Question

If $f\left( x \right) = \dfrac{{\sin 3x + A\sin 2x + B\sin x}}{{{x^5}}}$ for $x \ne 0$ is continuous at x = 0, then
A. $f\left( 0 \right) = 1$
B. $A = 1$
C. $A = - 4$
D. $B = - 5$

Answer 1

Hint- Here, it is a problem of continuity which tells us that any function on a graph which is smooth without holes, jumps or asymptotes is continuous while by function we can tell whether the function is continuous or not by following the following way .

Complete step-by-step answer:
Function must be continuous i.e. it must be defined at all points in its domain. The value of the function must exist as the limit approaches at any point c. The right and left limit of the function must exist and it cannot take jump at that point.

Here in this problem we need to find
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x + A\sin 2x + B\sin x}}{{{x^5}}}$
As we know $\left[ \ \sin 3x = 3\sin x - 4{\sin ^3}x \\\ \sin 2x = 2\sin x\cos x \\\ \right]$
Substituting these in the above function, we get
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{3\sin x - 4{{\sin }^3}x + A \times 2\sin x\cos x + B\sin x}}{{{x^5}}}$
Taking $\sin x$ common from the above equation
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin x\left( {3 - 4{{\sin }^2}x + 2A\cos x + B} \right)}}{{{x^5}}}} \right]$
As we know $\left[ {\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin x}}{x} \times \dfrac{{\left( {3 - 4{{\sin }^2}x + 2A\cos x + B} \right)}}{{{x^4}}}} \right]$
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {1 \times \dfrac{{\left( {3 - 4{{\sin }^2}x + 2A\left( {1 - 2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right) + B} \right)}}{{{x^4}}}} \right]$
For the limits to exist, the sum of constant must be equal to zero,
Therefore $2A + 3 + B = 0$
Now we are left with
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\left( { - 4{{\sin }^2}x - 4A{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right)}}{{{x^4}}}} \right]$
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{ - 4\left( {2{{\sin }^2}\left( {\dfrac{x}{2}} \right){{\cos }^2}\left( {\dfrac{x}{2}} \right) + A{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right)}}{{{x^4}}}} \right]$
Taking ${\sin ^2}\left( {\dfrac{x}{2}} \right)$ common

$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{{{\dfrac{x}{4}}^2} \times 4}} \times \dfrac{{ - 4\left( {2{{\cos }^2}\left( {\dfrac{x}{2}} \right) + A} \right)}}{{{x^4}}}} \right]$
After solving further the given equation we get
$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} - \left( {4 + A - 4 \times \dfrac{{{{\sin }^2}\dfrac{x}{2}}}{{{x^2}}}} \right) \\\ f\left( 0 \right) = 1 \\\$
For the limit to exist $A + 4 = 0$
Therefore
$A = - 4$
Substituting the value of A in $2A + 3 + B = 0$
$\- 8 + B + 3 = 0 \\\ B = 5 \\\$
Therefore A = -4 and B = 5 and the value of the limit is 1.
Hence, the correct option is A and C.

Note- In order to solve these types of questions, you must have a good concept of limits and derivatives. The above question can also be solved using L’ hospital rule. To solve this question, remember the limits of some special function or we can solve using series expansion and then substituting the value in the expansion. The expansion can be done by using Taylor series.