Match the following :
Column-IColumn-II(i) (x–2) is a common factor of expression x2 + ax + b & …

Question

Match the following :
Column-IColumn-II(i) (x–2) is a common factor of    expression x2 + ax + b & x2 + cx + dwhere a ≠≠ c, b ≠ d then  is equalto[A] 4(ii) If number of ways of arrangingletter of word CHEEKU is 3(k!) then kequals[B] 5(iii) Last non zero digit in 21! Is[C] 2(iv) If n∈N then remainder when (37)n + 2    + (16)n + 1 + (30)n  is dividedby 7 is    [D] 0

CGP EDU Academic Team · Accepted Answer

The correct answer is (i) [C]; (ii) [B]; (iii) [A]; (iv) [D]; (ii) (i). Ans.
(i) [C]
(ii) [B]
(iii) [A]
(iv) [D]
Sol. (i) Put x = 2 in both equation we get
4 + 2a + b = 0 ……..(i)
4 + 2c + d = 0 ……..(ii)
(i) – (ii)
⇒ 
(ii) No. of arrangements =  = 3. 5!
(iii) Last non zero digit in 21! can be obtained by using
exponent method
No. of 2 ′ ′ s =  = 18
No. of 5 ′ ′ s =  = 4
∴ No. of zeroes = 4
No. of 3 ′ s =  = 9
∴ Last non zero digit will be obtained
from 2 14 × 3 9 × 7 3 × 11 × 13 × 15 × 17 × 19 = 4
(iv) ∴ 37 n + 2 + 16 n + 1 + 30 n = (35 + 2)
n + 2 + (14 + 2) n + 1 + (28 + 2) n = 7k + 2
n+2 + 2 n+1 + 2 n = 7k + 2
n .7
∴ remainder = 0

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Column-I	Column-II
(i) (x–2) is a common factor of expression x2 + ax + b & x2 + cx + d where a ≠≠ c, b ≠ d then is equal to	[A] 4
(ii) If number of ways of arranging letter of word CHEEKU is 3(k!) then k equals	[B] 5
(iii) Last non zero digit in 21! Is	[C] 2
(iv) If n∈N then remainder when (37)n + 2 + (16)n + 1 + (30)n is divided by 7 is	[D] 0

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