Question

If the function $f\left(x\right)$, defined as $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{a(1-x\sin x)+b\cos x+5}{x^2},& x<0\end{array}\\ \begin{array}{cc}3,& x=0\end{array}\\ \begin{array}{cc}{\{1+\left(\frac{cx+d{x}^3}{x^2}\right)\}}^{\frac{1}{x}},& x>0\end{array}\end{array}$ is continuous at $x=0$, then

• A. $a=-1$
• B. $b=-4$
• C. $c=0$
• D. $d={\log }_{e}3$

Answer 1

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

The correct options are
A $a=-1$
B $c=0$
C $d={\log }_{e}3$
D $b=-4$

Since, $f\left(x\right)$ is continuous at $x=0$ so at $x=0$ both left and right limits must exist and both must be equal to $3.$

Now, $\frac{a(1-x\sin x)+b\cos x+5}{x^2}=\frac{(a+b+5)+(-a-\frac{b}{2})x^2+...}{x^2}=3$

$[$By expansion of $\sin x$ and $\cos x]$

If ${lim}_{x\to 0^{-}}f\left(x\right)=3$ exists, then $a+b+5=0$ and, $-a-\frac{b}{2}=3\Rightarrow a=-1$ and $b=-4$

Since, $\underset{x\to 0^{+}}{lim}{(1+\left(\frac{cx+{dx}^3}{x^2}\right))}^{\frac{1}{x}}$ exists.

$\Rightarrow \underset{x\to 0^{+}}{lim}\frac{cx+{dx}^3}{x^2}=0\Rightarrow c=0$

Now, $\underset{x\to 0^{+}}{lim}{(1+dx)}^{\frac{1}{x}}=\underset{x\to 0^{+}}{lim}{\left[{(1+dx)}^{\frac{1}{dx}}\right]}^d=e^d$

So, $e^d=3\Rightarrow d={\log }_{e}3$

Hence, $a=-1,b=-4,c=0$ and $d={\log }_{e}3$