Question

If $\displaystyle \lim_{x \to 0}{{\left( \cos x+a\sin bx \right)}^{\dfrac{1}{x}}}={{e}^{2}}$ then the possible values of ‘a’ and ‘b’ are:
(This question has multiple correct options)
(a) a = 1, b = 2
(b) a = 2, b = 1
$\left( \text{c} \right)\text{ }a=3,b=\dfrac{2}{3}$
$\left( \text{d} \right)\text{ }a=\dfrac{2}{3},b=3$

Answer 1

To solve the given question, we will assume that the value of $\displaystyle \lim_{x \to 0}{{\left( \cos x+a\sin bx \right)}^{\dfrac{1}{x}}}$ is y. Then we will take the natural logarithm on both the sides of the equation:
$\ln y=\displaystyle \lim_{x \to 0}\ln {{\left( \cos x+a\sin bx \right)}^{\dfrac{1}{x}}}$
Then we will solve the limit using L – Hospital’s rule. After solving, we will get the answer in the form of ‘a’ and ‘b’. We will then put the values of ‘a’ and ‘b’ from the options and then we will check which option satisfies the answer.

Complete step by step answer:
Before solving this question, we will check the indeterminant form we will get, then we will put x = 0 in the term after the limit. At x = 0, cosine function becomes 1 and sine function becomes 0. Thus, the indeterminant form we will get is of the form ${{1}^{\infty }}.$ Hence, to solve the question, we will assume that the value of the limit is y. Thus, we get the following equation.
$y=\displaystyle \lim_{x \to 0}{{\left( \cos x+a\sin bx \right)}^{\dfrac{1}{x}}}......\left( i \right)$
Now, we will take a natural logarithm on both sides of the equation (i). After doing this, we will get the following equation.
$\ln y=\displaystyle \lim_{x \to 0}\left[ \ln {{\left( \cos x+a\sin bx \right)}^{\dfrac{1}{x}}} \right]......\left( ii \right)$
Here, we are going to use the following identity:
$\ln {{x}^{a}}=a\ln x$
Thus, after applying the above identity in equation (ii), we get,
$\Rightarrow \ln y=\displaystyle \lim_{x \to 0}\left[ \dfrac{\ln \left( \cos x+a\sin bx \right)}{x} \right].....\left( iii \right)$
Now, to solve this, we will apply the L – Hospital’s theorem. According to L – Hospital’s rule, the limit of the indeterminant form $\dfrac{0}{0}$ can be calculated as follows.
$\displaystyle \lim_{x \to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to 0}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}$
where f(0) = 0 and g(0) = 0. As the equation (iii) is also of the form $\dfrac{0}{0},$ we can apply L – Hospital’s theorem here.
$\Rightarrow \ln y=\displaystyle \lim_{x \to 0}\left[ \dfrac{\dfrac{d}{dx}\left[ \ln \left( \cos x+a\sin bx \right) \right]}{\dfrac{d}{dx}\left( x \right)} \right]$
$\Rightarrow \ln y=\displaystyle \lim_{x \to 0}\left[ \dfrac{\dfrac{-\sin x+ab\cos bx}{\cos x+a\sin bx}}{1} \right].....\left( iv \right)$
In equation (iv), we have used some differentiation formulas.
$\dfrac{d}{dx}\ln x=\dfrac{1}{x}$
$\dfrac{d}{dx}\left( x \right)=1$
$\dfrac{d}{dx}\sin x=\cos x$
$\dfrac{d}{dx}\cos x=-\sin x$
$\dfrac{d}{dx}f\left( g\left( x \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)$
Thus, now we will put the value x = 0 in the equation (iv).
$\ln y=\displaystyle \lim_{x \to 0}\left[ \dfrac{\dfrac{-\left( 0 \right)+ab\left( 1 \right)}{1+0}}{1} \right]$
$\ln y=ab$.
$\Rightarrow y={{e}^{ab}}......\left( v \right)$
But in the question, we are given that
$y={{e}^{2}}....\left( vi \right)$
From (v) and (vi), we have,
${{e}^{ab}}={{e}^{2}}$
$\Rightarrow ab=2$
Now, we will check the options.
Option (a): a = 1, b = 2
Here,
$ab=1\times 2=2$
Hence, this option is correct.
Option (b): a = 2, b = 1
Here,
$ab=2\times 1=2$
Hence, this option is correct.
Option (c): $a=3,b=\dfrac{2}{3}$
Here,
$ab=3\times \dfrac{2}{3}=2$
Hence, this option is correct.
Option (d): $a=\dfrac{2}{3},b=3$
Here,
$ab=\dfrac{2}{3}\times 3=2$
Hence, this option is correct.

So, the correct answers are “Option A, B, C and D”.

Note: The above question can also be solved as follows:
$y=\displaystyle \lim_{x \to 0}{{\left( 1+\cos x+a\sin bx-1 \right)}^{\dfrac{1}{x}}}$
$\Rightarrow y=\displaystyle \lim_{x \to 0}{{\left[ 1+\left( \cos x+a\sin bx-1 \right) \right]}^{\dfrac{1}{x}}}$
We know that, ${{\left( 1+f\left( x \right) \right)}^{\dfrac{1}{x}}}=\dfrac{f\left( x \right)}{x}.$ Thus,
$\Rightarrow y=\displaystyle \lim_{x \to 0}\left( \dfrac{\cos x+a\sin bx-1}{x} \right)$
On applying L – Hospital’s rule,
$\Rightarrow y=\displaystyle \lim_{x \to 0}\dfrac{\left( -\sin x+ab\cos x-0 \right)}{1}$
$\Rightarrow y=\dfrac{ab}{1}$
$\Rightarrow y=ab$