Problem 145

Question

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 $y=x^{2}-x+1$

Step-by-Step Solution

Verified

Answer

The values of $x$ for which the curve has a negative gradient are $x < \frac{1}{2}$.

1Step 1: Understanding the problem

We need to determine the values of $x$ for which the curve given by $y=x^2-x+1$ has a negative gradient when it touches the line $y=x$ at the point $(1,1)$.

2Step 2: Confirm the point of tangency

Verify that the quadratic curve $y = x^2 - x + 1$ and the line $y = x$ intersect at the point $(1, 1)$. This requires substituting $x = 1$ into both equations and checking that they yield $y=1$.

3Step 3: Calculate the derivative

Find the derivative of the quadratic curve $y = x^2 - x + 1$ to determine the gradient. The derivative is calculated as $\frac{dy}{dx} = 2x - 1$.

4Step 4: Determine where the gradient is negative

Set the derivative $2x - 1$ less than 0 to find where the gradient is negative: $2x - 1 < 0$. Solve to find $x < \frac{1}{2}$.

Key Concepts

Understanding Derivatives: The BasicsExploring Gradients: Why They MatterPoint of Tangency: Where Two Lines Touch

Understanding Derivatives: The Basics

In calculus, the derivative of a function provides insight into the rate at which that function is changing at any given point. More precisely, the derivative at a particular value of x gives the slope of the tangent line to the graph at that point. This means the change in y for a very small change in x around that point. For the quadratic function given by the equation $ y = x^2 - x + 1 $, its derivative is $ \frac{dy}{dx} = 2x - 1 $. This tells us how steep the curve is at any point x. You differentiate the equation by applying the power rule: bringing the power of x to the front and reducing the power by one.

The derivative $ 2x - 1 $ consists of two terms: the $ 2x $ shows a linear increase, while $ -1 $ shifts the slope downward by one unit across all x.
The line $ y = x $ also has a derivative, which is 1, indicating a constant slope, and it plays a pivotal role in determining where and how curves intersect or touch this line.

By understanding derivatives, you can determine how the function behaves - whether it is increasing (positive derivative), decreasing (negative derivative), or stationary (zero derivative). In our case, it's the negative gradients that interest us, allowing us to find where the curve slopes downward.

Exploring Gradients: Why They Matter

A gradient is essentially the measure of steepness or inclination of a curve at a particular point. When discussing gradients, it's often in terms of the slope of a line, or more generally, a curve. In the context of the function $ y = x^2 - x + 1 $, after finding its derivative $ 2x - 1 $, you can analyze how the gradient changes as x varies:

When $ 2x - 1 = 0 $, the gradient is zero, indicating a flat or horizontal tangent, which usually corresponds to a maximum or minimum point on a curve.
When it is greater than zero ($ 2x - 1 > 0 $), the curve is rising, suggesting increasing function behavior.
When it is less than zero ($ 2x - 1 < 0 $), the curve is falling, indicating decreasing function behavior.

For the problem at hand, determining when the gradient is negative tells us where on the x-axis the curve is sloping downward to the left. Solving $ 2x - 1 < 0 $, we find this "negative gradient region" occurs when $ x < \frac{1}{2} $. This means any x-value less than this will have the curve descending.

Point of Tangency: Where Two Lines Touch

The point of tangency is a critical concept, especially when dealing with curves and lines. It describes a single point where a curve and a line barely meet with the same slope, meaning the line touches the curve without crossing it at that point. In our task looking at the curve $ y = x^2 - x + 1 $, and the line $ y = x $, this point of tangency occurs at $ (1, 1) $. At this point:

The y-values from both the line and the curve are equal (both are 1 when $ x = 1 $).
The derivatives (gradients) are also equal. The curve's derivative at $ x = 1 $ is equal to the line's constant gradient of 1.

Thus, the algebra confirms that at the point $ (1, 1) $, not only do the two equations share an x and y value, they are also sharing the same slope or gradient. This equality in gradients is what signifies a true point of tangency. Both visualizing and calculating these touching points help us understand how curves and lines relate and intersect in coordinate geometry.

Problem 143

Problem 146

Other exercises in this chapter

Problem 141

$\frac{\sin \alpha-\sin \beta}{\cos \beta-\cos \alpha}=F(\alpha)$, where \(0

View solution

Problem 143

$$ \begin{array}{ll} \text { Column-I } & \text { Column-II } \\ \hline \text { I. Let } f(x)=\left(1+b^{2}\right) x^{2}+2 b x+1 & \text { (A) }(0,1] \\ \text {

View solution

Problem 146

Assertion: The function $f(x)=\frac{\sin x}{x}$ is decreasing in the interval $\left(0, \frac{\pi}{2}\right)$ Reason: $\tan x>x$ for \(0

View solution

Problem 149

Assertion: If $0\frac{\alpha}{\beta}$ Reason: $x \tan x$ is increasing for \(0

View solution