JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The gradient of the curve passing through

(4,0)

is given by

\frac{dy}{dx}-\frac{y}{x}+\frac{5x}{\left( x+2 \right)\left( x-3 \right)}=0

, if the point

(5,a)

lies on the curve, then the value of

a

View answer and explanation

Correct option: C

5\,\log \frac{7}{12}

The correct answer is D (

5\,\log \frac{12}{7}

) because it is derived by solving the given differential equation and applying the provided conditions.

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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