Let the domain of the function f(x)= _2 _4 _6 (3+4 x-x^2) be (a, ~b) . If ∫_0^ b-a [x^2] dx=p - q - r , p, q, r ∈ ℕ, gcd(p, q, r)=1, where [ ·] is the greatest integer function, then p+q+r is equal to

Correct answer: A. 10 aligned & _4 _6(3+4 x-x^2) > 0 & _6(3+4 x-x^2) > 1 aligned aligned & 3+4 x-x^2 > 6 & x^2-4 x+3 < 0 & (x-1)(x-3) < 0 & x ∈(1,3) & so a=1 & b=3 & ∫_0^2[x^2] d x=? & I=∫_0^1[x^2] d x+∫_1^ 2 [x^2] d x+∫_ 2 ^ 3 [x^2] d x+∫_ 3 ^ 4 [x^2] d x & =0+|x|_1^ 2 +2|x|_ 2 ^ 3 +3|x|_ 3 ^ 4 aligned aligned & =( 2 -1)+2( 3 - 2 )+3(2- 3 ) & =5- 2 - 3 p+q+r=10 aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Definite Integration

Let the domain of the function f(x)= _2 _4 _6 (3+4 x-x^2) be (a,...

Let the domain of the function

f(\mathrm{x})=\log _2 \log _4 \log _6

\left(3+4 x-x^2\right)

be

(\mathrm{a}, \mathrm{~b})

. If

\int_0^{\mathrm{b}-\mathrm{a}}\left[\mathrm{x}^2\right]

\mathrm{dx}=\mathrm{p}

-\sqrt{\mathrm{q}}-\sqrt{\mathrm{r}}, \mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbb{N}, \operatorname{gcd}(\mathrm{p}, \mathrm{q}, \mathrm{r})=1,

where [

\cdot] is the greatest integer function, then

\mathrm{p}+\mathrm{q}+\mathrm{r}

is equal to

Options

A $10$
B $8$
C $11$
D $9$

Correct answer

A. $10$

Step-by-step solution

\begin{aligned} & \log _4 \log _6\left(3+4 x-x^2\right) > 0 \\ & \log _6\left(3+4 x-x^2\right) > 1\end{aligned}

\begin{aligned} & 3+4 x-x^2 > 6 \\ & x^2-4 x+3 < 0 \\ & (x-1)(x-3) < 0 \\ & x \in(1,3) \\ & \text{ so } a=1 & b=3 \\ & \Rightarrow \int_0^2\left[x^2\right] d x=? \\ & I=\int_0^1\left[x^2\right] d x+\int_1^{\sqrt{2}}\left[x^2\right] d x+\int_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right] d x+\int_{\sqrt{3}}^{\sqrt{4}}\left[x^2\right] d x \\ & =0+|x|_1^{\sqrt{2}}+2|x|_{\sqrt{2}}^{\sqrt{3}}+3|x|_{\sqrt{3}}^{\sqrt{4}}\end{aligned}

\begin{aligned} & =(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3}) \\ & =5-\sqrt{2}-\sqrt{3} \Rightarrow \mathrm{p}+\mathrm{q}+\mathrm{r}=10\end{aligned}

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