If the tangent to the curve y=x+sin y at a po | vedclass

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

Question

(C) The slope of the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$ is $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \f

Vedclass · Accepted Answer

$|b-a|=1$

Vedclass · Answer

(C) The slope of the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$ is $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$. Since the tangent at $(a, b)$ is parallel to this line,the slope of the tangent $\left. \frac{dy}{dx} \right|_{(a, b)} = 1$. Differentiating the curve $y = x + \sin y$ with respect to $x$,we get $\frac{dy}{dx} = 1 + \cos y \cdot \frac{dy}{dx}$. Substituting the point $(a, b)$ and the slope $1$,we have $1 = 1 + \cos b \cdot (1)$. This implies $\cos b = 0$,which means $\sin b = \pm 1$. Since $(a, b)$ lies on the curve,$b = a + \sin b$. Therefore,$b - a = \sin b$. Taking the absolute value on both sides,$|b - a| = |\sin b| = 1$.

If the tangent to the curve $y=x+\sin y$ at a point $(a, b)$ is parallel to the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right),$ then

Similar Questions

The equation of the normal to the curve $xy + 7 = 0$ is $Ax + By + C = 0$. Which of the following is true?

If the two curves $x=y^2$ and $xy=k$ cut each other at right angles, then a possible value of $8k^2$ is

The angle between the tangent lines to the graph of the function $f(x) = \int_{2}^{x} (2t - 5) \, dt$ at the points where the graph cuts the $x$-axis is

The equation of the normal to the curve $y=x \log x$ parallel to $2x-2y+3=0$ is

At which point does the line $\frac{x}{a} + \frac{y}{b} = 1$ touch the curve $y = be^{-x/a}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module