VITEEE Maths Test - 1 - JEE MCQ

The line which makes equal distance from the two fixed points will definitely pass through the midpoint of line joining the two points and will definitely perpendicular to the line formed by joining the two points.

Let's first find the probability of not getting a club card. There are 52 cards in a deck, with 13 clubs and 39 non-clubs:

P(not getting a club) = 39/52 = 3/4.

Now, we're looking for the number of times a card must be drawn so that the probability of getting at least one club is greater than 3/4.

Let n be the number of times a card is drawn. The probability of not getting a club in n draws is (3/4)^n. Since we want the probability of getting at least one club, we will consider the complementary probability, which is 1 - (3/4)^n.

We want this probability to be greater than 3/4:

1 - (3/4)^n > 3/4.

Now, we can solve for n:

(3/4)^n < 1/4.

Since (3/4)^n is a decreasing function, we can find the smallest integer n that satisfies the inequality by trial and error:

n = 1: (3/4)^1 = 3/4 (not less than 1/4)
n = 2: (3/4)^2 = 9/16 (not less than 1/4)
n = 3: (3/4)^3 = 27/64 (not less than 1/4)
n = 4: (3/4)^4 = 81/256 < 1/4.

Therefore, the smallest number of times a card must be drawn so that the probability of getting at least one club is greater than 3/4 is n = 4. The answer is D. 4.

ANSWER :- b

Solution :- f′(1-0)= lim h→0 f(1-0-h)−f(1-0)

= lim h→0 (sinπ[1-0-h]−sinπ[1-0])/h

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= lim h→0 (sinπ−sinπ)/h=0

m[ -3 4 ] + n[ 4 -3 ] = [ -3m 4m ] + [ 4n -3n ]

= [ (-3m)+4n 4m+(-3n) ] = [ 10 -11 ]

• we can compare both sides,

-3m + 4n = 10 .... (1)

And 4m - 3n = -11 .... (2)

• Solving both equations

multiply equation (1) by 4 and equation (2) by 3

-12m + 14n = 40

and 12m - 9n = -33

• Add equation (1) and (2)

-12m + 14n + 12m - 9n = 40 - 33

5n = 7

n = 7/5

• Substitute value of n in equation (1)

-3m + 4×(7/5) = 10

-3m + 28/5 = 10

-15m + 28 = 50

-15m = 50 - 28

m = -22/15

• Substitute value of m and n in 3m + 7n

3(-22/15) + 7(7/5) = -22/5 + 49/5

= 27/5

• Hence, value of 3m + 7n is 27/5

Between (n + 1) white balls there are (n + 2) gaps in which (n + 1) black ball are to arranged.
No. of reqd arrangements = (n + 1)! (n + 1)! = [(n + 1)!]²
Now between (n + 1) black balls (n + 1) white balls are to be filled no. of ways
= (n + 1)! (n + 1)!
∴ Reqd ways = 2[(n + 1)!]²

Let's denote the height of the pole as h and the horizontal distance from the bottom of the pole to the point where the rope touches the ground as b. We can use trigonometry in the right-angled triangle formed by the pole, the ground, and the rope.

The given angle is 30°, so we can use the sine and cosine functions to find the height (h) and base (b) of the triangle.

sin(30°) = h/5
h = 5 * sin(30°)
h = 5 * 1/2
h = 5/2 m

So, the height of the pole is 5/2 m. The correct answer is A. 5/2m.