Question

For a certain value of $c$, $\underset{x\to -\infty }{lim}[{(x^5+7x^4+2)}^c-x]=\lambda$, is finite and non-zero. Then, the value of $3c+\lambda$ is $\dots \dots \dots$.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

$\underset{x\to \infty }{lim}(x^{5c}{(1+\frac{7}{x}+\frac{2}{x^5})}^c-x)=\underset{x\to \infty }{lim}(x^{5c-1}{(1+\frac{7}{x}+\frac{2}{x^5})}^c-1)$
This will be of the form $\infty \times 0$ only, if
$5c-1=0⟹c=\frac{1}{5}$ substituting $c=\frac{1}{5}$, $\lambda$ becomes
Thus, $\lambda =\underset{x\to -\infty }{lim}x[{(1+x)}^{\frac{1}{5}}-1]$ where $x=\frac{7}{x}+\frac{2}{x^5}$
$=\underset{x\to -\infty }{lim}x[1+\frac{x}{5}+\cdots -1]=\underset{x\to -\infty }{lim}x(\frac{7}{x}+\frac{2}{x^5}).\frac{1}{5}=\frac{7}{5}$
Therefore, $c=\frac{1}{5}$ and $\lambda =\frac{7}{5}$
$⟹3c+\lambda =2$
Hence, option 'A' is correct.