Aspire Faculty ID #18024 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

Let $I(x) = \int \dfrac{3dx}{(4x+6)\sqrt{4x^2 + 8x + 3}}$ and $I(0) = \dfrac{\sqrt{3}}{4} + 20$. If $I\left(\dfrac{1}{2}\right) = \dfrac{a\sqrt{2}}{b} + c$, where $a,b,c \in \mathbb{N}$, $\gcd(a,b)=1$, then $a + b + c$ is equal to:

Solution

Let $4x+6 = \dfrac{1}{t}$

$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + c$

$I(0) = \frac{\sqrt{3}}{4} + c \Rightarrow c = 20$

$I(x) = \frac{3}{4}\sqrt{\frac{4x+2}{4x+6}} + 20$

$I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{8}{8}} + 20 = \frac{3\sqrt{2}}{8} + 20$

$\Rightarrow a + b + c = 3 + 8 + 20 = 31$

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