VITEEE 2018 :: Mathematics :: General questions

• Mathematics - General questions

The focus of the curve y² + 4x- 6y + 13 = 0 is

Answer:

Explanation:

The given equation of curve is:

y² + 4x- 6y + 13 = 0

which can be written as: y² - 6y + 9 + 4x + 4 = 0

→ (y² - 6Y + 9) = -4(x+1)

→ (y - 3)² = -4(x+1)

put Y = y - 3 and X = x + 1

On comparing Y² = 4aX 

Length of focus from vertex, a = -1

At focus X = a and Y = 0 → x+ 1 = - 1 

x = -2

.'. y - 3 = 0 → y = 3

.'. Focus is (- 2,3).

If A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16 then,

I    P(not A) = 0.58

II   P(not B) = 0.52

III   P(A or B) = 0.47

Answer:

Explanation:

I .  P(notA) = 1 - 0.42 = 0.58

II .  P(not B) = 1 - P(B) = 1 - 0.48 = 0.52

III .  P(A or B) = P(A υ B) = P(A) + P(B) - P(A ∩ B) = 0.42 + 0.48 - 0.16 = 0.74

The value of $\lim_{x \rightarrow 0}\frac{x^{3}\cot x}{1 - \cos x}$ is ;

Answer:

Explanation:

$\lim_{x \rightarrow 0}\frac{x^{3}\times\cot x}{1-\cos x}$

= $\lim_{x \rightarrow 0}\left(\frac{x^{3}\times\cot x}{1-\cos x}\times\frac{1+\cos x}{1+\cos x}\right)$

= $\lim_{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{3}\times\lim_{x \rightarrow 0}\cos x$

$\times\lim_{x \rightarrow 0}1+\cos x$ = 2

Amplitude of $\frac{1+\sqrt{3i}}{\sqrt{3}+1}$ is :

Answer:

Explanation:

Let r(cos θ + isin θ) = $\frac{1+i\sqrt{3}}{\sqrt{3}+1}$

=$\frac{1}{\sqrt{3}+1}+i\frac{\sqrt{3}}{\sqrt{3}+1}$

 r cos θ = $\frac{1}{\sqrt{3}+1}$ ; r sin θ = $\frac{\sqrt{3}}{\sqrt{3}+1}$

→ tan θ = √3 

→ θ = $\frac{\pi}{3}$

The no. of points of discontinuity of the function f (x) = x - [x] in the interval (0, 7) are

Answer:

Explanation:

The graph of the function f (x) = x - [x] for the interval (0, 7) is shown below :

It is obvious from the above graph that the function x - [x] is discontinuous at the points x = 1,2,3,4, 5,6. Therefore no. of points of discontinuity of the given function in ths given interval are 6.

• Mathematics - General questions