VITEEE 2017 :: Mathematics :: General questions

• Mathematics - General questions

The solution of the differential equation $\log_{}{}x\frac{dy}{dx}+\frac{y}{x} = \sin2x is$

Answer:

Explanation:

$\frac{dy}{dx}+\frac{y}{x\log_{}{}x} =\frac{\sin2x}{\log_{}{}x}$

I.F. = $e^{\int_{}^{}\frac{dx}{x\log_{}{}x} }$

.'. I.F. = $e^{\int_{}^{}\frac{1}{t}dt}$ = $e^{\log_{}{}t}$ = t = log |x|

solution is given by

y(I.F.) = $\int_{}^{}Q(I.F.)dx + C$

y log |x| = $\int_{}^{}\frac{sin 2x}{log |x|}(log|x|)dx +C = -\frac{Cos 2x}{2}+C $

The value of x obtained from the equation

$\begin{vmatrix}x+\alpha & \beta & \gamma \\\gamma & x+\beta & \alpha \\\alpha & \beta& x+\gamma \end{vmatrix} =0$

will be

Answer:

Explanation:

Given $\begin{vmatrix}x+\alpha & \beta & \gamma \\\gamma & x+\beta & \alpha \\\alpha & \beta& x+\gamma \end{vmatrix} =0$

Operate C₁ → C₁ + C₂ + C₃

$\begin{vmatrix}x+\alpha+\beta+\gamma & \beta & \gamma \\x+\alpha+\beta+\gamma & x+\beta & \alpha \\x+\alpha+\beta+\gamma & \beta& x+\gamma \end{vmatrix} =0$

$(x+\alpha+\beta+\gamma )\begin{vmatrix}1& \beta & \gamma \\1 & x+\beta & \alpha \\1 & \beta& x+\gamma \end{vmatrix} =0$

x = -($\alpha+\beta+\gamma$)

Again if 

$\begin{vmatrix}1& \beta & \gamma \\1 & x+\beta & \alpha \\1 & \beta& \gamma \end{vmatrix} =0$→

$\begin{vmatrix}1& \beta & \gamma \\0 & x & \alpha-\gamma \\0 & 0& x \end{vmatrix}$=0

x² = 0 → x = 0

.'. Solutions of the equation x = 0, -($\alpha+\beta+\gamma$)

The area bounded by y - 1 = lxl, y = 0 and  |x| = 1/2 will be

Answer:

Explanation:

The given lines are,

y - 1 = x, x ≥ 0;

y - 1 = -x, x<0

y = 0, x = -1/2, x < 0; x = 1/2, x ≥ 0

so that the area bounded is as shown in the figure.

Required area = 2$\int_{0}^{1/2} (1+x).dx$ = 2$\left(x+\frac{x^{2}}{2}\right)_0^\frac{1}{2}$

= 2 (1/2 + 1/8) = 5/4

The range of the function f (x) = $\frac{1}{2-cos3x}$ is

Answer:

Explanation:

We have - 1≤ cos 3x≤ 1 = - 1 ≤-cos 3x≤ 1

Add '2' on both sides

1≤ 2 - cos 3x ≤ 3

1 ≥ $\frac{1}{2-cos3x}$ ≥ 1/3

Value of $\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cos x}}dx$ is

Answer:

Explanation:

Let $I =\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cos x}}dx$ ...(1)

Then $I =\int_{0}^{\pi/2} \frac{\sqrt{sin(\pi/2-x)}}{\sqrt{sin(\pi/2-x)}+\sqrt{cos (\pi/2-x)}}dx$

$I =\int_{0}^{\pi/2} \frac{\sqrt{cos x}}{\sqrt{cos x}+\sqrt{sin x}}dx$...(2)

Adding (1) and (2) we get

2I = $\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{cos x}+\sqrt{sinx}}dx$ + $\int_{0}^{\pi/2} \frac{\sqrt{cos x}}{\sqrt{sin x}+\sqrt{cosx}}dx$

= $\int_{0}^{\pi/2} \frac{\sqrt{cos x}+\sqrt{sin x}}{\sqrt{cos x}+\sqrt{sin x}}dx$

= $\int_{0}^{\pi/2}1.dx$=$[x]_0^\left(\pi/2\right)$ = $\pi/2-0$

I = $\pi/4$ = $\int_{0}^{\pi/2} \frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cos x}}dx$ = $\pi/4$

• Mathematics - General questions