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Parabola Determined by a Sum of Squares Condition | JEE Advanced Parabola

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

Question
A parabola y=f(x)y = f(x) has a directrix of slope 00. If
f2(0)+f2(1)+f2(2)+45=10f(0)+8f(1)+4f(2),f^{2}(0) + f^{2}(1) + f^{2}(2) + 45 = 10f(0) + 8f(1) + 4f(2),
 then
Athe graph of y=f(x)y = f(x) is symmetric about x=−12x = -\dfrac12correct
Bf(x)<0  ∀ x∈Rf(x) < 0 \ \ \forall\, x \in \mathbb{R}
Cℓn∣f(x)∣\ell n\left|f(x)\right| is a one-one function
Dthe range of f(x)f(x) is (−∞, 418]\left(-\infty,\, \dfrac{41}{8}\right]correct
Solution
Step 1: A directrix of slope 00 is horizontal, so the axis is vertical and y=f(x)y=f(x) is a quadratic, f(x)=ax2+bx+cf(x) = ax^{2}+bx+c with a≠0a \ne 0. Step 2: Splitting 45=25+16+445 = 25 + 16 + 4,
(f2(0)−10f(0)+25)+(f2(1)−8f(1)+16)+(f2(2)−4f(2)+4)=0\left(f^{2}(0) - 10f(0) + 25\right) + \left(f^{2}(1) - 8f(1) + 16\right) + \left(f^{2}(2) - 4f(2) + 4\right) = 0
(f(0)−5)2+(f(1)−4)2+(f(2)−2)2=0\left(f(0)-5\right)^{2} + \left(f(1)-4\right)^{2} + \left(f(2)-2\right)^{2} = 0
 Step 3: A sum of three real squares is zero only if each is zero, so
f(0)=5,f(1)=4,f(2)=2f(0) = 5, \qquad f(1) = 4, \qquad f(2) = 2
 Step 4:
c=5;a+b+5=4⇒a+b=−1;4a+2b+5=2⇒4a+2b=−3c = 5; \qquad a + b + 5 = 4 \Rightarrow a+b = -1; \qquad 4a + 2b + 5 = 2 \Rightarrow 4a+2b = -3
 From the last two, 4a+2(−1−a)=−3⇒2a=−1⇒a=−124a + 2(-1-a) = -3 \Rightarrow 2a = -1 \Rightarrow a = -\tfrac12, and then b=−12b = -\tfrac12:
f(x)=−x22−x2+5f(x) = -\frac{x^{2}}{2} - \frac{x}{2} + 5
 Step 5: The axis of symmetry is
x=−b2a=−−1/22(−1/2)=−12x = -\frac{b}{2a} = -\frac{-1/2}{2(-1/2)} = -\frac12
 so (1) is true. Step 6: f(0)=5>0f(0) = 5 > 0, so (2) is false. Step 7: a<0a<0, so ff attains a maximum at x=−12x=-\tfrac12:
f(−12)=−12⋅14+14+5=−18+28+5=418f\left(-\tfrac12\right) = -\frac12\cdot\frac14 + \frac14 + 5 = -\frac18 + \frac28 + 5 = \frac{41}{8}
 and the range is (−∞, 418]\left(-\infty,\ \dfrac{41}{8}\right], so (D) is true. Step 8: f ⁣(−12+t)=f ⁣(−12−t)f\!\left(-\tfrac12+t\right) = f\!\left(-\tfrac12-t\right) for every tt, so the two points −12±t-\tfrac12 \pm t lie in the domain of ℓn∣f∣\ell n|f| together, and taking the modulus and then the logarithm keeps the equality. Hence ℓn∣f(x)∣\ell n|f(x)| takes the same value at two different points and (3) is false. Answer: (1) and (4).
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