Free Objective Test 01 Practice Test - 11th and 12th

Question 1

The general solution of the differential equation $\frac{d y}{d x} = y t a n x - y^{2} s e c x$ is

A. tan x = (c + sec x)y

B. sec y = (c + tan y )x

C. sec x = (c + tan x)y

D. None of these

SOLUTION

Solution : C

We have $\frac{d y}{d x} = y t a n x - y^{2} s e c x \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x$
Putting $\frac{1}{y} = v \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x}$ , we obtain
$\frac{d v}{d x} + t a n x . v = s e c x$ which is linear
$I . F = e^{\int t a n x d x} = e^{l o g s e c x} = s e c x$
$∴$ The solution is
$v s e c x = \int s e c^{2} x d x + c \Rightarrow \frac{1}{y} s e c x = t a n x + c$
$\Rightarrow s e c x = y (c + t a n x)$

Question 2

If integrating factor of $x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0$ is $e^{\int P d x}$ , then P is equal to

\frac{2 x^{2} - a x^{3}}{x (1 - x^{2})}

(2 x^{2} - 1)

\frac{2 x^{2} - 1}{a x^{3}}

\frac{(2 x^{2} - 1)}{x (1 - x^{2})}

SOLUTION

Solution : D

$x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0 \frac{d y}{d x} + \frac{(2 x^{2} - 1)}{x (1 - x^{2})} y = \frac{a x^{2}}{(1 - x^{2})}, ∴ P = \frac{2 x^{2} - 1}{x (1 - x^{2})}$ .

Question 3

Solution of the equation $x d y = (y + x \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x$

f (\frac{x}{y}) = c y

f (\frac{y}{x}) = c x

f (\frac{y}{x}) = c x y

f (\frac{y}{x}) = 0

SOLUTION

Solution : B

We have, $x d y = (y + \frac{x f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x$
$\Rightarrow \frac{d y}{d x} = \frac{y}{x} + \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}$ which is homogenous
Putting $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ , we obtain
$v + x \frac{d v}{d x} = v + \frac{f (v)}{f^{'} (v)} \Rightarrow \frac{f^{'} (v)}{f (v)} d v = \frac{d x}{x}$
Integrating, we get log $f (v) = l o g x + l o g c$
$\Rightarrow l o g f (v) = l o g c x \Rightarrow f (\frac{y}{x}) = c x$

Question 4

The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

2 x e^{t a n^{- 1} y}, = e^{2 t a n^{- 1} y} + k

x e^{t a n^{- 1} y}, = e^{t a n^{- 1} y} + k

x e^{2 t a n^{- 1} y}, = e^{- t a n^{- 1} y} + k

(x - 2) k e^{t a n^{- 1} y}

SOLUTION

Solution : A

$\frac{d x}{d y} + \frac{1}{1 + y^{2}} x = \frac{1}{1 + y^{2}} e^{t a n^{- 1} y}$
$I . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{t a n^{- 1} y}$
$∴$ Solution is $x . e^{t a n^{- 1}} y$
$= \int e^{t a n^{- 1} y} . \frac{1}{1 + y^{2}} e^{t a n^{- 1}} d y = \frac{1}{2} e^{2 t a n^{- 1} y} + \frac{1}{2} k \Rightarrow 2 x e^{t a n^{- 1} y} = e^{2 t a n^{- 1} y} + k$

Question 5

The orthogonal trajectories of the family of curves $a^{n - 1} y = x^{n}$ are given by

x^{n} + n^{2} y

=constant

n y^{2} + x^{2}

=constant

n^{2} x + y^{n}

=constant

n^{2} x - y^{n}

=constant

SOLUTION

Solution : B

Differentiating, we have $a^{n - 1} \frac{d y}{d x} = n x^{n - 1} \Rightarrow a^{n - 1} = n x^{n - 1} \frac{d x}{d y}$
Putting this value in the given equation, we have $n x^{n - 1} \frac{d x}{d y} y = x^{n}$
Replacing $\frac{d y}{d x}$ by $- \frac{d y}{d y}$ , we have $n y = - x \frac{d x}{d y}$
$\Rightarrow n y d y + x d x = 0 \Rightarrow n y^{2} + x^{2}$ =constant. Which is the required family of orthogonal trajectories.

Question 6

The solution of $(y (1 + x^{- 1}) + s i n y) d x + (x + l o g x + x c o s y) d y = 0$ is

(1 + y^{- 1} s i n y) + x^{- 1} l o g x = C

(y + s i n y) + x y l o g x = C

x y + y l o g x + x s i n y = C

N o n e o f t h e s e

SOLUTION

Solution : C

The given equation can be written as $y (1 + x^{- 1}) d x + (x + l o g x) d y + s i n y d x + x c o s y d y = 0$
$\Rightarrow d (y (x + l o g x)) + d (x s i n y) = 0 \to y (x + l o g x) + x s i n y = C$

