The intercepts on the x-axis made by tangents | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the curve $y = \int_{0}^{x} |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $\frac{dy}{dx} = |x|$.Since the tang

Vedclass · Accepted Answer

$\pm 1$

Vedclass · Answer

(A) Given the curve $y = \int_{0}^{x} |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $\frac{dy}{dx} = |x|$.Since the tangent is parallel to the line $y = 2x$,its slope must be $2$. Thus,$|x| = 2$,which gives $x = 2$ or $x = -2$.For $x = 2$,$y = \int_{0}^{2} |t| dt = \int_{0}^{2} t dt = [\frac{t^2}{2}]_{0}^{2} = 2$. The point is $(2, 2)$.The equation of the tangent at $(2, 2)$ is $y - 2 = 2(x - 2) \Rightarrow y = 2x - 2$. The $x$-intercept is found by setting $y = 0$,giving $2x = 2$,so $x = 1$.For $x = -2$,$y = \int_{0}^{-2} |t| dt = \int_{0}^{-2} (-t) dt = [-\frac{t^2}{2}]_{0}^{-2} = -2$. The point is $(-2, -2)$.The equation of the tangent at $(-2, -2)$ is $y - (-2) = 2(x - (-2)) \Rightarrow y + 2 = 2x + 4 \Rightarrow y = 2x + 2$. The $x$-intercept is found by setting $y = 0$,giving $2x = -2$,so $x = -1$.Thus,the $x$-intercepts are $\pm 1$.

The intercepts on the $x$-axis made by tangents to the curve $y = \int_{0}^{x} |t| dt, x \in R$ which are parallel to the line $y = 2x$ are equal to:

Similar Questions

If the curves $y = \frac{\ln x}{x}$ and $y = \lambda x^2$ (where $\lambda$ is a constant) touch each other,then $\lambda$ is

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where the gradient is $3$. Then the values of $a, b, c$ are

$x_1, x_2 \in N$. If a line having slope $2$ is a tangent to the curve $y=x^4-6x^3+13x^2-10x+5$ at points $P(x_1, y_1)$ and $Q(x_2, y_2)$,then $x_1x_2+y_1y_2=$

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to:

The triangle formed by the tangent to the curve $f(x) = x^2 + bx - b$ at the point $(1, 1)$ and the coordinate axes lies in the first quadrant. If its area is $2$,then the value of $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module