SSC CHSL 19 Mar 2020 Shift 1 Paper

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Question : 73 of 100

Marks: +1, -0

If

k x^{3}+4 x^{2}+3 x-4

and

x^{3}-4 x+k

leave the same remainder when divided by

(x-3)

, then the value of

k

is

2
$-1$
1
0

Solution:

According to the question,
Put

x=3

(3)^{3} \×\K+4 \×(3)^{2}+3 \×3-4

=(3)^{3}-4 \×3+\K

⇒

27 \K+36+9-4=27-12+\K

⇒

27 \K-\K=15-41

⇒

26 \K=-26

∴

\K={-26}/{26}=-1

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