VITEEE 2018 :: Mathematics :: General questions

• Mathematics - General questions

The value of λ, does the line Y = x + λ touches the ellipse 9x² + 16y² = 144 is/are

Answer:

Explanation:

 Equation of ellipse is 9x² + 16y² = 144 or  $\frac{x^{2}}{16}+\frac{y^{2}}{9} = 1$

Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$

then we get a² = 16 and b² = 9 and comparing the line  y = x+λ, with y = mx+c

.'. m = 1 and c = λ

If the line y = x + 1, touches the eilipse 9x² + 16y² = 144, then c² = a²m²+b²

→ λ² = 16×1² +9 →  λ² = 25 .'. λ =± 5

One of the values of $\left(\frac{1+i}{\sqrt{2}}\right)^{\frac{2}{3}}$ is

Answer:

Explanation:

$\left(\frac{1+i}{\sqrt{2}}\right)^{2/3}$ = $\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}i}\right)^{2/3}$

= (cos 45^º + i sin45^º)^2/3 = (cos 2/3 × 45° + i sin 2/3 × 45°)

= (cos 30° + i sin 30°)

= $\frac{\sqrt{3}}{2} + i \times\frac{1}{2}$ = 1/2 × (√3 +i)

The radius of the sphere  $x^{2} +y^{2} +z^{2} =49,2x+3y-z-5\sqrt{14} =0$ is

Answer:

Explanation:

The sphere, $x^{2} +y^{2} +z^{2} =49$ has centre at the origin (0, 0, 0) and radius 7.

 DisTance of the plane $2x+3y-z-5\sqrt{14} =0$

from the origin

$\frac{|2(0)+3(0)-(0)-5\sqrt{14}|}{\sqrt{2^{2}+3^{2}+(-1)^{2}}}$

= $\frac{|-5\sqrt{14}|}{\sqrt{14}}$ = $\frac{5\sqrt{14}}{\sqrt{14}}$ = 5

Thus in Figure ; OP = 7, ON = 5

NP² = OP² - ON² = 7² - 5² = 49 - 25 = 24

.'. NP = $2\sqrt{6}$ Hence the radius of the circle = NP = $2\sqrt{6}$

What is the length of the projection of  $3\hat{i}+4\hat{j}+5\hat{k}$ on the xy - plane?

Answer:

Explanation:

xy-plane is perpendicular to z - axis. Let the vector

$\vec{a}$ = 3i + 4j + 5k make angle θ with z - axis, then it makes 90 - θ with xy-plane. unit vector along z-axis is k.

So, cos θ = $\frac{\vec{a}\hat{k}}{|\vec{a}|.|\hat{k}|}$

= $\frac{(3i + 4j + 5k).k}{|3i+4j+5k|}$ = $\frac{5}{5\sqrt{2}}$ = $\frac{1}{\sqrt{2}}$ →θ = $\frac{\pi}{4}$

Hence angle with xy - plane $\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

projection of $\vec{a}$ on xy-plane $|\vec{a}|.\cos\frac{\pi}{4}$ = $5\sqrt{2}\times\frac{1}{\sqrt{2}}=5$

 If the I.F. of the differential equation $\frac{dy}{dx}+5y = \cos x is \int_{}^{} e^{Adx}$ then A =

Answer:

Explanation:

The I.F. of the differential equation

$\frac{dy}{dx}+Py = Q$ is $e^{\int_{}^{} pdx}$ Here P = 5 therefore I.F = $e^{\int_{}^{} 5dx}$

Hence A = 5

• Mathematics - General questions