Line $y = x$ and curve $y = x^{2} + b x + c$ touches at $(1, 1)$ then __________.

b = - 1, c = 1

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b = 1, c = 2

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b = 1, c = 1

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b = 0, c = 1

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Solution

The correct option is A $b = - 1, c = 1$
Given line $l : y = x$
$l : x - y = 0$
$∴$ slope of line $l = 1$
Now $y = x^{2} + b x + c$
$∴ \frac{d y}{d x} = 2 x + b$
$∴ {(\frac{d y}{d x})}_{P (1, 1,)} = 2 (1) + b$
$= 2 + b$
But slope of tangent of the curve $= 1$ .
$∴ 2 + b = 1$
$∴ b = - 1$
Now $(1, 1) \in y = x^{2} + b x + c$
$∴ 1 = 1 + b (1) + c$
$∴ b + c = 0$
$∴ c = - b$
$c = - (- 1)$
$∴ c = 1$
$∴$ We have $b = - 1$ and $c = 1$ .

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Line y=x and curve y=x2+bx+c touches at (1,1) then __________.

The correct option is A b=−1,c=1Given line l:y=xl:x−y=0∴ slope of line l=1Now y=x2+bx+c∴dydx=2x+b∴(dydx)P(1,1,)=2(1)+b=2+bBut slope of tangent of the curve =1.∴2+b=1∴b=−1Now (1,1)∈y=x2+bx+c∴1=1+b(1)+c∴b+c=0∴c=−bc=−(−1)∴c=1∴ We have b=−1 and c=1.

Line $y = x$ and curve $y = x^{2} + b x + c$ touches at $(1, 1)$ then __________.