Application of Derivatives · Mathematics · WB JEE

A particle is projected vertically upwards and is at a height h after t₁ seconds and again after t₂ seconds then

The equation of the tangent to the conic $${x^2} - {y^2} - 8x + 2y + 11 = 0$$ at (2, 1) is

A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by x = t $$-$$ 6t² + t³. Its acceleration will be zero at

The Rolle's theorem is applicable in the interval $$-$$1 $$\le$$ x $$\le$$ 1 for the function

The distance covered by a particle in t seconds is given by x = 3 + 8t $$-$$ 4t². After 1 second its velocity will be

If the rate of increase of the radius of a circle is 5 cm/sec., then the rate of increase of its area, when the radius is 20 cm, will be

Angle between y² = x and x² = y at the origin is

If the normal to the curve y = f(x) at the point (3, 4) makes an angle 3$$\pi$$/4 with the positive x-axis, then f'(3) is

The equation of normal of $${x^2} + {y^2} - 2x + 4y - 5 = 0$$ at (2, 1) is

The point in the interval [0, 2$$\pi$$], where $$f(x) = {e^x}\sin x$$ has maximum slope, is

The co-ordinates of the point on the curve $$y = {x^2} - 3x + 2$$ where the tangent is perpendicular to the straight line y = x are

The acceleration of a particle starting from rest moving in a straight line with uniform acceleration is 8m/sec². The time taken by the particle to move the second metre is

The function $f(x)=2 x^3-3 x^2-12 x+4, x \in \mathbb{R}$ has

Let $\phi(x)=f(x)+f(2 a-x), x \in[0,2 a]$ and $f^{\prime \prime}(x)>0$ for all $x \in[0, a]$. Then $\phi(x)$ is

Let $f$ be a function which is differentiable for all real $x$. If $f(2)=-4$ and $f^{\prime}(x) \geq 6$ for all $x \in[2,4]$, then

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x,(x \neq 0)$, then

Let $f(\theta)=\left|\begin{array}{ccc}1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|$. Suppose $A$ and $B$ are respectively maximum and minimum values of $f(\theta)$.Then $(A,B)$ is equal to

The maximum number of common normals of $y^2=4 a x$ and $x^2=4 b y$ is equal to

$$f(x)=\cos x-1+\frac{x^2}{2!}, x \in \mathbb{R}$$ Then $$\mathrm{f}(x)$$ is

Let $$\mathrm{y}=\mathrm{f}(x)$$ be any curve on the $$\mathrm{X}-\mathrm{Y}$$ plane & $$\mathrm{P}$$ be a point on the curve. Let $$\mathrm{C}$$ be a fixed point not on the curve. The length $$\mathrm{PC}$$ is either a maximum or a minimum, then

If a particle moves in a straight line according to the law $$x=a \sin (\sqrt{\lambda} t+b)$$, then the particle will come to rest at two points whose distance is [symbols have their usual meaning]

Let $$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$$ be given by $$\mathrm{f}(x)=\left|x^2-1\right|$$, then

Consider the function $$\mathrm{f}(x)=x(x-1)(x-2) \ldots(x-100)$$. Which one of the following is correct?

A missile is fired from the ground level rises x meters vertically upwards in t sec, where $$x = 100t - {{25} \over 2}{t^2}$$. The maximum height reached is

The portion of the tangent to the curve $${x^{{2 \over 3}}} + {y^{{2 \over 3}}} = {a^{{2 \over 3}}},a > 0$$ at any point of it, intercepted between the axes

Given $$f(x) = {e^{\sin x}} + {e^{\cos x}}$$. The global maximum value of $$f(x)$$

A particle moving in a straight line starts from rest and the acceleration at any time t is $$a - k{t^2}$$ where a and k are positive constants. The maximum velocity attained by the particle is

A train moving with constant acceleration takes t seconds to pass a certain fixed point and the front and back end of the train pass the fixed point with velocities u and v respectively. Show that the length of the train is $${1 \over 2}(u + v)t$$.

Prove that the centre of the smallest circle passing through origin and whose centre lies on y = x + 1 is $$\left( { - {1 \over 2},{1 \over 2}} \right)$$.

If the area of a rectangle is 64 square units, find the minimum value possible for its perimeter.

Let $f(x)=x^3, x \in[-1,1]$. Then which of the following are correct?

The acceleration f $$\mathrm{ft} / \mathrm{sec}^2$$ of a particle after a time $$\mathrm{t}$$ sec starting from rest is given by $$\mathrm{f}=6-\sqrt{1.2 \mathrm{t}}$$. Then the maximum velocity $$\mathrm{v}$$ and time $$\mathrm{T}$$ to attend this velocity are

A balloon starting from rest is ascending from ground with uniform acceleration of 4 ft/sec$$^2$$. At the end of 5 sec, a stone is dropped from it. If T be the time to reach the stone to the ground and H be the height of the balloon when the stone reaches the ground, then

If $$f(x) = 3\root 3 \of {{x^2}} - {x^2}$$, then

From a balloon rising vertically with uniform velocity v ft/sec a piece of stone is let go. The height of the balloon above the ground when the stone reaches the ground after 4 sec is [g = 32 ft/sec²]