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

Find the slope of the tangent to the curve $y = x^{3} - x + 1$ at the point where the $x$-coordinate is $2$.

Find the equation of the tangent to the curve $y + \frac{2}{x - 3} = 0$ which has a slope of $2$.

The area of the triangle formed by the coordinate axes and a tangent to the curve $xy = a^2$ at the point $(x_1, y_1)$ on it is

Let $a, b, c \in \mathbb{R}$ be such that $2a + 3b + 6c = 0$ and $g(x)$ be the antiderivative of $f(x) = ax^2 + bx + c$. If the slopes of the tangents drawn to the curve $y = g(x)$ at $(1, g(1))$ and $(2, g(2))$ are equal,then

If $y = 4x - 6$ is a tangent to the curve $y^2 = ax^4 + b$ at $(3, 6)$,then the values of $a$ and $b$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module