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IPMAT Indore 2023 (SA) PYQs

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IPMAT Indore 2023

Number System
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Remainder

Easy

The polynomial 4x10−x9+3x11−5x7+cx6+2x5−x4+x3−4x2+6x−24x ^ {10} - x ^ 9 + 3x ^ {11} - 5x ^ 7 + c x ^ 6 + 2x ^ 5 - x ^ 4 + x ^ 3 - 4x ^ 2 + 6x - 2 when divided by x−1x - 1 leaves a remainder 2.2. Then the value of c+6c + 6 is

Entered answer:

✅ Correct Answer: 5
When a polynomial P(x)P(x) is divided by (x−k)(x-k), the remainder equals P(k)P(k).
P(x)=4x10−x9+3x11−5x7+cx6+2x5−x4+x3−4x2+6x−2P(x) = 4x ^ {10} - x ^ 9 + 3x ^ {11} - 5x ^ 7 + \boxed{c} x ^ 6 + 2x ^ 5 - x ^ 4 + x ^ 3 - 4x ^ 2 + 6x - 2
This is divided by (x−1)(x-1)
Hence, the remainder is P(1)P(1) or 22 (as per question).
∴P(1)=2\therefore P(1) = 2
⇒P(1)=4−1+3−5+c+2−1+1−4+6−2=2\Rightarrow P(1) = 4 - 1 + 3 - 5 + \boxed{c} + 2 - 1 + 1 - 4 + 6 - 2 = 2
⇒P(1)=1+c+2=2\Rightarrow P(1) = 1 + \boxed{c} + 2= 2
⇒P(1)=c=2−3\Rightarrow P(1) = \boxed{c}= 2-3
⇒P(1)=c=−1\Rightarrow P(1) = \boxed{c}= -1
c+6=−1+6=5c+6 = -1 +6 = 5
Hence, the answer is 5.

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