Question

Let $f\left(x\right)=\frac{\sin \left\{x\right\}}{x^2+ax+b}$. If $f\left(5^{+}\right)$ & $f\left(3^{+}\right)$ exists finitely and are not zero, then the value of

$(a+b)$ is (where $\{\cdot \}$ represents fractional part function).

• A. $7$
• B. $10$
• C. $11$
• D. $20$

Answer 1

The correct option is A $7$

$\underset{x\to 5^{+}}{lim}f\left(x\right)=\underset{x\to 5^{+}}{lim}\frac{\sin \left\{x\right\}}{x^2+ax+b}$

$=\frac{\sin \left\{5\right\}}{25+5a+b}=\frac{0}{25+5a+b}=0$

But given it is non-zero

$\Rightarrow 25+5a+b=\mathrm{0.....................................1}$

So that we get $\frac{0}{0}$ form

$\underset{x\to 3^{+}}{lim}f\left(x\right)=\frac{\sin \left\{3\right\}}{9+3a+b}=\frac{0}{9+3a+b}$

It is not zero

$\Rightarrow 9+3a+b=\mathrm{0......................................}2$

From $1$ and $2$

$16+2a=0\Rightarrow a=-8$

$b=15$

$a+b=7$