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Functions question

2003 · Shift 0 · Q69
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Functions question

2003 · Shift 0 · Q69

JEE MainMathematicsFunctionsMCQ+4 / −1
Domain of definition of the function f(x) = 34−x2{3 \over {4 - {x^2}}}4−x23​+log⁡10(x3−x){\log _{10}}\left( {{x^3} - x} \right)log10​(x3−x), is
  1. A
    (-1, 0) ∪\cup∪(1, 2) ∪\cup∪(2, ∞\infty∞)
  2. B
    (1, 2)
  3. C
    (-1, 0) ∪\cup∪ (1, 2)
  4. D
    (1, 2) ∪\cup∪(2, ∞\infty∞)
View written solutionFree

Correct answer: A

  1. We need the domain of f(x)=34−x2+log⁡10(x3−x).f(x)=\frac{3}{4-x^2}+\log_{10}(x^3-x).f(x)=4−x23​+log10​(x3−x).

For the function to be defined, both parts must be defined simultaneously.

  1. Condition from the rational term: 34−x2\frac{3}{4-x^2}4−x23​ is defined when the denominator is nonzero.

So, 4−x2≠04-x^2\neq 04−x2=0 x2≠4x^2\neq 4x2=4 x≠±2.x\neq \pm 2.x=±2.

  1. Condition from the logarithmic term: log⁡10(x3−x)\log_{10}(x^3-x)log10​(x3−x) is defined only when x3−x>0.x^3-x>0.x3−x>0.

Factorize: x3−x=x(x2−1)=x(x−1)(x+1).x^3-x=x(x^2-1)=x(x-1)(x+1).x3−x=x(x2−1)=x(x−1)(x+1).

So we solve x(x−1)(x+1)>0.x(x-1)(x+1)>0.x(x−1)(x+1)>0.

Critical points are: −1, 0, 1.-1,\ 0,\ 1.−1, 0, 1.

Now check signs in intervals:

  • For x∈(−∞,−1)x\in(-\infty,-1)x∈(−∞,−1), take x=−2x=-2x=−2: (−)(−)(−)=−(-)(-)(-) = -(−)(−)(−)=− so negative.

  • For x∈(−1,0)x\in(-1,0)x∈(−1,0), take x=−12x=-\tfrac12x=−21​: (−)(−)(+)=+(-)(-)(+) = +(−)(−)(+)=+ so positive.

  • For x∈(0,1)x\in(0,1)x∈(0,1), take x=12x=\tfrac12x=21​: (+)(−)(+)=−(+)(-)(+) = -(+)(−)(+)=− so negative.

  • For x∈(1,∞)x\in(1,\infty)x∈(1,∞), take x=2x=2x=2: (+)(+)(+)=+(+)(+)(+) = +(+)(+)(+)=+ so positive.

Hence, x3−x>0⇒x∈(−1,0)∪(1,∞).x^3-x>0 \quad \Rightarrow \quad x\in(-1,0)\cup(1,\infty).x3−x>0⇒x∈(−1,0)∪(1,∞).

  1. Combine both conditions:

From log term: x∈(−1,0)∪(1,∞).x\in(-1,0)\cup(1,\infty).x∈(−1,0)∪(1,∞).

From rational term: x≠−2, 2.x\neq -2,\ 2.x=−2, 2.

Now −2-2−2 is not in (−1,0)∪(1,∞)(-1,0)\cup(1,\infty)(−1,0)∪(1,∞), so it does not matter. But 2∈(1,∞)2\in(1,\infty)2∈(1,∞) must be excluded.

Therefore domain is (−1,0)∪(1,2)∪(2,∞).(-1,0)\cup(1,2)\cup(2,\infty).(−1,0)∪(1,2)∪(2,∞).

  1. Compare with options:
  • A: (−1,0)∪(1,2)∪(2,∞)(-1,0)\cup(1,2)\cup(2,\infty)(−1,0)∪(1,2)∪(2,∞) ✓
  • B: (1,2)(1,2)(1,2) ✗
  • C: (−1,0)∪(1,2)(-1,0)\cup(1,2)(−1,0)∪(1,2) ✗
  • D: (1,2)∪(2,∞)(1,2)\cup(2,\infty)(1,2)∪(2,∞) ✗

So the correct option is A.

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