MHT CET · Maths · Application of Derivatives
Let \(\mathrm{f}(x)=\int \frac{x^2-3 x+2}{x^4+1} \mathrm{~d} x\), then function decreases in the interval
- A \((-\infty,-2)\)
- B \((-2,-1)\)
- C \((1,2)\)
- D \((2, \infty)\)
Answer & Solution
Correct Answer
(C) \((1,2)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(x)=\int \frac{x^2-3 x+2}{x^4+1} \mathrm{~d} x \)
\( \Rightarrow \mathrm{f}^{\prime}(x)=\frac{x^2-3 x+2}{x^4+1} \)
For \(\mathrm{f}(x)\) to be decreasing,
\(\mathrm{f}^{\prime}(x) < 0 \)
\( \Rightarrow \frac{x^2-3 x+2}{x^4+1} < 0 \)
\( \Rightarrow \frac{(x-1)(x-2)}{x^4+1} < 0 \)
\( \Rightarrow(x-1)(x-2) < 0 \)
\( \Rightarrow x \in(1,2) \)
\( \Rightarrow \mathrm{f}^{\prime}(x)=\frac{x^2-3 x+2}{x^4+1} \)
For \(\mathrm{f}(x)\) to be decreasing,
\(\mathrm{f}^{\prime}(x) < 0 \)
\( \Rightarrow \frac{x^2-3 x+2}{x^4+1} < 0 \)
\( \Rightarrow \frac{(x-1)(x-2)}{x^4+1} < 0 \)
\( \Rightarrow(x-1)(x-2) < 0 \)
\( \Rightarrow x \in(1,2) \)
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