JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(y=f(x)\) be a thrice differentiable function in \((-5,5)\). Let the tangents to the curve \(y=f(x)\) at \((1, \mathrm{f}(1))\) and \((3, \mathrm{f}(3))\) make angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\), respectively with positive \(x\)-axis. If \(27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}\) where \(\alpha, \beta\) are integers, then the value of \(\alpha+\beta\) equals
- A \(-14\)
- B \(26\)
- C \(-16\)
- D \(36\)
Answer & Solution
Correct Answer
(B) \(26\)
Step-by-step Solution
Detailed explanation
\(y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x)\) \( \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \)…
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