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Definite Integration: Roots Can

JEE Maths question with a full step-by-step solution.

Question
If 3f(x)+4f(3−x)=3x2−83f(x)+4f(3-x) = 3x^2-8 and ∫αβf(x) dx≤0\displaystyle\int_{\alpha}^{\beta}f(x)\,dx \le 0, where α\alpha, β\beta (α<β)\left(\alpha<\beta\right) are roots of x2−3x+a=0x^2-3x+a = 0, then 'aa' can be
A−1101\dfrac{-1}{101}
B1101\dfrac{1}{101}
C2301\dfrac{2}{301}
D401301\dfrac{401}{301}correct
Solution
Step 1: From x2−3x+a=0x^2-3x+a = 0,
α+β=3,αβ=a\alpha+\beta = 3 ,\qquad \alpha\beta = a
 The functional equation involves f(3−x)f(3-x) and 3=α+β3 = \alpha+\beta, so x↦α+β−xx \mapsto \alpha+\beta-x maps [α,β]\left[\alpha,\beta\right] onto itself. Step 2: Writing I=∫αβf(x) dxI = \displaystyle\int_{\alpha}^{\beta}f(x)\,dx, the king property gives
I=∫αβf(α+β−x)dx=∫αβf(3−x)dxI = \int_{\alpha}^{\beta}f\left(\alpha+\beta-x\right)dx = \int_{\alpha}^{\beta}f\left(3-x\right)dx
 Step 3: Integrating the functional equation over [α,β]\left[\alpha,\beta\right],
∫αβ[3f(x)+4f(3−x)]dx=∫αβ(3x2−8)dx\int_{\alpha}^{\beta}\left[3f(x)+4f(3-x)\right]dx = \int_{\alpha}^{\beta}\left(3x^2-8\right)dx
3I+4I=7I=[x3−8x]αβ3I+4I = 7I = \left[x^3-8x\right]_{\alpha}^{\beta}
 Step 4:
β3−α3−8(β−α)=(β−α)(β2+αβ+α2−8)\beta^3-\alpha^3-8\left(\beta-\alpha\right) = \left(\beta-\alpha\right)\left(\beta^2+\alpha\beta+\alpha^2-8\right)
β−α=(α+β)2−4αβ=9−4a (>0)\beta-\alpha = \sqrt{\left(\alpha+\beta\right)^2-4\alpha\beta} = \sqrt{9-4a} \ (>0)
α2+αβ+β2=(α+β)2−αβ=9−a\alpha^2+\alpha\beta+\beta^2 = \left(\alpha+\beta\right)^2-\alpha\beta = 9-a
 so
7I=9−4a(9−a−8)=9−4a(1−a)7I = \sqrt{9-4a}\left(9-a-8\right) = \sqrt{9-4a}\left(1-a\right)
 Step 5: α<β\alpha<\beta, so the roots are real and distinct and 9−4a>0\sqrt{9-4a}>0. Hence I≤0I \le 0 gives
1−a≤0⟹a≥11-a \le 0 \quad\Longrightarrow\quad a \ge 1
 Step 6: Real roots need 9−4a≥09-4a \ge 0, i.e. a≤94a \le \dfrac94. Hence
1≤a≤94=2.251 \le a \le \frac94 = 2.25
 with a=94a = \dfrac94 itself ruled out, since α<β\alpha<\beta needs the two roots distinct. Step 7:
−1101≈−0.0099 ✗,1101≈0.0099 ✗,2301≈0.0066 ✗,401301≈1.332-\frac{1}{101} \approx -0.0099 \ \text{✗},\qquad \frac{1}{101} \approx 0.0099 \ \text{✗},\qquad \frac{2}{301} \approx 0.0066 \ \text{✗},\qquad \frac{401}{301} \approx 1.332
 Answer: (4) 401301\dfrac{401}{301}
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