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Chapter 14: Problem 110

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) \(2 f^{\prime}(0)-f(0)=1\)

Short Answer

Expert verified
The correct answer is (B) \(2 f(0)+f^{\prime}(0)=1\).

Step by step solution

01

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.
02

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 \(f(x) = ax^2 + bx + c\).
03

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: \(a + b + c = 1\).
04

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: \(2a + b = 1\).
05

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 \(c = a\).
06

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 \(f'(0) = 1 - 2a\).
07

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
Understanding this equation is crucial when working with parabolas, as it allows us to explore other interesting properties, such as the axis of symmetry and vertex location, and solve a variety of graph-related problems.
Derivative
The derivative of a function is a fundamental concept in calculus that measures change or the rate of change of the function鈥檚 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 \(x = 0\).
The derivative provides instant insights into the original function鈥檚 behavior, especially critical points where \(f'(x) = 0\) denote minima, maxima, or points of inflection.
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 \(f'(x_0)\).
Understanding the tangent condition is critical for many calculus problems, as it relates to optimizing functions and understanding instantaneous rates of change.
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.
Grasping slopes of tangents helps in visualizing and solving real-world problems where the notion of rate of change is important, such as physics for velocity and acceleration. It is directly tied to the underlying geometry of the curve, providing a link between algebraic expressions and graphical depictions.

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Most popular questions from this chapter

If \(\alpha\) and \(\beta(\alpha<\beta)\) be two different real roots of the equation \(a x^{2}+b x+c=0\), then (A) \(\alpha>-\frac{b}{2 a}\) (B) \(\beta<-\frac{b}{2 a}\) (C) \(\alpha<-\frac{b}{2 a}<\beta\) (D) \(\beta<-\frac{b}{2 a}<\alpha\)

The range of values of \(a\) so that the equation \(x^{3}-3 x+\) \(a=0\) has three real and distinct roots is (A) \((-\infty,-2) \cup(2, \infty)\) (B) \((-2,0)\) (C) \((-2,0)\) (D) \((-2,2)\)

Assertion: If a quadratic curve touches the line \(y=x\) at the point \((1,1)\), then the values of \(x\) for which the curve has a negative gradient are \(x<\frac{1}{2}\) Reason: The equation of the curve is \(y=x^{2}-x+1\)

How many real solutions does the equation \(x^{7}+14 x^{5}\) \(+16 x^{3}+30 x-560=0\) have? \(\quad\) [2008] (A) 7 (B) 1 (C) 3 (D) 5

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