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Definite Integration question

2021 · 16 Mar · Shift 2 · Q32
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  5. /2021 · 16 Mar · Shift 2 · Q32

Definite Integration question

2021 · 16 Mar · Shift 2 · Q32

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let P(x) = x2 + bx + c be a quadratic polynomial with real coefficients such that ∫01P(x)dx\int_0^1 {P(x)dx}∫01​P(x)dx= 1 and P(x) leaves remainder 5 when it is divided by (x −-− 2). Then the value of 9(b + c) is equal to :
  1. A
    9
  2. B
    11
  3. C
    7
  4. D
    15
View written solutionFree

Correct answer: C

  1. Let P(x)=x2+bx+c.P(x)=x^2+bx+c.P(x)=x2+bx+c.

We are given two conditions.

  1. First condition: ∫01P(x) dx=1.\int_0^1 P(x)\,dx=1.∫01​P(x)dx=1. So, ∫01(x2+bx+c) dx=1.\int_0^1 (x^2+bx+c)\,dx=1.∫01​(x2+bx+c)dx=1.

Evaluate term by term: ∫01x2 dx=13,∫01bx dx=b⋅12=b2,∫01c dx=c.\int_0^1 x^2\,dx=\frac13, \qquad \int_0^1 bx\,dx=b\cdot \frac12=\frac b2, \qquad \int_0^1 c\,dx=c.∫01​x2dx=31​,∫01​bxdx=b⋅21​=2b​,∫01​cdx=c. Hence, 13+b2+c=1.\frac13+\frac b2+c=1.31​+2b​+c=1. So, b2+c=23.(1)\frac b2+c=\frac23. \qquad (1)2b​+c=32​.(1)

  1. Second condition: when P(x)P(x)P(x) is divided by (x−2)(x-2)(x−2), the remainder is 555.

By the Remainder Theorem, P(2)=5.P(2)=5.P(2)=5. Now, P(2)=22+2b+c=4+2b+c.P(2)=2^2+2b+c=4+2b+c.P(2)=22+2b+c=4+2b+c. Thus, 4+2b+c=5,4+2b+c=5,4+2b+c=5, which gives 2b+c=1.(2)2b+c=1. \qquad (2)2b+c=1.(2)

  1. Solve equations (1) and (2):

From (1), c=23−b2.c=\frac23-\frac b2.c=32​−2b​. Substitute into (2): 2b+(23−b2)=1.2b+\left(\frac23-\frac b2\right)=1.2b+(32​−2b​)=1. Multiply by 666: 12b+4−3b=6,12b+4-3b=6,12b+4−3b=6, 9b=2,9b=2,9b=2, b=29.b=\frac29.b=92​.

Then from (2), 2(29)+c=1,2\left(\frac29\right)+c=1,2(92​)+c=1, 49+c=1,\frac49+c=1,94​+c=1, c=59.c=\frac59.c=95​.

  1. Now compute: 9(b+c)=9(29+59)=9(79)=7.9(b+c)=9\left(\frac29+\frac59\right)=9\left(\frac79\right)=7.9(b+c)=9(92​+95​)=9(97​)=7.

  2. Therefore, the correct option is 7\boxed{7}7​ which is Option C.

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