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Definite Integration question

2025 · 3 Apr · Shift 1 · Q33

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JEE Main
Mathematics
Definite Integration
2025 · 3 Apr · Shift 1 · Q33

Definite Integration question

2025 · 3 Apr · Shift 1 · Q33

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

Let the domain of the function

f(x)=\log _2 \log _4 \log _6\left(3+4 x-x^2\right)

be

(a, b)

. If

\int_0^{b-a}\left[x^2\right] d x=p-\sqrt{q}-\sqrt{r}, p, q, r \in \mathbb{N}, \operatorname{gcd}(p, q, r)=1

, where

[\cdot]

is the greatest integer function, then

p+q+r

is equal to

A
10
B
11
C
9
D
8

View written solutionFree

Correct answer: A

Find the domain of $f(x)=\log_2\log_4\log_6(3+4x-x^2).$

For the outermost logarithm $\log_2(\cdot)$ to be defined, we need $\log_4\log_6(3+4x-x^2)>0.$

Since base $4>1$ , this means $\log_6(3+4x-x^2)>1.$

Again, since base $6>1$ , this gives $3+4x-x^2>6.$
Solve the inequality $3+4x-x^2>6$ $-x^2+4x-3>0$ $x^2-4x+3<0$ $(x-1)(x-3)<0.$

Hence, $1<x<3.$

Therefore, the domain is $(a,b)=(1,3).$

So, $b-a=3-1=2.$
Evaluate the integral $\int_0^{b-a}[x^2]dx=\int_0^2 [x^2]dx.$

Now split according to where $[x^2]$ changes value:
- For $0\le x<1$ , $0\le x^2<1$ , so $[x^2]=0$ .
- For $1\le x<\sqrt2$ , $1\le x^2<2$ , so $[x^2]=1$ .
- For $\sqrt2\le x<\sqrt3$ , $2\le x^2<3$ , so $[x^2]=2$ .
- For $\sqrt3\le x<2$ , $3\le x^2<4$ , so $[x^2]=3$ .
- At isolated points $x=1,\sqrt2,\sqrt3,2$ , values do not affect the integral.
Therefore,
=\int_0^1 0\,dx+\int_1^{\sqrt2}1\,dx+\int_{\sqrt2}^{\sqrt3}2\,dx+\int_{\sqrt3}^{2}3\,dx.$$ Compute: $$=0+(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3).$$ Simplify: $$=\sqrt2-1+2\sqrt3-2\sqrt2+6-3\sqrt3$$ $$=5-\sqrt2-\sqrt3.$$ Thus, $$p=5,\quad q=2,\quad r=3.$$
Find $p+q+r$ $p+q+r=5+2+3=10.$
Compare with stored answer

Derived answer: 10

Stored correct answer: A (which is 10)

So the derived answer agrees with the stored answer.

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