1.

Let R1 and R2 be the remainders when the polynomials x3 + 2x2 – 5ax – 7 and x2 + ax2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. If 2R1 + R2 = 6, the value of a is :(a) –2 (b) 1 (c) –1 (d) 2

Answer»

(d) 2

Let f(x) = x3 + 2x2 – 5ax – 7

∴ R1 = f(–1) = (–1)3 + 2. (–1)2 – 5.a. (–1) –7 

= –1 + 2 + 5a –7 = 5a – 6 

g(x) = x3 + ax2 – 12x + 6 

R2 = g(2) = (2)3 + a.(2)2 – 12.(2) + 6 

= 8 + 4a – 24 + 6 = 4a – 10 

Given, 2R1 + R2 = 6 ⇒ 2(5a – 6) + (4a – 10) = 6 

⇒ 10a – 12 + 4a – 10 = 6 

⇒ 14a – 22 = 6 ⇒ 14a = 28 ⇒ a = 2.



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