If the curve y = a x ^ 2 + bx + c, x ∈ R, passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are:

Correct answer: C. a = 1, b = 1, c = 0 a + b + c = 2 (1) and (dy)/(dx) | _ (0, 0) = 1 2ax + b | _ (0, 0) = 1 b = 1 Curve passes through origin So, c = 0 and a = 1

JEE Main 202124 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Application of Derivatives

If the curve y = a x ^ 2 + bx + c, x ∈ R, passes through the point...

If the curve

y = a{x}^{2} + bx + c, x \in R,

passes through the point

\left(1, 2\right)

and the tangent line to this curve at origin is

y = x,

then the possible values of

a, b, c

are:

Options

A $a = - 1, b = 1, c = 1$
B $a = 1, b = 0, c = 1$
C $a = 1, b = 1, c = 0$
D $a = \dfrac{1}{2}, b = \dfrac{1}{2}, c = 1$

Correct answer

C. $a = 1, b = 1, c = 0$

Step-by-step solution

$a + b + c = 2 \ldots \left(1\right)$

and $\dfrac{dy}{dx}{|}_{\left(0, 0\right)} = 1$

$2ax + b{|}_{\left(0, 0\right)} = 1$

$b = 1$

Curve passes through origin

So, $c = 0$

and $a = 1$

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