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Inverse Trigonometric Functions question

2023 · 10 Apr · Shift 2 · Q37
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Inverse Trigonometric Functions question

2023 · 10 Apr · Shift 2 · Q37

JEE MainMathematicsInverse Trigonometric FunctionsNumerical+4 / −1
If the domain of the function f(x)=sec⁡−1(2x5x+3)f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)f(x)=sec−1(5x+32x​) is [α,β)U(γ,δ][\alpha, \beta) \mathrm{U}(\gamma, \delta][α,β)U(γ,δ], then ∣3α+10(β+γ)+21δ∣|3 \alpha+10(\beta+\gamma)+21 \delta|∣3α+10(β+γ)+21δ∣ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 24

  1. For the inverse secant function y=sec⁡−1(t)y=\sec^{-1}(t)y=sec−1(t) to be defined, its argument must satisfy t∈(−∞,−1]∪[1,∞).t\in (-\infty,-1]\cup[1,\infty).t∈(−∞,−1]∪[1,∞). So we need 2x5x+3≤−1or2x5x+3≥1,\frac{2x}{5x+3}\le -1 \quad \text{or} \quad \frac{2x}{5x+3}\ge 1,5x+32x​≤−1or5x+32x​≥1, with also 5x+3≠0⇒x≠−35.5x+3\ne 0 \Rightarrow x\ne -\frac35.5x+3=0⇒x=−53​.

  2. Solve 2x5x+3≥1.\frac{2x}{5x+3}\ge 1.5x+32x​≥1. Bring to one side: 2x−(5x+3)5x+3≥0\frac{2x-(5x+3)}{5x+3}\ge 05x+32x−(5x+3)​≥0 −3x−35x+3≥0\frac{-3x-3}{5x+3}\ge 05x+3−3x−3​≥0 x+15x+3≤0.\frac{x+1}{5x+3}\le 0.5x+3x+1​≤0. Critical points are x=−1x=-1x=−1 and x=−35x=-\frac35x=−53​. By sign analysis, x+15x+3≤0⇒x∈[−1,−35).\frac{x+1}{5x+3}\le 0 \quad \Rightarrow \quad x\in[-1,-\tfrac35).5x+3x+1​≤0⇒x∈[−1,−53​).

  3. Solve 2x5x+3≤−1.\frac{2x}{5x+3}\le -1.5x+32x​≤−1. Bring to one side: 2x+(5x+3)5x+3≤0\frac{2x+(5x+3)}{5x+3}\le 05x+32x+(5x+3)​≤0 7x+35x+3≤0.\frac{7x+3}{5x+3}\le 0.5x+37x+3​≤0. Critical points are x=−37x=-\frac37x=−73​ and x=−35x=-\frac35x=−53​. By sign analysis, 7x+35x+3≤0⇒x∈(−35,−37].\frac{7x+3}{5x+3}\le 0 \quad \Rightarrow \quad x\in(-\tfrac35,-\tfrac37].5x+37x+3​≤0⇒x∈(−53​,−73​].

  4. Therefore the domain is [−1,−35)∪(−35,−37].[-1,-\tfrac35)\cup(-\tfrac35,-\tfrac37].[−1,−53​)∪(−53​,−73​]. So, α=−1,β=−35,γ=−35,δ=−37.\alpha=-1,\quad \beta=-\frac35,\quad \gamma=-\frac35,\quad \delta=-\frac37.α=−1,β=−53​,γ=−53​,δ=−73​.

  5. Compute 3α+10(β+γ)+21δ3\alpha+10(\beta+\gamma)+21\delta3α+10(β+γ)+21δ =3(−1)+10(−35−35)+21(−37)=3(-1)+10\left(-\frac35-\frac35\right)+21\left(-\frac37\right)=3(−1)+10(−53​−53​)+21(−73​) =−3+10(−65)−9=-3+10\left(-\frac65\right)-9=−3+10(−56​)−9 =−3−12−9=−24.=-3-12-9=-24.=−3−12−9=−24. Hence, ∣3α+10(β+γ)+21δ∣=24.\left|3\alpha+10(\beta+\gamma)+21\delta\right|=24.∣3α+10(β+γ)+21δ∣=24.

So the required integer is 242424.

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