MHT CET 2021 20th September Morning Shift | Application of Derivatives Question 72 | Mathematics

A particle is moving on a straight line. The distance $$\mathrm{S}$$ travelled in time $$\mathrm{t}$$ is given by $$\mathrm{S=a t^2+b t+6}$$. If the particle comes to rest after 4 seconds at a distance of $$16 \mathrm{~m}$$. from the starting point, then the acceleration of the particle is.

$$\frac{-3}{4} \mathrm{~m} / \mathrm{sec}^2$$

$$\frac{-1}{2} \mathrm{~m} / \mathrm{sec}^2$$

$$-1 \mathrm{~m} / \mathrm{sec}^2$$

$$\frac{-5}{4} \mathrm{~m} / \mathrm{sec}^2$$

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

The equation of the tangent to the curve $$y = 4x{e^x}$$ at $$\left( { - 1,{{ - 4} \over e}} \right)$$ is

$$6x - {e \over 4}y = - 5$$

$$x - {e \over 4}y = 0$$

$$x = - 1$$

$$y = {{ - 4} \over e}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

If $$f(x)=\log (\sin x), x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right]$$, then value of '$$c$$' by applying LMVT is

$$\frac{\pi}{2}$$

$$\frac{2 \pi}{3}$$

$$\frac{3 \pi}{4}$$

$$\frac{\pi}{4}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

The equation of tangent at $$P(-4,-4)$$ on the curve $$x^2=-4 y$$ is

$$2 x-y+4=0$$

$$2 x+y-4=0$$

$$3 x-y+8=0$$

$$2 x+y+4=0$$

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