Question

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

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

Answer 1

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

The correct options are
A $k=5$
B $\frac{{\sin }^{10}\alpha +{\cos }^5\beta }{{\sin }^2\alpha +\cos \beta }=1$
C ${\cos }^2\beta +{\sin }^4\alpha =2$

$f\left(x\right)=1+\ln x\cdot \ln (k^2+25)$ and $g\left(x\right)=\frac{1}{|x|-1}$
$\underset{x\to 1}{lim}f(x{)}^{g\left(x\right)}$
$=\underset{x\to 1}{lim}{(1+\ln x\cdot \ln (k^2+25\left)\right)}^{\frac{1}{|x|-1}}$
$=\underset{x\to 1}{lim}{(1+\ln x\cdot \ln (k^2+25\left)\right)}^{\frac{1}{x-1}}\left[1^{\infty }\text{form}\right]$
$=\exp \left(\underset{x\to 1}{lim}\frac{\ln x\cdot \ln (k^2+25)}{x-1}\right)$
$=e^{\ln (k^2+25)}=k^2+25$

$\therefore k^2+25=k(2{\sin }^2\alpha +3\cos \beta +5)$
$\Rightarrow 2{\sin }^2\alpha +3\cos \beta +5=k+\frac{25}{k}\cdots \left(1\right)$
$\frac{k+\frac{25}{k}}{2}\ge \sqrt{k\cdot \frac{25}{k}}$
$\Rightarrow k+\frac{25}{k}\ge 10$
Maximum value of $2{\sin }^2\alpha +3\cos \beta +5$ is $10$ if $\alpha =(2n+1)\frac{\pi }{2}$ and $\beta =2n\pi$
So, for equation $\left(1\right)$, L.H.S. $\le 10$
So, both L.H.S. and R.H.S. should be equal to $10$.
$k+\frac{25}{k}=10$
$\Rightarrow k^2-10k+25=0$
$\Rightarrow (k-5{)}^2=0$
$\Rightarrow k=5$

${\sin }^2\alpha =1$ and $\cos \beta =1$
Hence, $\frac{{\sin }^{10}\alpha +{\cos }^5\beta }{{\sin }^2\alpha +\cos \beta }=\frac{1+1}{1+1}=1$