The number of discontinuities in $R$ for the function $f(x)=\frac{x-1}{x^3+6 x^2+11 x+6}$ is
The number of discontinuities in $R$ for the function $f(x)=\frac{x-1}{x^3+6 x^2+11 x+6}$ is
- 3
- 2
- 1
- 0
Solution
Given,
$
\begin{aligned}
& f(x)=\frac{x-1}{x^3+6 x^2+11 x+6} \\
& f(x)=\frac{x-1}{(x+1)\left(x^2+5 x+6\right)} \\
& f(x)=\frac{x-1}{(x+1)(x+2)(x+3)}
\end{aligned}
$
For is discontinuous, hence denominator is 0 .
$
\begin{array}{ll}
\Rightarrow & (x+1)(x+2)(x+3)=0 \\
\Rightarrow & x=-1,-2,-3
\end{array}
$
Hence, number of discontinuous in $R$ is 3
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
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