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Quadratic Equation and Inequalities question

2017 · 8 Apr · Shift 1 · Q32
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Quadratic Equation and Inequalities question

2017 · 8 Apr · Shift 1 · Q32

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x −-− 1 and it leaves remainder 6 when divided by x + 1; then :
  1. A
    p(2) = 11
  2. B
    p(2) = 19
  3. C
    p(−-− 2) = 19
  4. D
    p(−-− 2) = 11
View written solutionFree

Correct answer: C

  1. Since p(x)p(x)p(x) is a quadratic polynomial and p(0)=1p(0)=1p(0)=1, let p(x)=ax2+bx+1.p(x)=ax^2+bx+1.p(x)=ax2+bx+1.

  2. By the Remainder Theorem:

  • When divided by x−1x-1x−1, the remainder is p(1)=4p(1)=4p(1)=4.
  • When divided by x+1x+1x+1, the remainder is p(−1)=6p(-1)=6p(−1)=6.

So we get: p(1)=a+b+1=4⇒a+b=3p(1)=a+b+1=4 \Rightarrow a+b=3p(1)=a+b+1=4⇒a+b=3 p(−1)=a−b+1=6⇒a−b=5p(-1)=a-b+1=6 \Rightarrow a-b=5p(−1)=a−b+1=6⇒a−b=5

  1. Solve these two equations: a+b=3a+b=3a+b=3 a−b=5a-b=5a−b=5 Adding, 2a=8⇒a=42a=8 \Rightarrow a=42a=8⇒a=4 Then, b=3−a=3−4=−1.b=3-a=3-4=-1.b=3−a=3−4=−1.

Thus, p(x)=4x2−x+1.p(x)=4x^2-x+1.p(x)=4x2−x+1.

  1. Now compute the required values: p(2)=4(2)2−2+1=16−2+1=15p(2)=4(2)^2-2+1=16-2+1=15p(2)=4(2)2−2+1=16−2+1=15 So options A and B are false.

p(−2)=4(−2)2−(−2)+1=16+2+1=19p(-2)=4(-2)^2-(-2)+1=16+2+1=19p(−2)=4(−2)2−(−2)+1=16+2+1=19 So option C is true and option D is false.

  1. Therefore, the correct option is: C: p(−2)=19\boxed{\text{C: } p(-2)=19}C: p(−2)=19​
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