VITEEE 2019 :: Mathematics :: General questions

• Mathematics - General questions

If sin y = x sin(a+y), then $\frac{dy}{dx}$ is equal to

Answer:

Explanation:

Given, sin y = x sin (a + y)

$x = \frac{\sin y}{\sin(a+y)}$

$\frac{dx}{dy} = \frac{d}{dy}\left[\frac{\sin y}{\sin(a+y)}\right]$

 = $\frac{\sin(a+y)\cos y - \sin y\cos(a+y)}{\sin^{2}(a+y)}$

= $\frac{\sin a}{\sin^{2}(a+y)}$

$\frac{dy}{dx}= \frac{\sin^{2}(a+y)}{\sin a}$

The equation $y^{2}+3=2(2x+y)$ represents a parabola with the vertex at

Answer:

Explanation:

The given equation can be rewritten as

$(y-1)^{2}=4\left(x - \frac{1}{2}\right)$ which is a parabola with its vertex (1/2,1)  axis along the line y = 1, hence axis parallel to x-axis. Its focus is (3/2,1).

The solution of sin x = $-\frac{\sqrt{3}}{2}$ is

Answer:

Explanation:

We have, sin x = $-\frac{\sqrt{3}}{2}$

= -sin$\frac{\pi}{3}$

Hence, sin x = $\frac{4\pi}{3}$  which gives

$x = n\pi+ (-1)^{n}\frac{4\pi}{3}, n \epsilon Z$

 If a, b, c are in A. P, then the value of $\begin{vmatrix}x+1& x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}$ is?

Answer:

Explanation:

Given a, b, c are in A.P.

.'. 2b = a+c

Now,$\begin{vmatrix}x+1& x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}$

[Applying R₂ → 2R₂]

= $\frac{1}{2}\begin{vmatrix}x+1& x+2 & x+a \\ 2x+4 & 2x+6 & 2x+2b \\ x+3 & x+4 & x+c \end{vmatrix}$

= $\frac{1}{2}\begin{vmatrix}x+1& x+2 & x+a \\ 2x+4 & 2x+6 & 2x+(a+c) \\ x+3 & x+4 & x+c \end{vmatrix}$

[Using equation(i)]

=$\frac{1}{2}\begin{vmatrix}x+1& x+2 & x+a \\ 0 & 0 & 0 \\ x+3 & x+4 & x+c \end{vmatrix}$

=$\frac{1}{2}.0$ =0

[Applying R₂ → R₂ - (R₁+R₃)]

The radius of a right circular cylinder increases at the rate of 0.1 cm/min, and the height decreases at the rate of 0.2 cm/min. The rate of change of the volume of the cylinder, in cm³/min, when the radius is 2 cm and the height is 3 cm is

Answer:

Explanation:

$V=\pi r^{2}h$

Differentiating both sides, we get

$\frac{dV}{dt}=\pi\left(r^{2}\frac{dh}{dt}+2r\frac{dr}{dt}h\right)$

= $\pi r\left(r\frac{dh}{dt}+2h\frac{dr}{dt}\right)$

$\frac{dr}{dt}=\frac{1}{10}and\frac{dh}{dt}=-\frac{2}{10}$

$\frac{dV}{dt}=\pi r\left(r\left(\frac{-2}{10}+2h\frac{1}{10}\right)\right)$

= $\frac{\pi\ r}{5}\left(-r+h\right)$

Thus, when r = 2 and h = 3,

 $\frac{dV}{dt}=\frac{\pi (2)}{5}(-2+3)=\frac{2\pi}{5}$

• Mathematics - General questions