MHT CET · Maths · Limits
Given \(f(x)=\frac{a x+b}{x+1}, \lim _{x \rightarrow \infty} f(x)=1 \quad\) and
\(\lim _{x \rightarrow 0} f(x)=2\), then \(f(-2)\) is
- A 0
- B 1
- C \(\overline{2}\)
- D 3
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
Given, \(\lim _{x \rightarrow \infty} f(x)=1\)
\(\Rightarrow \lim _{x \rightarrow \infty} \frac{a x+b}{x+1}=1\)
\(\Rightarrow \lim _{x \rightarrow \infty} \frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1\)
\(\Rightarrow a=1\)
\(\Rightarrow \lim _{x \rightarrow 0} f(x)=2\)
\(\Rightarrow \lim _{x \rightarrow 0} \frac{a x+b}{x+1}=2\)
\(\Rightarrow b=2\)
\(\Rightarrow[\sin (\log x)+\cos (\log x)] d x \)
\(=\int \frac{d}{d x}\{x \sin (\log x)\} d x \)
\( =x \sin (\log x)+c\)
\(\Rightarrow \lim _{x \rightarrow \infty} \frac{a x+b}{x+1}=1\)
\(\Rightarrow \lim _{x \rightarrow \infty} \frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1\)
\(\Rightarrow a=1\)
\(\Rightarrow \lim _{x \rightarrow 0} f(x)=2\)
\(\Rightarrow \lim _{x \rightarrow 0} \frac{a x+b}{x+1}=2\)
\(\Rightarrow b=2\)
\(\Rightarrow[\sin (\log x)+\cos (\log x)] d x \)
\(=\int \frac{d}{d x}\{x \sin (\log x)\} d x \)
\( =x \sin (\log x)+c\)
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