NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

WB JEE 2009

MCQ (Single Correct Answer)

The slope at any point of a curve y = f(x) is given by $${{dy} \over {dx}} = 3{x^2}$$ and it passes through ($$-$$1, 1). The equation of the curve is

A
y = x3 + 2
B
y = $$-$$x3 + 4
C
y = 3x2 + 4
D
y = $$-$$x3 $$-$$ 2

Explanation

$$\int {dy = \int {3{x^2}dx \Rightarrow y = {x^3} + c} } $$

$$\because$$ The curve passes through ($$-$$1, 1)

$$1 = - 1 + c \Rightarrow c = 2$$

$$y = {x^3} + 2$$.

2

WB JEE 2008

MCQ (Single Correct Answer)

The differential equation of the family of curves $$y = {e^{2x}}(a\cos x + b\sin x)$$, where a and b are arbitrary constants, is given by

A
y2 $$-$$ 4y1 + 5y = 0
B
2y2 $$-$$ y1 + 5y = 0
C
y2 + 4y1 $$-$$ 5y = 0
D
y2 $$-$$ 2y1 + 5y = 0

Explanation

Given equation of curve is

$$y = {e^{2x}}(a\cos x + b\sin x)$$ ..... (i)

taking derivative w.r.t. x

$${y_1} = {e^{2x}}( - a\sin x + b\cos x) + (a\cos x + b\sin x)2{e^{2x}}$$

or, $${y_1} = {e^{2x}}( - a\sin x + b\cos x) + 2y$$ ..... (ii) (using (i))

Taking derivative again

$${y_2} = 2{e^{2x}}( - a\sin x + b\cos x) + {e^{2x}}( - a\cos x - b\sin x) + 2{y_1}$$

$$ \Rightarrow {y_2} = 2{e^{2x}}( - a\sin x + b\cos x) - y + 2{y_1}$$ (using (i))

$$ \Rightarrow {y_2} = 4{y_1} - 5y$$

or, $${y_2} - 4{y_1} + 5y = 0$$ (using (ii))

3

WB JEE 2008

MCQ (Single Correct Answer)

The differential equation of the family of circles passing through the fixed points (a, 0) and ($$-$$a, 0) is

A
y1(y2 $$-$$ x2) + 2xy + a2 = 0
B
y1y2 + xy + a2x2 = 0
C
y1(y2 $$-$$ x2 + a2) + 2xy = 0
D
y1(y2 + x2) $$-$$ 2xy + a2 = 0

Explanation

Let the equation of circle is

x2 + y2 + 2gx + 2fy + c = 0 ..... (i)

Circle passes through (a, 0) and ($$-$$a, 0)

so a2 + 2ga + c = 0 ..... (ii)

a2 $$-$$ 2ga + c = 0 ...... (iii)

from (ii) and (iii) we get, g = 0 and c = $$-$$a2

$$\therefore$$ x2 + y2 + 2 . 0 . x + 2fy $$-$$ a2 = 0 from (i)

or x2 + y2 + 2fy $$-$$ a2 = 0

or, $$f = {{{a^2} - {x^2} - {y^2}} \over y}$$

Differentiating both sides w.r.t. x, we get

$$0 = {{y( - 2x - 2y'y) - ({a^2} - {x^2} - {y^2})y'} \over {{y^2}}}$$

$$ \Rightarrow ({x^2} + {y^2} - {a^2})y' - 2xy - 2{y^2}y' = 0$$

$$ \Rightarrow ({x^2} - {y^2} - {a^2})y' - 2xy = 0$$

$$ \Rightarrow y'({y^2} - {x^2} + {a^2}) + 2xy = 0$$

$$ \Rightarrow {y_1}({y^2} - {x^2} + {a^2}) + 2xy = 0$$ ($$\because$$ $${y_1} = y'$$)

4

WB JEE 2008

MCQ (Single Correct Answer)

The order and degree of the following differential equation $${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/2}} = {{{d^3}y} \over {d{x^3}}}$$ are respectively

A
3, 2
B
3, 10
C
2, 3
D
3, 5

Explanation

Given differential equation is

$${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/2}} = {{{d^3}y} \over {d{x^3}}}$$

Squaring both sides

$${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^5} = {\left( {{{{d^3}y} \over {d{x^3}}}} \right)^2}$$

$$\therefore$$ order = 3, degree = 2.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12