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Functions question

2024 · 9 Apr · Shift 1 · Q31
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Functions question

2024 · 9 Apr · Shift 1 · Q31

JEE MainMathematicsFunctionsMCQ+4 / −1
If the domain of the function f(x)=sin⁡−1(x−12x+3)f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)f(x)=sin−1(2x+3x−1​) is R−(α,β)\mathbf{R}-(\alpha, \beta)R−(α,β), then 12αβ12 \alpha \beta12αβ is equal to :
  1. A
    40
  2. B
    36
  3. C
    24
  4. D
    32
View written solutionFree

Correct answer: D

  1. For the function f(x)=sin⁡−1(x−12x+3)f(x)=\sin^{-1}\left(\frac{x-1}{2x+3}\right)f(x)=sin−1(2x+3x−1​) to be defined, the argument of sin⁡−1\sin^{-1}sin−1 must satisfy −1≤x−12x+3≤1,-1 \le \frac{x-1}{2x+3} \le 1,−1≤2x+3x−1​≤1, and also 2x+3≠0⇒x≠−32.2x+3 \ne 0 \quad \Rightarrow \quad x\ne -\frac32.2x+3=0⇒x=−23​.

  2. So we solve the double inequality: −1≤x−12x+3≤1.-1 \le \frac{x-1}{2x+3} \le 1.−1≤2x+3x−1​≤1. This is equivalent to solving both:

  • x−12x+3≥−1\frac{x-1}{2x+3} \ge -12x+3x−1​≥−1
  • x−12x+3≤1\frac{x-1}{2x+3} \le 12x+3x−1​≤1
  1. First inequality: x−12x+3≥−1\frac{x-1}{2x+3} \ge -12x+3x−1​≥−1 x−12x+3+1≥0\frac{x-1}{2x+3}+1 \ge 02x+3x−1​+1≥0 x−1+2x+32x+3≥0\frac{x-1+2x+3}{2x+3} \ge 02x+3x−1+2x+3​≥0 3x+22x+3≥0.\frac{3x+2}{2x+3} \ge 0.2x+33x+2​≥0. Critical points are: x=−23,x=−32.x=-\frac23,\quad x=-\frac32.x=−32​,x=−23​. Using sign analysis, 3x+22x+3≥0⇒x∈(−∞,−32)∪[−23,∞).\frac{3x+2}{2x+3} \ge 0 \Rightarrow x\in (-\infty,-\tfrac32) \cup \left[-\tfrac23,\infty\right).2x+33x+2​≥0⇒x∈(−∞,−23​)∪[−32​,∞).
  1. Second inequality: x−12x+3≤1\frac{x-1}{2x+3} \le 12x+3x−1​≤1 x−12x+3−1≤0\frac{x-1}{2x+3}-1 \le 02x+3x−1​−1≤0 x−1−(2x+3)2x+3≤0\frac{x-1-(2x+3)}{2x+3} \le 02x+3x−1−(2x+3)​≤0 −x−42x+3≤0.\frac{-x-4}{2x+3} \le 0.2x+3−x−4​≤0. Multiplying numerator and denominator by −1-1−1, x+42x+3≥0.\frac{x+4}{2x+3} \ge 0.2x+3x+4​≥0. Critical points are: x=−4,x=−32.x=-4,\quad x=-\frac32.x=−4,x=−23​. Using sign analysis, x+42x+3≥0⇒x∈(−∞,−4]∪(−32,∞).\frac{x+4}{2x+3} \ge 0 \Rightarrow x\in (-\infty,-4] \cup \left(-\tfrac32,\infty\right).2x+3x+4​≥0⇒x∈(−∞,−4]∪(−23​,∞).
  1. Intersect the two solution sets: [(−∞,−32)∪[−23,∞)]∩[(−∞,−4]∪(−32,∞)].\left[(-\infty,-\tfrac32) \cup \left[-\tfrac23,\infty\right)\right] \cap \left[(-\infty,-4] \cup \left(-\tfrac32,\infty\right)\right].[(−∞,−23​)∪[−32​,∞)]∩[(−∞,−4]∪(−23​,∞)]. This gives (−∞,−4]∪[−23,∞).(-\infty,-4] \cup \left[-\tfrac23,\infty\right).(−∞,−4]∪[−32​,∞).

So the domain is R−(−4,−23).\mathbb R - (-4,-\tfrac23).R−(−4,−32​). Hence, α=−4,β=−23.\alpha=-4,\quad \beta=-\frac23.α=−4,β=−32​.

  1. Now compute: 12αβ=12(−4)(−23)=12⋅83=32.12\alpha\beta=12\left(-4\right)\left(-\frac23\right)=12\cdot \frac{8}{3}=32.12αβ=12(−4)(−32​)=12⋅38​=32.

Therefore, the correct answer is: 32\boxed{32}32​ which corresponds to option D.

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