The curves y=4x²+2x-8 and y=x³-x+13 touch each other at the point

Correct Answer is : None of the above

Correct Answer is : None of the above

Test Index

AP EAMCET Engineering 18 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 3 of 160

Marks: +1, -0

The curves

y=4x^2+2x-8

y=x^3-x+13

touch each other at the point

$(34,3)$
$(3,34)$
$(-3,34)$
$(-34,3)$
None of the above

Solution:

v=4x^{2}+2x-8

differentiate w.r.t.

x'

on both sides,

\frac{dy}{dx}=4(2x)+2(1)

\frac{dy}{dx}=8x+2

...(i)

y=x^{3}-x+13

differentiate w.r.t. '

x'

on both sides,

\frac{dy}{dx}=3x^{2}-1

...(ii)

Since, curves are touch each other

\Rightarrow 8x+2=3x^{2}-1

\Rightarrow 3x^{2}-8x-3=0

\Rightarrow (x-3)(3x+1)=0

\Rightarrow x=3

(or)

\frac{-1}{3}

Put,

x=3

y=x^{3}-x+13

y=3^{3}-3+13

y=37

\therefore

Point of contact

P=(3,37)

Go to Question:

Prev Question

Next Question