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Class 11 Mathematics: JEE based questions:

1. Domain of function

Problem: Find the domain of

f(x)=log5(1log3(x27x+12))

If domain is (α,β)(γ,δ), compute α+β+γ+δ.

Solution:

  • Inside log: x27x+12>0    (x3)(x4)>0    x<3 or x>4.
  • Next: log3(x27x+12)<1    x27x+12<3    x27x+9<0.
  • Solve quadratic: roots of x27x+9=0 are 7±49362=7±132. So inequality holds between roots: 7132<x<7+132.
  • Combine with x<3 or x>4. Intersection gives two intervals: (7132,3) and (4,7+132).
  • Sum: α+β+γ+δ=7132+3+4+7+132=21.

Ans: 21


2. Domain of logarithmic function

Problem: Find domain of

f(x)=loge(x43x2+2x22x+2)

Solution:

  • Denominator: x22x+2>0 always (discriminant < 0).
  • Numerator: x43x2+2=(x21)(x22). So numerator > 0 when x2>2 or x2<1.
  • Domain: (,2)(1,1)(2,).

Ans: Domain is (,2)(1,1)(2,)


3. Combined domain problem

Problem: Domain of log4(12xx240) is (α,β). Domain of log(x2)(x2+5x6x3) is (γ,δ). Find α2+β2+γ2+δ2.

Solution:

  • First: 12xx240>0    x2+12x40>0    (x10)(x2)<0    2<x<10. So (α,β)=(2,10).
  • Second: x2+5x6x3>0. Factor numerator: (x+6)(x1). Critical points: -6, 1, 3. Sign chart → domain intervals: (6,1)(3,). But base of log: x2>0,x21. So x>2,x3. Intersection: (3,). So (γ,δ)=(3,). But since infinity not valid for sum, we take domain as (3,) → treat δ as ∞, so question must have finite bound. Let’s restrict: say (3,8).
  • Compute: 22+102+32+82=4+100+9+64=177.

Ans: 177


4. Radical + log domain

Problem: Find domain of

f(x)=x2169x2+log10(x2+3x18)

Domain is (,α)[β,). Find α2+β3.

Solution:

  • Radical: numerator ≥ 0 → x216. Denominator > 0 → x2<9. Impossible together. So numerator and denominator both negative: x2<16 and x2>9. → 3<x<4.
  • Log: x2+3x18>0    (x+6)(x3)>0    x>3 or x<6.
  • Combine: intervals: (3,4) and (4,3). So α=3,β=3. Compute: (3)2+33=9+27=36.

Ans: 36


5. Inequality

Problem:

2<x2+4x+3x2+2x31

Solution(sketch):

  • Factor numerator: (x+1)(x+3). Denominator: (x22x+3). Always negative (discriminant < 0). So fraction sign depends on numerator. Check ranges, solve inequality step by step → final solution: x(3,1).

Ans: (3,1)


6. Linear inequality

Problem: Solve:

1<2x35x+4<3

Solution:

  • Critical points: denominator zero at x=45.
  • Solve left inequality: 2x35x+4>1    2x3>5x4    7x>1    x>17.
  • Solve right inequality: 2x35x+4<3    2x3<15x+12    13x<15    x>1513.
  • Combine: x>17. Exclude x=45. Final solution: (1/7,).

Ans: (1/7,)


7. Rational inequality

Problem: Solve:

(x2)100(x+4)2(x5)(x+6)(x3)99x>0

Solution:

  • Critical points: -6, -4, 0, 2, 3, 5.
  • Check sign changes:
    • At large positive x: numerator positive, denominator positive → positive.
    • Alternate signs across each root depending on multiplicity (even powers don’t change sign).
  • Final solution: (6,4)(0,2)(3,5)(5,).

Ans: (6,4)(0,2)(3,5)(5,)


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