The slope of the graph as shown in the figure | vedclass

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

Question

(B) The slope of a curve at any point is given by the tangent of the angle $	heta$ that the tangent line at that point makes with the positive $x$-ax

Vedclass · Accepted Answer

$m_1 < m_2 < m_3$

Vedclass · Answer

(B) The slope of a curve at any point is given by the tangent of the angle $	heta$ that the tangent line at that point makes with the positive $x$-axis,i.e.,$m = 	an 	heta$.As we move along the curve from point $1$ to point $3$,the tangent line becomes steeper.This means the angle $	heta$ that the tangent makes with the positive $x$-axis increases as we move from point $1$ to point $3$ (i.e.,$	heta_1 Since the function $	an 	heta$ is an increasing function for $0^\circ Therefore,$m_1 < m_2 < m_3$.

The slope of the graph as shown in the figure at points $1, 2$ and $3$ is $m_1, m_2$ and $m_3$ respectively. Then:

Similar Questions

The greatest value of the function $-5\, \sin\, \theta + 12\, \cos\, \theta$ is

$\int\limits_0^{\frac{\pi }{4}} {\sin 2x} \,dx$ is equal to

As $\theta$ increases from $0^{\circ}$ to $90^{\circ}$,the value of $\cos \theta$ :-

$A$ particular straight line passes through the origin and a point whose abscissa is double the ordinate of the point. The equation of such a straight line is:

If $x^3+3xy+y^3=1$,then the correct option$(s)$ is/are:-
$(A) \left(\frac{dy}{dx}\right)_{(1,1)}=-1$
$(B) \left(\frac{dy}{dx}\right)_{(1,1)}=-2$
$(C) \left(\frac{dy}{dx}\right)_{(1,0)}=-1$
$(D) \left(\frac{dy}{dx}\right)_{(1,0)}=-3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module