VITEEE 2018 :: Mathematics :: General questions

• Mathematics - General questions

$f(x)= \begin{cases} x \sin1/x& x \neq 0\\0& x = 0\end{cases} at x = 0 is$

Answer:

Explanation:

For function to be continuous :

f(0+h) = f(0-h) = f(0)

f(0 + h) = $\lim_{h \rightarrow 0}h\sin1/h$ = 0 x (a finite quantity) = 0

f (0-h) = $\lim_{h \rightarrow 0}-h\sin1/-h$ = 0 x (a finite quantity) = 0
 

$\lim_{x \rightarrow 0}x\sin1/x$ = 0 x (a finite quantity) = 0

function is continuous at x = 0

For function to be differentiable :

f' (0 + h) = f'(0 - h)

$\frac{f(0+h)-f(0)}{h}$ = $\lim_{h \rightarrow 0}\frac{h \sin1/h - 0}{h}$

= $\lim_{h \rightarrow 0}\sin\left(\frac{ 1}{h}\right)$

which does not exist

The equation of the hyperbola with vertices (3 , 0), (-3 , 0) and semi-latus rectum 4 is given by :

Answer:

Explanation:

We have a = 3 and $\frac{b^{2}}{a}$ = 4

→ b² = 12

Hence, the equation of the hyperbola is

$\frac{x^{2}}{9}-\frac{y^{2}}{12}=1$

$4x^{2}-3y^{2}= 36$ = $4x^{2}-3y^{2}-36=0$

The eccentricity of the ellipse whose major axis is three times the minor axis is:

Answer:

Explanation:

Let a be the major axis and b, the minor axis of the ellipse, then 3 minor axis = major axis

→ 3b = a

Eccentricity is given by : b² = a²(1 - e²)

→  b² = 9b² (1 - e²)

$\frac{1}{9}= (1-e^{2})\Rightarrow e^{2}=\frac{8}{9}\Rightarrow e= \frac{2\sqrt{2}}{3}$

The interval in which the function f(x) = $\frac{4x^{2}+1}{x}$ is decreasing is :

Answer:

Explanation:

Given f(x) = $\frac{4x^{2}+1}{x}$

Thus f'(x) = 4 - $\frac{1}{x^{2}}$

f(x) will be decreasing if f '(x) < 0

Thus 4 - $\frac{1}{x^{2}}$ < 0

→ $\frac{1}{x^{2}}$ > 4 → $-\frac{1}{2}<x<\frac{1}{2}$

Thus interval in which f(x) is decreasing, is $\left(-\frac{1}{2},\frac{1}{2}\right)$

A football is inflated by pumping air in it. When it acquires spherical shape its radius increases at the rate of 0.02 cm/s. The rate of increase of its volume when the radius is 10 cm is ___ π cm/s

Answer:

Explanation:

V = $\frac{4}{3}\pi r^{3}$

$\frac{dv}{dt}=\frac{4}{3}\pi\frac{d}{dt}r^{3}$

= $\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}$ = $4\pi r^{2}\frac{dr}{dt}$

When r = 10 cm;

$\frac{dv}{dt}=4\pi (10)^{2}.(0.02) = 8\pi cm^{3}/s$.

• Mathematics - General questions