Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the parabola \(y=f(x)\), having axis parallel to the \(y\)-axis, touches the line \(y=x\) at \((1,1)\), then (A) \(2 f^{\prime}(0)+f(0)=1\) (B) \(2 f(0)+f^{\prime}(0)=1\) (C) \(2 f(0)-f^{\prime}(0)=1\) (D) \(2 f^{\prime}(0)-f(0)=1\)

Step-by-Step Solution

Answer

The correct option is (B): \(2f(0) + f'(0) = 1\).

Step 1: Understanding the Problem

A parabola with a vertical axis touches the line \(y=x\) at the point \((1,1)\) means that the parabola passes through this point and has the same slope as the line at this point.

Step 2: Set Up the Parabola Equation

Assume a general form of the parabola as \(y = ax^2 + bx + c\). Since the parabola touches the line \(y = x\) at \((1, 1)\), we have \(ax^2 + bx + c = x\) at \(x = 1\). So, \(1 = a(1)^2 + b(1) + c = a + b + c\).

Step 3: Equating Derivatives at the Touching Point

The derivative of the parabola \(f(x) = ax^2 + bx + c\) is \(f'(x) = 2ax + b\). At \(x = 1\), the derivative should equal the slope of the line \(y = x\), which is 1. Therefore, \(2a(1) + b = 1\), or \(2a + b = 1\).

Step 4: Expressing in Terms of \(f(0)\) and \(f'(0)\)

From our assumptions, we can write \(f(0) = c\) and \(f'(0) = b\). Solve the equations from Steps 2 and 3:\- From Step 2: \(a + b + c = 1\)\- From Step 3: \(2a + b = 1\)\Substitute \(a = \frac{1 - b}{2}\) from Step 3 into Step 2:\\(\frac{1-b}{2} + b + c = 1\).

Step 5: Simplifying Expressions

Rewriting Step 4 with all terms in terms of \(b\):\\(1-b + 2b + 2c = 2\), which simplifies to \(b + 2c = 1\). \So, replace \(b\) with \(f'(0)\) and \(c\) with \(f(0)\). The equation becomes \(f'(0) + 2f(0) = 1\).

Step 6: Identifying the Correct Option

From Step 5, our final expression \(f'(0) + 2f(0) = 1\) matches option (B) once rearranged. Thus, the correct answer is (B): \(2f(0) + f'(0) = 1\).

Key Concepts

Equation of Parabola

To understand the specific characteristics of a parabola, we need to know its equation. A common form for the equation of a parabola with a vertical axis is \(y = ax^2 + bx + c\). This equation is quadratic, meaning it includes an \(x^2\) term, crucial for the curve's U-shape appearance.
In our exercise, the parabola is required to touch a line at a specific point, indicating a particular setup of its equation. As the parabola touches the line \(y = x\) at \((1,1)\), it implies that if you substitute \(x=1\) into the equation, you should get \(y=1\).
Thus, for the equation \(y = ax^2 + bx + c\), follows that \(1 = a(1)^2 + b(1) + c = a + b + c\). This equation is vital for deriving further information about the parabola's properties at this particular point. You'll often hear it referred to as the point of tangency.

Derivative of a Parabola

The derivative of a function gives us its slope, which tells us how steep the parabola is at any given point. For a parabola defined by the equation \(f(x) = ax^2 + bx + c\), the derivative is found using the formula \(f'(x) = 2ax + b\).
In our case, the derivative indicates that at \(x = 1\), the slope (or steepness) of the parabola should match the slope of the line \(y = x\), which is 1. Hence, setting \(2a(1) + b = 1\) to ensure they touch at \(x = 1\).
Through this condition, we create another vital equation \(2a + b = 1\). This derivative aspect links closely with the idea of tangency, ensuring that at the touchpoint \((1,1)\), both the parabola and line have exactly the same slope.

Parabola and Line Tangency

When a parabola is tangent to a line, it means they just barely meet - touching at a single point with a matching slope. In our exercise, a parabola touches the line \(y = x\) at \((1,1)\). This tells us a few critical things.

• Firstly, each set of coordinates in their equations at this point must be equal.
• Second, the derivative of the parabola must equal the slope of the line.

To nail down these conditions mathematically, we first ensure that the parabola's equation at \(x=1\) equates with \(y=1\). This gives us \(a + b + c = 1\).
Next, we ensure the slope matches by aligning derivatives, yielding the second useful equation from this situation: \(2a + b = 1\). Together, these equations allow you to express other terms, like \(f(0)\) and \(f'(0)\), linking them back to the touchpoint. This exploration helps identify the specific characteristics and expressions relating to their tangency.