MHT CET · Maths · Limits
Let \(\mathrm{f}(\mathrm{x})=5-|\mathrm{x}-2|\) and \(\mathrm{g}(\mathrm{x})=|\mathrm{x}+1|, \mathrm{x} \in \mathrm{R}\). If \(\mathrm{f}(\mathrm{x})\) attains maximum value at \(\alpha\) and \(g(x)\) attains minimum value at \(\beta\), then \(\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{\left(x^2-6 x+8\right)}\) is equal to
- A \(\frac{1}{2}\)
- B \(\frac{3}{2}\)
- C \(-\frac{3}{2}\)
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(\mathrm{x})=5-|\mathrm{x}-2|\) is maximum at \(\mathrm{x}=2 \Rightarrow \alpha=2\)
\(\mathrm{g}(\mathrm{x})=|\mathrm{x}+1|\) is minimum at \(\mathrm{x}=-1 \Rightarrow \beta=-1\)
Now \(\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{\left(x^2-6 x+8\right)}=\lim _{x \rightarrow-(2)(-1)} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}\)
\(\begin{aligned} & =\lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4} \\ & =\frac{(2-1)(2-3)}{2-4} \\ & =\frac{1}{2}\end{aligned}\)
\(\mathrm{g}(\mathrm{x})=|\mathrm{x}+1|\) is minimum at \(\mathrm{x}=-1 \Rightarrow \beta=-1\)
Now \(\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{\left(x^2-6 x+8\right)}=\lim _{x \rightarrow-(2)(-1)} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}\)
\(\begin{aligned} & =\lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4} \\ & =\frac{(2-1)(2-3)}{2-4} \\ & =\frac{1}{2}\end{aligned}\)
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