Question

Number of solutions of $f\left(x\right)+g\left(x\right)=0$ is

• A. $2$
• B. $4$
• C. $0$
• D. $1$

Answer 1

The correct option is A $2$

$\underset{n\to \infty }{lim}{\left(\cos \frac{x}{\sqrt{n}}\right)}^n\left(1^{\infty }\right)$ form

$\Rightarrow$ If $\underset{n\to \infty }{lim}{f\left(x\right)}^{g\left(x\right)}$ is such that $\underset{n\to \infty }{lim}f\left(x\right)\to 1$ and $\underset{n\to \infty }{lim}g\left(x\right)\to \infty$

Then $\underset{n\to \infty }{lim}{f\left(x\right)}^{g\left(x\right)}=\underset{n\to \infty }{lim}{e}^{(f\left(x\right)-1).g\left(x\right)}$

$\Rightarrow \underset{n\to \infty }{lim}{e}^{\left(\cos \frac{x}{\sqrt{n}}-1\right).n}$

$\Rightarrow e^{-\left(2{\sin }^2\frac{x}{2\sqrt{n}}\right).n}=e^{-\left(2{\sin }^2\frac{x}{2\sqrt{n}}\right)/(\frac{x^2}{4n}\times \frac{4}{x^2})}$

$\Rightarrow e^{-x^2/2}$
$\underset{x\to \infty }{lim}(\sqrt{x^2+x+1}-\sqrt{x^2+1})$
On rationalising, $b=\underset{x\to \infty }{lim}\frac{(x^2+x+1-x^2+1)}{(\sqrt{x^2+x+1}+\sqrt{x^2+1})}$
Taking highest power of $x$ common out from both numerator and denominator
$b=\underset{x\to \infty }{lim}\frac{x(1+\frac{2}{x})}{x(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x^2}})}=\frac{1}{2}$

Thus $f\left(x\right)+g\left(x\right)=0$ has 2 solutions as it is $2^{nd}$ order equation.