Problem 110
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
Verified Answer
The correct answer is (B) \(2 f(0)+f^{\prime}(0)=1\).
1Step 1: Understand the Touching Point Condition
The parabola touches the line \(y=x\) at the point \((1,1)\). This means \(f(1) = 1\) and the derivative \(f'(1) = 1\) as the slope of the tangent to a curve at the point of contact must equal the slope of the line.
2Step 2: Represent the Parabola Equation
A parabola with its axis parallel to the \(y\)-axis can be represented as \(y = ax^2 + bx + c\). This implies \(f(x) = ax^2 + bx + c\).
3Step 3: Use the Touching Point to Find Conditions
Since \(f(1) = 1\), substituting \(x=1\) gives us: \ \(a(1)^2 + b(1) + c = 1\) \ which simplifies to: \(a + b + c = 1\).
4Step 4: Differentiate and Apply the Slope Condition
Differentiate \(f(x)\) to get \(f'(x) = 2ax + b\). Substitute \(x=1\) to apply the condition \(f'(1) = 1\): \ \(2a(1) + b = 1\) \ which simplifies to: \(2a + b = 1\).
5Step 5: Solve for \(b\) and \(c\) in terms of \(a\)
From the equation \(2a + b = 1\), solve for \(b\): \ \(b = 1 - 2a\). \ Substitute into \(a + b + c = 1\): \ \(a + (1-2a) + c = 1\) \ which simplifies to: \ \(-a + c = 0\) or \(c = a\).
6Step 6: Use Derived Conditions to Find the Expression at \(x=0\)
Plugging \(x=0\) into \(f(x) = ax^2 + (1-2a)x + a\), gives \(f(0) = a\). \ The derivative \(f'(x) = 2ax + 1 - 2a\) evaluated at \(x=0\) gives \(f'(0) = 1 - 2a\).
7Step 7: Identify the Correct Option
Using \(f(0) = a\) and \(f'(0) = 1 - 2a\), test the options. \ Substitute into the expressions: \ (A) \(2(1-2a) + a = 1\) which simplifies to \(2 - 4a + a = 1\) \ (B) \(2a + (1-2a) = 1\) which simplifies to \(1 = 1\), a true statement. \ Therefore, option \(B\) matches the conditions.
Key Concepts
Parabola EquationDerivativeTangent Line ConditionSlope of Tangent
Parabola Equation
A parabola is a symmetrical curve, and when its axis is parallel to the y-axis, it can be expressed in a simple quadratic form. The general expression for such a parabola is \(y = ax^2 + bx + c\). Here, \(a\), \(b\), and \(c\) are constants that determine the shape and position of the parabola on the graph.
- The value of \(a\) affects the "width" and "direction" of the parabola. A positive \(a\) makes the parabola open upwards, while a negative one opens it downwards.
- Coefficient \(b\) influences the horizontal placement and tilt, and \(c\) sets the vertical position.
Derivative
The derivative of a function is a fundamental concept in calculus that measures change or the rate of change of the function’s output with respect to change in its input. When you take the derivative of a parabola, \(f(x) = ax^2 + bx + c\), you get \(f'(x) = 2ax + b\). This expression shows how the slope of the curve changes at different points along the curve.
- The first term \(2ax\) changes linearly with \(x\), indicating that the slope becomes steeper or more shallow as you move along the x-axis.
- The constant term \(b\) in the derivative represents the initial slope when \(x = 0\).
Tangent Line Condition
A tangent line to a curve at a specific point touches the curve without crossing it. It provides a "snapshot" of the curve's behavior at that point. For a curve represented by \(y = f(x)\), the line \(y = mx + c\) will be tangent if it has the same slope as the curve at that point.
- The touching condition at a point \( (x_0, y_0) \) means \(f(x_0) = y_0\) and \(f'(x_0) = m\), where \(m\) is the slope of the line.
- This is because the slope of the tangent, \(m\), is derived from the rate of change of the curve \(f'(x_0)\).
Slope of Tangent
The slope of the tangent line is a key consideration when analyzing curves in calculus. It is determined using the derivative of the function at the given point. For the parabola \(f(x) = ax^2 + bx + c\), its derivative \(f'(x) = 2ax + b\) tells us the slope of the tangent line at any point \(x\).
- If the problem specifies a tangent at a specific point, you will substitute that \(x\)-value into \(f'(x)\).
- A larger absolute value of the slope indicates a steeper line.
Other exercises in this chapter
Problem 108
The curves \(x^{2}-4 y^{2}+c=0\) and \(y^{2}=4 x\) will cut orthogonally for (A) \(c \in(0,16)\) (B) \(c \in(-3,4)\) (C) \(c \in(3,4)\) (D) None of these
View solution Problem 109
Which of the following is not true? The function \(f(x)=x^{2}+\frac{\lambda}{x}\) has a (A) minimum at \(x=2\) if \(\lambda=16\) (B) maximum at \(x=2\) if \(\la
View solution Problem 111
The angle between the tangents at any point \(P\) and the line joining \(P\) to the origin \(O\), where \(P\) is a point on the curve \(\ln \left(x^{2}+y^{2}\ri
View solution Problem 112
If the equation \(a x^{2}+b x+c=0\) has two distinct positive roots, then the equation \(a x^{2}+(b+6 a) x+\) \((c+3 b)=0\) has (A) two positive roots (B) exact
View solution