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Find a, b, c if y=2x is Tangent to y=ax²+bx+c at x=1 | JEE

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

Question
If at x=1x = 1 the line y=2xy = 2x is the tangent to the parabola y=ax2+bx+cy = ax^2 + bx + c, then the respective possible values of aa, bb and cc are:
A12, 1, 12\dfrac{1}{2},\ 1,\ \dfrac{1}{2}correct
B1, 12, 121,\ \dfrac{1}{2},\ \dfrac{1}{2}
C12, 12, 1\dfrac{1}{2},\ \dfrac{1}{2},\ 1
D−12, 1, 32-\dfrac{1}{2},\ 1,\ \dfrac{3}{2}
Solution
Step 1: At x=1x = 1 the contact point is (1, a+b+c)(1,\ a + b + c), and the tangent (via TT) simplifies to
y=(2a+b)x+(c−a).y = (2a + b)x + (c - a).
 Step 2: This equals y=2xy = 2x, so 2a+b=22a + b = 2 and c=ac = a, i.e. b=2(1−a)b = 2(1 - a). Only a=12, b=1, c=12a = \dfrac{1}{2},\ b = 1,\ c = \dfrac{1}{2} fits. Correct answer: (1)
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