The value of \alpha for which 4\alpha \int\limits_ - 1^2 e^ - \alpha | x |dx = 5 , is:

Definite Integration PYQ — JEE Main 2020 · 7 Jan · Shift 2 · Q24 with Solution

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Correct answer: A

We need to solve $4\alpha \int_{-1}^{2} e^{-\alpha |x|}\,dx = 5.$
Split the integral at $x=0$ because of $|x|$ : $\int_{-1}^{2} e^{-\alpha |x|}\,dx = \int_{-1}^{0} e^{-\alpha(-x)}\,dx + \int_{0}^{2} e^{-\alpha x}\,dx.$ So, $= \int_{-1}^{0} e^{\alpha x}\,dx + \int_{0}^{2} e^{-\alpha x}\,dx.$
Evaluate both integrals: $\int_{-1}^{0} e^{\alpha x}\,dx = \left[\frac{e^{\alpha x}}{\alpha}\right]_{-1}^{0} = \frac{1-e^{-\alpha}}{\alpha},$

and $\int_{0}^{2} e^{-\alpha x}\,dx = \left[-\frac{e^{-\alpha x}}{\alpha}\right]_{0}^{2} = \frac{1-e^{-2\alpha}}{\alpha}.$

Hence,

= \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha}.$$ 4. Substitute into the given equation: $$4\alpha \cdot \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha} = 5.$$ So, $$4(2 - e^{-\alpha} - e^{-2\alpha}) = 5.$$ $$8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5.$$ $$4e^{-\alpha} + 4e^{-2\alpha} = 3.$$ $$e^{-\alpha} + e^{-2\alpha} = \frac{3}{4}.$$ 5. Let $$t = e^{-\alpha} > 0.$$ Then $$t + t^2 = \frac{3}{4}.$$ $$t^2 + t - \frac{3}{4} = 0.$$ Multiply by 4: $$4t^2 + 4t - 3 = 0.$$ Factorize: $$4t^2 + 4t - 3 = (2t-1)(2t+3)=0.$$ So, $$t=\frac{1}{2} \quad \text{or} \quad t=-\frac{3}{2}.$$ Since $t>0$, we take $$t=\frac{1}{2}.$$ 6. Therefore, $$e^{-\alpha} = \frac{1}{2} \implies \alpha = \log_e 2.$$ 7. Checking options: - A: $\log_e 2$ ✅ - B: $\log_e \sqrt{2}$ ❌ - C: $\log_e\left(\frac{4}{3}\right)$ ❌ - D: $\log_e\left(\frac{3}{2}\right)$ ❌ Hence the correct answer is **A**.

View written solutionFree

Correct answer: A

We need to solve $4\alpha \int_{-1}^{2} e^{-\alpha |x|}\,dx = 5.$
Split the integral at $x=0$ because of $|x|$ : $\int_{-1}^{2} e^{-\alpha |x|}\,dx = \int_{-1}^{0} e^{-\alpha(-x)}\,dx + \int_{0}^{2} e^{-\alpha x}\,dx.$ So, $= \int_{-1}^{0} e^{\alpha x}\,dx + \int_{0}^{2} e^{-\alpha x}\,dx.$
Evaluate both integrals: $\int_{-1}^{0} e^{\alpha x}\,dx = \left[\frac{e^{\alpha x}}{\alpha}\right]_{-1}^{0} = \frac{1-e^{-\alpha}}{\alpha},$

and $\int_{0}^{2} e^{-\alpha x}\,dx = \left[-\frac{e^{-\alpha x}}{\alpha}\right]_{0}^{2} = \frac{1-e^{-2\alpha}}{\alpha}.$

Hence,

= \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha}.$$ 4. Substitute into the given equation: $$4\alpha \cdot \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha} = 5.$$ So, $$4(2 - e^{-\alpha} - e^{-2\alpha}) = 5.$$ $$8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5.$$ $$4e^{-\alpha} + 4e^{-2\alpha} = 3.$$ $$e^{-\alpha} + e^{-2\alpha} = \frac{3}{4}.$$ 5. Let $$t = e^{-\alpha} > 0.$$ Then $$t + t^2 = \frac{3}{4}.$$ $$t^2 + t - \frac{3}{4} = 0.$$ Multiply by 4: $$4t^2 + 4t - 3 = 0.$$ Factorize: $$4t^2 + 4t - 3 = (2t-1)(2t+3)=0.$$ So, $$t=\frac{1}{2} \quad \text{or} \quad t=-\frac{3}{2}.$$ Since $t>0$, we take $$t=\frac{1}{2}.$$ 6. Therefore, $$e^{-\alpha} = \frac{1}{2} \implies \alpha = \log_e 2.$$ 7. Checking options: - A: $\log_e 2$ ✅ - B: $\log_e \sqrt{2}$ ❌ - C: $\log_e\left(\frac{4}{3}\right)$ ❌ - D: $\log_e\left(\frac{3}{2}\right)$ ❌ Hence the correct answer is **A**.