Let Px - x2 - bx - c be a quadratic polynomial with real coefficients such that -01 Px dx -1 and Px leaves remainder 5 when it is divided by x - 2- Then the value of 9b-c is equal to -

(x 2)Q(x) + 5 = x^2 + bx + c Put x = 2 5 = 2b + c + 4 …(1) ∫_0^1 (x^2 + bx + c)dx =1 ⇒13 + b2 + c = 1 ⇒b2 + c = 23 ...(2) Solving (1) and (2), we g ...

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Byju's Answer

Standard IX

Mathematics

Polynomial and Its General Form

Let Px = x2 +...

Question

Let $P (x) = x^{2} + b x + c$ be a quadratic polynomial with real coefficients such that $1 \int 0 P (x) d x = 1$ and $P (x)$ leaves remainder $5$ when it is divided by $(x - 2) .$ Then the value of $9 (b + c)$ is equal to :

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Solution

The correct option is A $7$
$(x - 2) Q (x) + 5 = x^{2} + b x + c$
Put $x = 2$
$5 = 2 b + c + 4 \dots (1)$

$1 \int 0 (x^{2} + b x + c) d x = 1$
$\Rightarrow \frac{1}{3} + \frac{b}{2} + c = 1$
$\Rightarrow \frac{b}{2} + c = \frac{2}{3} . . . (2)$

Solving $(1)$ and $(2),$ we get
$b = \frac{2}{9}$ and $c = \frac{5}{9}$
$9 (b + c) = 7$

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Let P(x)=x2+bx+c be a quadratic polynomial with real coefficients such that 1∫0P(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 :

The correct option is A 7(x−2)Q(x)+5=x2+bx+c Put x=2 5=2b+c+4 …(1) 1∫0(x2+bx+c)dx=1 ⇒13+b2+c=1 ⇒b2+c=23 ...(2) Solving (1) and (2), we get b=29 and c=59 9(b+c)=7

Let $P (x) = x^{2} + b x + c$ be a quadratic polynomial with real coefficients such that $1 \int 0 P (x) d x = 1$ and $P (x)$ leaves remainder $5$ when it is divided by $(x - 2) .$ Then the value of $9 (b + c)$ is equal to :