Problem 1 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(y=f(x)\) be a parabola, having its axis parallel to \(y\)-axis, which is touched by the line \(y=x\) at \(x=1\), then (A) \(f^{\prime}(0)=f^{\prime}(1)\) (B) \(2 f(0)=1-f^{\prime}(0)\) (C) \(f^{\prime}(1)=1\) (D) \(f(0)+f^{\prime}(0)+f^{\prime \prime}(0)=1\)

Step-by-Step Solution

Answer

(C) is the correct choice: \(f'(1) = 1\).

Step 1: Express the Parabola Equation

Since the parabola has its axis parallel to the y-axis, it has the form \(y = f(x) = ax^2 + bx + c\). This is a standard equation for a parabola oriented vertically.

Step 2: Determine Conditions Based on the Tangency

The line \(y = x\) touches the parabola at \(x=1\). This means at \(x=1\), the parabola \(f(x)\) satisfies both the condition \(f(1) = 1\) and the slope condition \(f'(1) = 1\).

Step 3: Use Condition f(1) = 1

Substitute \(x = 1\) into the parabola equation: \(f(1) = a(1)^2 + b(1) + c = a + b + c = 1\). This gives us the first equation: \(a + b + c = 1\).

Step 4: Use Condition f'(1) = 1

The derivative of the parabola \(f'(x) = 2ax + b\). At \(x = 1\), \(f'(1) = 2a(1) + b = 2a + b = 1\). This provides the second equation: \(2a + b = 1\).

Step 5: Solve the System of Equations

Use the two equations: 1. \(a + b + c = 1\) 2. \(2a + b = 1\)Subtract the second equation from the first: \(a - c = 0\). Hence, \(a = c\).

Step 6: Determine Specific Answers for the Options

Evaluate based on the expressions derived and conditions:- (A) With \(f'(0) = b\) and the above findings, we found \(b = 1 - 2a\), and \(f'(1) = 1\) from step 4, so \(f'(0)\) isn't necessarily \(f'(1)\).- (B) We have \(f(0) = c\), and from (a-c) = 0, \(f(0) + f'(0) = a + (1-2a) = 1-a+f''(0)\). Solving \(2c = 1 - (1-2a)\) implies that the choice might be true with the trial solutions implying 1-2c.- (C) From Step 4, \(f'(1) = 1\) is verified.- (D) For this, \(f(0) + f'(0) + f''(0) = 1-c + b + 2a\), translating the choice condition for cross-checking with initial conditions.Thus, choice (C) is confirmed.

Key Concepts

Derivative

The derivative is a fundamental concept in calculus, representing the rate of change or slope of a function at any given point. For our specific parabola described by the equation \(y = ax^2 + bx + c\), the derivative function \(f'(x)\) is calculated using the rules of differentiation.

• The general derivative for \(ax^2\) is \(2ax\).
• The derivative of \(bx\) is simply \(b\), since the derivative of \(x\) is 1.
• The constant \(c\) has a derivative of 0.

Thus, for the equation \(f(x) = ax^2 + bx + c\), its derivative is \(f'(x) = 2ax + b\). This derivative function helps find the slope of the parabola at any specific x-value. For instance, to find \(f'(1)\), substitute \(x = 1\) into the derivative equation, giving us \(f'(1) = 2a(1) + b\), which simplifies to \(f'(1) = 2a + b\). This result is crucial in determining where a tangent line touches the parabola and is used extensively for understanding tangency conditions.

Equation of a Line

Understanding the equation of a line is key when examining how straight lines interact with curves like parabolas. In the context of our problem, the line in question is \(y = x\), which is a simple diagonal line passing through the origin with a slope of 1.

• A line's equation is typically expressed as \(y = mx + c\), where \(m\) is the slope, and \(c\) is the y-intercept.
• For the line \(y = x\), it's obvious that \(m = 1\) and \(c = 0\), forming a neat, straight line that makes a 45-degree angle with both the x-axis and y-axis.

In this problem, the line \(y = x\) acts as a tangent to the parabola at the point \(x = 1\). This tangency condition implies that at this specific point, the slope of the parabola must equal that of the line, both being 1.

Tangency Condition

The tangency condition is essential for understanding when and where a line touches a curve exactly at one point without crossing it. For a line to be tangent to a parabola, two conditions must be fulfilled:

• The function value of the parabola must equal the line's value at the tangency point. For our problem, this condition is \(f(1) = 1\).
• The slope of the parabola at the tangency point must equal the slope of the line. Hence, \(f'(1) = 1\) matches the slope of line \(y = x\).

These conditions ensure that the line \(y = x\) just "touches" the parabola at \(x = 1\) and does not intersect it any further. When these conditions are applied, they lead to the formation of simultaneous equations that can be solved to fulfill both position and slope requirements at the tangent point. For instance, knowing \(2a + b = 1\) and \(a + b + c = 1\) helps create a system to find the specific coefficients \(a, b,\) and \(c\) that satisfy the equation of tangency.