JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(I=\int\limits_{1}^{2} \frac{d x}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}},\) then
- A \(\frac{1}{9} < I^{2} < \frac{1}{8}\)
- B \(\frac{1}{3} < I^{2} < \frac{1}{2}\)
- C \(\frac{1}{9} < I < \frac{1}{8}\)
- D \(\frac{1}{3} < I < \frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{9} < I^{2} < \frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\frac{1}{\sqrt{2 x^{3}-9 x^{2}+12 x+4}}\) \(f^{\prime}(x)=\frac{-6(x-1)(x-2)}{2\left(2 x^{3}-9 x^{2}+12 x+4\right)^{3 / 2}}\) \(\therefore f(\mathrm{x})\) is decreasing in \((1,2)\) \(f(1)=\frac{1}{3} ; f(2)=\frac{1}{\sqrt{8}}\)…
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