta }^2$

$\Rightarrow x=2$