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 tangent

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 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 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

The extreme values of $4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$ over $\mathbb{R}$ are

The $P-P-P$ bond angle in white phosphorus is

Vegetative propagation in $Pistia$ occurs by:

If the mean and variance of a binomial variate $X$ are $8$ and $4$ respectively,then $P(X < 3)$ equals to

$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module