JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The curve

y=a x^3+b x^2+c x+5

touches the

\mathrm{x}

-axis at

\mathrm{P}(-2,0)

and cut the

\mathrm{y}

-axis at a point

\mathrm{Q}

where its gradient is 3 . Then,

a+b+c=

View answer and explanation

Correct option: B

7 / 4

Option B (

7 / 4

) is correct because it is obtained by using the given conditions to find the coefficients of the cubic polynomial and then calculating

a+b+c

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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