CTET Solved Paper MCQ - CTET Feburary 2016 Paper2

141

x⁴ + 4 = (x²)² + (2)²
= (x² + 2)² - 2.(x²).(2) [As a² + b² = (a+b)² - 2ab]
= (x² + 2)² - 4x²
= (x² + 2)² - (2x)²
= (x² + 2 + 2x)(x² + 2 - 2x) [As a² - b² = (a+b)(a-b)]
∴ (x² - 2x + 2) is a factor of x⁴ + 4.

143 The mean of the five observations x, x + 2, x + 4, x + 6, x + 8 is 11. Then, the mean of the first three observations is

Given, the mean of the 5 observations x, x + 2, x + 4, x + 6, x + 8 is 11.
ctetfeb16 q56_d

144 A bag has 5 red marbles, 4 green marbles and 3 blue marbles. All marbles are identical in all respects other than colour. A marble is taken out from the bag without looking into it. What is the probability that it is a non-green marble?

Given, a bag contains 5 red marbles, 4 green marbles and 3 blue marbles.
Total number of marbles = 12.
Total number of non-green marbles =8
Here, total number of cases =12 and number of favourable cases = 8
.'. Probability of getting a non-green marble =
Number of favourable cases/Total number of cases.
= 8/12. = 2/3.

145 If AB x BA = BCB, where A, B and C stand for just one digit and A ≠ B ≠ C, then the value of A + B + C is

Given, AB x BA = BCB.
Here, we cannot take AB as 11 because from question A ≠ B ≠ C.
If we take A = 1 and B = 2, then 12x21 = 252.
From, here we can say that A = 1, B = 2 and C = 5
.'. A + B + C = 1 + 2 + 5 = 8.