The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5,6}
Number of possible outcomes of a dice = 6
(i) Prime numbers on a dice are 2, 3, and 5.
Total prime numbers on a dice = 3
Probability of getting a prime number = 3/6 = 1/2
(ii) Numbers lying between 2 and 6 = 3, 4, 5
Total numbers lying between 2 and 6 = 3
Probability of getting a number lying between 2 and 6 = 3/6 = 1/2
(iii) Odd numbers on a dice = 1, 3, and 5
Total odd numbers on a dice = 3
Probability of getting an odd number = 3/6 = 1/2

(i) Total number of cards = 5
Total number of queens = 1
P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 1/5
(ii) When the queen is drawn and put aside, the total number of remaining cards will be 4.
(a) Total number of aces = 1
P(getting Ace) = (Number of favourable outcomes) = 1/4
(b) As queen is already drawn, therefore, the number of queens will be 0.
P(getting a queen) = (Number of favourable outcomes)/(Number of total possible outcomes) = 0/4 = 0

(i) Total number ofbulbs = 20
Total number of defective bulbs =4
P(getting a defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 4/20 = 1/5
(ii) Remaining total number of bulbs = 19
Remaining total number of non-defective bulbs = 16 – 1 = 15
P(getting a not defective bulb) = (Number of favourable outcomes)/(Number of total possible outcomes) = 15/19

(i) False,
Let A = 30° and B = 60°
Therefore, LHS = sin(A + B) = sin(30° + 60°) = sin 90° 1 and
RHS = sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = (1+√3)/2 ≠ 1
Hence, sin (A + B) ≠ sin A + sin B

(ii) True,
As we know that sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2 and sin 90° = 1
Hence, for the increasing values of θ, sin θ is also increasing.

(iii) False,
As we know that cos 0° = 1, cos 30° = √3/2, cos 45° = 1√2, cos 60° = 1/2 and cos 90° = 0
Hence, for the increasing values of θ, cos θ is decreasing.

(iv) False,
∵ cos 30° = √3/2, but sin 30° = 1/2.

(v) True,
∵ tan 0° = 0, we know that cot 0° = 1/tan 0° = 1/0, which is not defined.

See the explanation video of the above solution here✌

6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – 9x + 2x -3
= 3x(2x – 3) + 1(2x -3)
= (3x +1)(2x -3)
The value of 6x² – 7x – 3 is zero if 3x + 1 = 0 or 2x -3 = 0.
⇒ x = -1/3 or x = 3/2,
Therefore, the zeroes of 6x² – 7x – 3 are -1/3 and 3/2.
Now, Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(7)/6 = -(Cofficient of x)/Cofficient of x²
Product of zeroes = (-1/3) x 3/2 = -1/2 = (-3)/6 = Constant term/Cofficient of x².

See here 😎