VITEEE 2017 :: Mathematics :: General questions

• Mathematics - General questions

The conditional (p∧q) ⇒ p is

Answer:

Explanation:

∴ (p∧q) ⇒ p is a  tautology.

If the eccentricity of the hyperbola

x²-y² cosec²α= 25 is $\sqrt{5}$    times the 

eccentricity of the elipse x² cosec²α+y²=5 then α is equal to;

Answer:

Explanation:

Eccentrcity of $\frac{x^{2}}{25}+\frac{y^{2}}{25\sin^{2}\alpha}=1$   is

$\sqrt{1+\sin^{2}\alpha}$

Eccentrcity of $\frac{x^{2}}{5\sin^{2}\alpha}+\frac{y^{2}}{5}=1$     is

$\sqrt{1-\sin^{2}\alpha}$

Given, $\sqrt{1+\sin^{2}\alpha}=\sqrt{5}\sqrt{1-\sin^{2}\alpha} $

⇒ $\sin^{2}\alpha=\frac{2}{3}$

⇒ $\alpha=\sin^{-1}\sqrt{\frac{2}{3}}=\tan^{-1}\sqrt{2}$

The tangent lines to the curve $y^{2}=4ax$ at points

where x= a, are

Answer:

Explanation:

The given equation of the curve is $y^{2}=4ax$           ...(1)

Differentiating both sides of (1) with respect to x, we get

$2y\frac{dy}{dx}=4a;\Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=
\frac{2a}{y}$                    ...(2)

If ψ be the angle which the tangent to the curve at (x,y) makes with the positive direction of x-axis then   $\tan\psi=\frac{\text{d}y}{\text{d}x}$ or

 $\tan\psi= \frac{2a}{y}$         ...(3),     [using (2)]

At $x=a$, then from (1), 

$y^{2}=4a$.$ a=4a^{2}\Rightarrow y=±2a$

Hence, we get two points (a, 2a) and (a, -2a) on the curve

At $(a, 2a)$, $x =a$, $y=2a$ and let $\psi=\psi_{1}$_. 

∴ from (3), $\tan\psi_{1}=\frac{2a}{2a}=1=\tan45°$  

                                  $\Rightarrow\psi_{1}=45°$

At (a,−2a), x=a, y=−2a  and let  $\psi=\psi_{2}$_.

∴ from (3), $\tan\psi_{2}=\frac{2a}{-2a}=-1=\tan135°$  

                                   or    $\psi_{2}=135°$

Hence the required angle between tangents to (1) at $(a,2a)$ and   $(a,-2a)= \psi_{2}-\psi_{1}= 135°- 45°=90°$

This shows that the tangent lines to (1) at $(a,2a)$ and $(a,-2a)$ are perpendicular to each other.

If $f(x)=\begin{vmatrix}\cos x & 1 & 0 \\1 & 2\cos x & 1 \\0 & 1 & 2\cos x \end{vmatrix}$

$\int_{0}^{\pi/2}f(x)dx$ is equal to

Answer:

Explanation:

f(x)= 4cos³ x - cos x - 2cos x = cos 3x

      (Expansion of determinant)

$\therefore\int_{0}^{\pi/2}f(x)dx =\frac{\sin3x}{3}]^{{\pi/2}}_{0}=-\frac{1}{3}$

The distance of the point ( 1, -2, 3) from the plane x - y + z =5, measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is

Answer:

Explanation:

Equation of the line through (1, -2, 3)

parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$ (say)           ...(1)

Then any point on (1) is (2r + 1, 3r - 2, -6r+3)

If this point lies on the plane x - y + z = 5
then

(2r+1)-(3r-2)+(-6r+3)= 5 ⇒ r = $\frac{1}{7}$

Hence the point is $(\frac{9}{7},-\frac{11}{7},\frac{15}{7})$

Distance between (1,-2, 3 ) and $(\frac{9}{7},-\frac{11}{7},\frac{15}{7})$

$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1$

• Mathematics - General questions