Question

If the curve $y=a x^{2}+b x+c, x \in R,$ 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 :

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

Answer 1

The Correct Option is C

To solve the problem, we need to determine the coefficients \(a\), \(b\), and \(c\) for the quadratic curve equation \(y = ax^2 + bx + c\) 

We have the following conditions:

• The curve passes through the point (1, 2).
• The tangent line to the curve at the origin (0, 0) has the equation \(y = x\).

Step 1: Use the condition that the curve passes through (1, 2).

Substitute \(x = 1\) and \(y = 2\) in the curve equation:

\(2 = a \cdot 1^2 + b \cdot 1 + c\)

This simplifies to:

\(a + b + c = 2\) (Equation 1)

Step 2: Use the condition regarding the tangent line at the origin.

The derivative of the curve \(y = ax^2 + bx + c\) is:

\(\frac{dy}{dx} = 2ax + b\)

At the origin, substitute \(x = 0\):

\(\frac{dy}{dx}\bigg|_{x=0} = b\)

This derivative is equal to the slope of the tangent line at the origin, which is 1 (since \(y = x\) has a slope of 1).

Therefore:

\(b = 1\) (Equation 2)

Step 3: Solve for \(a\) and \(c\) using the equations.

Substitute \(b = 1\) into Equation 1:

\(a + 1 + c = 2\)

Simplifying gives:

\(a + c = 1\) (Equation 3)

We will test the given options to find which combination satisfies all the equations:

• For the option \((a = 1, b = 1, c = 0)\):
  • Substitute into Equation 3: \(1 + 0 = 1\), which is correct.

Thus, the correct values of \(a\), \(b\), and \(c\) are:

\(a = 1, b = 1, c = 0\).