VITEEE 2019 :: Mathematics :: General questions

• Mathematics - General questions

The domain of the function f(x) = $\sqrt{\frac{1}{|x-2|-(x-2)}}$ is:

Answer:

Explanation:

|x-2| = $\begin{cases}x-2 & x \geq 2\\2-x & x < 2\end{cases}$

|x-2|-(x-2) = $\begin{cases}0 & x \geq 2\\4-2x & x < 2\end{cases}$

given expression is defined for (-∞ ,2)

Value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\cot x}}dx $ is

Answer:

Explanation:

I = $ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\cot x}}dx$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ...(i)

Then I=

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}}dx $ 

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\cos x)}}{\sqrt{\cos x)}+\sqrt{\sin x)}}dx$ ...(ii)

Adding (i) and (ii) we get

2I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}+\sqrt{\cos x)}}{\sqrt{\cos x)}+\sqrt{\sin x)}}dx$

$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}1.dx =[x]_\frac{\pi}{6}^\frac{\pi}{3}$

= $\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$ 

I = $\frac{\pi}{12}$

The equation of the chord of the hyperbola 25x² - 16y2 = 400, that is bisected at point (5,3) is:

Answer:

Explanation:

The equation of the chord, having mid-point  as (x_1, y₁), of the hyperbola

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$ is given by T = S₁ ....(i)

Where, T = $\frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}-1$

and S₁ = $\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}-1$

According to the question,  (x_1, y₁) = (5, 3) and a² = 16 b² = 25

and 25x² - 16y² = 400,

$\frac{x^{2}}{16}-\frac{y^{2}}{25} =1$

$\frac{5x}{16}-\frac{3y}{25} =\frac{25}{16}-\frac{9}{25}$

                                                             [using (i)]

125x - 48y = 625 - 144 = 125x - 48y = 481

If $\int_{}^{}\frac{\sin x}{\sin (x - \alpha)}dx = Ax + B\log_{}{\sin (x - \alpha)}+C$ then value of (A,B) is

Answer:

Explanation:

$\int_{}^{}\frac{\sin x}{\sin (x - \alpha)}dx$

= $\int_{}^{}\frac{\sin (x-\alpha+\alpha)}{\sin (x - \alpha)}dx$

=$\int_{}^{}\frac{\sin (x-\alpha)\cos\alpha+\cos (x-\alpha)\sin\alpha}{\sin (x - \alpha)}dx$

=$\int_{}^{}\left\{\cos\alpha+\sin\alpha\cot(x - \alpha)\right\}dx$

= $(\cos\alpha)x+(\sin\alpha)\log_{}{\sin(x-\alpha)}+C$

A = $\cos\alpha$ B = $\sin\alpha$

A wire 34 cm long is to be bent in the form of a quadrilateral of which each angle is 90°. What is the maximum area which can be enclosed inside the quadrilateral?

Answer:

Explanation:

Let one side of quadrilateral be x and another side be Y so, 2(x + Y) = 34 or,

(x+y) = 17   .....(i)

We know from the basic principle that for a given perimeter square has the maximum area, so, x = y and putting this value in equation (i)

x = y = 17/2=8.5

Area = x.y = 8.5× 8.5 = $72.25cm^{2}$

• Mathematics - General questions