Question

Let $f\left(x\right)=1+e^{\ln \left(\ln x\right)}\ln (k^2+25)$ and $g\left(x\right)=\frac{1}{|x|-1}.$ If $\underset{x\to 1^{+}}{lim}\left(f\right(x){)}^{g\left(x\right)}=k(2{\sin }^2\alpha +3\cos \beta +5)$ for $k>0$ and $\alpha ,\beta \in R,$ then which of the following is (are) CORRECT?

• A. The value of $k$ is equal to $5$
• B. The value of $\frac{{\sin }^{10}\alpha +{\cos }^5\beta }{{\sin }^2\alpha +\cos \beta }$ is equal to $1$
• C. ${\cos }^2\beta +{\sin }^4\alpha =2$
• D. ${\sin }^2\alpha >\cos \beta$

Answer 1

Answer: (A), (B), (C)

${\cos }^2\beta +{\sin }^4\alpha =2$

$f\left(x\right)=1+\left(\ln x\right)\ln (k^2+25)$
$g\left(x\right)=\frac{1}{|x|-1}$

Now, $\underset{x\to 1^{+}}{lim}\left(f\right(x){)}^{g\left(x\right)}$
$=\underset{x\to 1^{+}}{lim}{(1+(\ln x\left)\ln \right(k^2+25\left)\right)}^{1/(x-1)}=\exp \left(\underset{x\to 1^{+}}{lim}\frac{\left(\ln x\right)\ln (k^2+25)}{(x-1)}\right)$
Let $x=1+t$
${lim}_{x\to 1^{+}}\left(f\right(x){)}^{g\left(x\right)}=\exp \left(\underset{t\to 0^{+}}{lim}\frac{\left(\ln \right(1+t\left)\right)\ln (k^2+25)}{t}\right)=e^{\ln (k^2+25)}=k^2+25$

Now, $\frac{k^2+25}{k}=2{\sin }^2\alpha +3\cos \beta +5$
$\Rightarrow k+\frac{25}{k}=2{\sin }^2\alpha +3\cos \beta +5$
$\Rightarrow {(\sqrt{k}-\frac{5}{\sqrt{k}})}^2+10=2{\sin }^2\alpha +3\cos \beta +5$
Above equation is true only when $k=5$ and ${\sin }^2\alpha =1$ and $\cos \beta =1$