JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

If the curve

y=x^2+b x+c

touches the line

y=x

at the point

(1,1)

, then the set of values of

x

for which the curve has a negative gradient is

View answer and explanation

Correct option: A

(-\infty, 1 / 2)

Option A

((-\infty, 1 / 2))

is correct because finding the derivative of the curve and setting it less than 0 gives the interval where the gradient is negative.

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z