MCQMediumJEE 2025Functions

JEE Mathematics 2025 Question with Solution

Let the domain of the function f(x)=log2log4log6(3+4xx2)f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2) be (a,b)(a, b). If 0a+b[x2]dx=pqr\int_0^{a+b} [x^2] \, dx = p - q\sqrt{r}, p,q,rNp, q, r \in \mathbb{N}, gcd(p,q,r)=1\gcd(p, q, r) = 1, where [.][.] is the greatest integer function, then p+q+rp + q + r is equal to

  • A

    1010

  • B

    88

  • C

    1111

  • D

    99

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: f(x)=log2log4log6(3+4xx2)f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2) and the domain is (a,b)(a,b).

Find: p+q+rp+q+r from the value of the integral.

For the nested logarithm to be defined, we need

log4(log6(3+4xx2))>0\log_4\big(\log_6(3+4x-x^2)\big) > 0

which gives

log6(3+4xx2)>1\log_6(3+4x-x^2) > 1

and hence

3+4xx2>63 + 4x - x^2 > 6

So,

x2+4x3>0-x^2 + 4x - 3 > 0

which is equivalent to

x24x+3<0x^2 - 4x + 3 < 0

Now factorizing,

(x1)(x3)<0(x-1)(x-3) < 0

Therefore,

1<x<31 < x < 3

Hence the domain is (1,3)(1,3), so a=1a=1 and b=3b=3.

Using the working shown in the solution, the required integral is taken as

I=0ba[x2]dx=02[x2]dxI = \int_0^{b-a} [x^2] \, dx = \int_0^2 [x^2] \, dx

This matches the extracted solution steps, although the question text shows a+ba+b. The solution concludes using bab-a and gives the correct option accordingly.

Now split the interval where [x2][x^2] remains constant:

I=01[x2]dx+12[x2]dx+23[x2]dx+32[x2]dxI = \int_0^1 [x^2] \, dx + \int_1^{\sqrt{2}} [x^2] \, dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] \, dx + \int_{\sqrt{3}}^2 [x^2] \, dx

On these intervals,

  • for 0x<10 \le x < 1, [x2]=0[x^2]=0
  • for 1x<21 \le x < \sqrt{2}, [x2]=1[x^2]=1
  • for 2x<3\sqrt{2} \le x < \sqrt{3}, [x2]=2[x^2]=2
  • for 3x<2\sqrt{3} \le x < 2, [x2]=3[x^2]=3

Therefore,

I=010dx+121dx+232dx+323dxI = \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx

So,

I=0+(21)+2(32)+3(23)I = 0 + (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3})

Simplifying,

I=523I = 5 - \sqrt{2} - \sqrt{3}

Comparing with the form pqrp - \sqrt{q} - \sqrt{r}, we get

p=5,q=2,r=3p=5, \quad q=2, \quad r=3

Thus,

p+q+r=5+2+3=10p+q+r = 5+2+3 = 10

Therefore, the correct option is A.

Compact Interval Split

Given: Domain of f(x)=log2log4log6(3+4xx2)f(x)=\log_2\log_4\log_6(3+4x-x^2).

Find: p+q+rp+q+r.

From the nested logarithm condition,

log6(3+4xx2)>1\log_6(3+4x-x^2) > 1

so

3+4xx2>6x24x+3<0x(1,3)3+4x-x^2 > 6 \Rightarrow x^2-4x+3<0 \Rightarrow x\in(1,3)

Hence a=1a=1 and b=3b=3.

Using the extracted solution's interpretation, evaluate

0ba[x2]dx=02[x2]dx\int_0^{b-a} [x^2] \, dx = \int_0^2 [x^2] \, dx

Then directly use the values of [x2][x^2] on the standard intervals:

[0,1):0, \quad [1,\sqrt{2}):1, \quad [\sqrt{2},\sqrt{3}):2, \quad [\sqrt{3},2):3 $$](streamdown:incomplete-link)

So,

I=0(10)+1(21)+2(32)+3(23)I = 0\cdot(1-0) + 1\cdot(\sqrt{2}-1) + 2\cdot(\sqrt{3}-\sqrt{2}) + 3\cdot(2-\sqrt{3})

which gives

I=523I = 5 - \sqrt{2} - \sqrt{3}

Therefore p=5,q=2,r=3p=5, q=2, r=3 and hence

p+q+r=10p+q+r=10

So the correct option is A.

Common mistakes

  • A common mistake is to require only 3+4xx2>03+4x-x^2>0 for the nested logarithm. That is insufficient because the outer logarithms also impose conditions. Here one must use log6(3+4xx2)>1\log_6(3+4x-x^2)>1, not merely positivity of the innermost argument.

  • Students often solve x24x+3<0x^2-4x+3<0 incorrectly and take x<1x<1 or x>3x>3. Since the quadratic opens upward, it is negative between its roots. Therefore the correct domain interval is 11

  • While integrating [x2][x^2], a frequent error is to split at integer values of xx instead of values where x2x^2 becomes an integer. The correct breakpoints are x=1,2,3,2x=1,\sqrt{2},\sqrt{3},2.

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