Question

Let $f\left(x\right)=\frac{x\left(1+a\cos x\right)-b\sin x}{x^3},x\ne 0andf\left(0\right)=1$, then values if 'a' and 'b' so that 'f' is continuous are

• A. $\frac{5}{2},\frac{3}{2}$
• B. $\frac{5}{2},\frac{-3}{2}$
• C. $-\frac{5}{2}-\frac{3}{2}$
• D. $\frac{1}{2}-\frac{3}{2}$

Answer 1

Answer: (C)

$-\frac{5}{2}-\frac{3}{2}$

Using expansions,

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}\cdot \cdot \cdot$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}\cdot \cdot \cdot$

We have,

${lim}_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}$

$={lim}_{x\to 0}\frac{x+ax(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdot \cdot \cdot )-b(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdot \cdot \cdot )}{x^3}$

$={lim}_{x\to 0}\frac{(1+a-b)x+(\frac{b}{6}-\frac{a}{2})x^3+(\frac{a}{24}-\frac{b}{120})x^4}{x^3}\cdot \cdot \cdot$

$={lim}_{x\to 0}\frac{(1+a-b)+(\frac{b}{6}-\frac{a}{2})x^2+(\frac{a}{24}-\frac{b}{120})x^4}{x^2}\cdot \cdot \cdot$

Since, $f\left(0\right)=1$, a finite quantity.

$\therefore$ We must have

$1+a-b=0$ - Equation 1

$\frac{b}{6}-\frac{a}{2}=1$ - Equation 2

$a-b=-1$

$-3a+b=6$

On adding, we get $-2a=5$

$a=\frac{-5}{2}$

Now, by putting the value of $a$ in Equation 1, we get

$a-b=-1$

$b=\frac{-3}{2}$

Hence, option C is correct.