VITEEE 2019 :: Mathematics :: General questions

• Mathematics - General questions

The integrating factor of $\frac{xdy}{dx}-y = x^{4}-3x $ is

Answer:

Explanation:

Since $\frac{xdy}{dx}-y = x^{4}-3x $ 

.'. $\frac{dy}{dx}-\frac{y}{x} = x^{3}-3 $

Hence 

IF = $e^{\int_{}^{}pdx }$ = $e^{-\int_{}^{}\frac{1}{x}dx }$

= $e^{-\log_{}{x} }$ = 1/x

The solution of the differential equation $\left\{1+x\sqrt{(x^{2}+y^{2})}\right\}dx + \left\{\sqrt{(x^{2}+y^{2})}-1\right\}ydy = 0 is ?

Answer:

Explanation:

Rearranging the equation, we have

dx - ydy + $\sqrt{(x^{2}+y^{2})}(xdx+ydy) = 0$

$dx - ydy+\frac{1}{2}\sqrt{(x^{2}+y^{2})}d(x^{2}+y^{2}) = 0$

On integrating, we get

$x-\frac{y^{2}}{2}+\frac{1}{2}\int_{}^{}\sqrt{t}dt = C$

$\left\{t =\sqrt{x^{2}+y^{2}}\right\}$

or $x-\frac{y^{2}}{2}+\frac{1}{3}(x^{2}+y^{2})^{3/2}$=C

If f(x) = $x^{3}+bx^{2}+cx+d$ and 0<b²<c, then in (-∞, ∞)

Answer:

Explanation:

f(x) = $x^{3}+bx^{2}+cx+d$ and 0<b²<c

.'. f'(x) = $3x^{2}+2bx+c$ 

Discriminant  = 4b² -12c = 4(b² - 3c)<0

f'(x) >0 $\forall$ xΕ R

Thus,f(x) is strictly increasing $\forall$ x Ε R

Let f be the function defined by

f(x) = $\begin{cases}\frac{x^{2}-1}{x^{2}-2|x-1|-1} & x \neq 1\\1/2 & x = 1\end{cases}$

Answer:

Explanation:

For x< 1, f(x) = $\frac{x^{2}-1}{x^{2}+2x-3}$

= $\frac{x+1}{x+3}$

.'. $\lim_{x \rightarrow 1^{-}}$f(x) = 1/2

For x>1, f(x) = $\frac{x^{2}-1}{x^{2}-2x+1} =\frac{x+1}{x-1}$

.'. $\lim_{x \rightarrow 1^{+}}$ f(x) = ∞ 

.'. the function is not continuous at x= 1

Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is

Answer:

Explanation:

Required number of ways

= $\frac{6!}{3!3!}$ = $\frac{720}{6\times6}$

= 20

• Mathematics - General questions