Question 7

The solution of $y_{2} - 7 y_{1} + 12 y = 0$ is

y = C_{1} e^{3 x} + C_{2} e^{4 x}

y = C_{1} x e^{3 x} + C_{2} e^{4 x}

y = C_{1} e^{3 x} + C_{2} x e^{4 x}

N o n e o f t h e s e

SOLUTION

Solution : A

The given equation can be written as $(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 . . . . (i)$
If $\frac{d y}{d x} - 4 y = u$ then (I) reduces to $\frac{d u}{d x} - 3 u = 0$
$\Rightarrow \frac{d u}{u} = 3 d x \Rightarrow u = C_{1} e^{3 x}$ . Therefore, we have $\frac{d y}{d x} - 4 y = C_{1} e^{3 x}$ which is a linear equation whose I.F.is $e^{- 4 x}$ . So $\frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x}$
$\Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2} \Rightarrow y = C_{1} e^{3 x} + C_{2} e^{4 x}$

Question 8

The solution lof $\frac{d y}{d x} - x t a n (y - x) = 1$

c e \frac{e^{x^{2}}}{2} = s i n (y - x)

c e \frac{e^{x^{2}}}{2} = s i n (y + x)

c e \frac{e^{x^{2}}}{2} = s i n (y - x)^{2}

c e \frac{e^{x^{2}}}{2} = c o s (y - x)

SOLUTION

Solution : A

Put y-x = z. Then $\frac{d y}{d x} - 1 = \frac{d z}{d x} \Rightarrow \frac{d y}{d x} = 1 + \frac{d z}{d x}$
Given $\frac{d y}{d x} - x t a n (y - x) = 1 \Rightarrow 1 + \frac{d z}{d x} - x t a n z = 1 \Rightarrow \frac{d z}{d x} = x t a n z \Rightarrow \frac{1}{t a n z} d z = x d x \Rightarrow \int c o t z d z = \int x d x$
$\Rightarrow l o g | s i n z | - l o g c = \frac{x^{2}}{2} \Rightarrow l o g | \frac{s i n z}{c} | = \frac{x^{2}}{2} = \frac{s i n z}{c} = \frac{e^{x^{2}}}{2} \Rightarrow s i n (y - x) = c \frac{e^{x^{2}}}{2}$
$∴$ The solution is $s i n (y - x) = c \frac{e^{x^{2}}}{2}$ , where c is arbitrary constant.

Question 9

Solution of the differential equation : $\frac{d y}{d x} = \frac{3 x^{2} y^{4} + 2 x y}{x^{2} - 2 x^{3} y^{3}}$ is

x^{2} y^{2} + \frac{x^{2}}{y} = c

x^{3} y^{2} + \frac{x^{2}}{y} = c

x^{3} y^{2} + \frac{y^{2}}{x} = c

x^{2} y^{3} + \frac{x^{2}}{y} = c

SOLUTION

Solution : B

$x^{2} d y - 2 x^{3} y^{3} d y = 3 x^{2} y^{4} d x + 2 x y d x \Rightarrow x^{2} d y - 2 x y d x = 3 x^{2} y^{4} d x + 2 x^{3} y^{3} d y \Rightarrow \frac{2 x y d x - x^{2} d y}{y^{2}} + 3 x^{2} y^{2} d x + 2 x^{3} y d y = 0 \Rightarrow d (\frac{x^{2}}{y}) + d (x^{3} y^{2}) = 0 \Rightarrow \frac{x^{2}}{y} + x^{3} y^{2} = C$

Question 10

The degree of the differential equation satisfying the relation $\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})$ is

A. 1

B. 2

C. 3

D. none of these

SOLUTION

Solution : A

On Putting $x = t a n A, y = t a n B$ we get
$s e c A + s e c B = λ (t a n A s e c B - t a n B s e c A) c o s A + c o s B = λ (s i n A - s i n B) t a n (\frac{A - B}{2}) = \frac{1}{λ} t a n^{- 1} x - t a n^{- 1} y = 2 t a n^{- 1} \frac{1}{λ}$
On differentiating $\frac{1}{1 + x^{2}} - \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0$

Question 11

The order of the differential equation of all tangent lines to the parabola $y = x^{2}$ is

A. 1

B. 2

C. 3

D. 4

SOLUTION

Solution : A

The parametric form of the given equation is $x = t, y = t^{2}$ .The equation of any tangent at t is $2 x t = y + t^{2}$ , Differentiating we get $2 t = y_{1} (= \frac{d y}{d x})$ putting this value in the above equation, we have $2 x \frac{y_{1}}{2} = y + {(\frac{y_{1}}{2})}^{2} \Rightarrow 4 x y_{1} = 4 y + y_{1}^{2}$
The order of this equation is 1
Hence (A) is the correct answer

Question 12

A function y =f(x) has a second order derivative f"=6(x-1). If its graph passes through the point(2,1) and at that point the tangent to the graph is y =3x -5, then the function is

(x - 1)^{2}

(x - 1)^{3}

(x + 1)^{2}

(x + 1)^{3}

SOLUTION

Solution : B

Since f"(x)=6(x-1)
$\Rightarrow f^{'} (x) = 3 (x - 1)^{2} + c$ (integrating) ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3
$\Rightarrow f^{'} (2) = 3$
$3 (2 - 1)^{3} + c = 3$ [from eq(i)
$\Rightarrow 3 + c = 3 \Rightarrow c = 0$
From Eq (i) we have
$f^{'} (x) = 3 (x - 1)^{2}$
$\Rightarrow f (x) = (x - 1)^{3} + k$ (Integrating)----(ii)
$∴ 1 = (2 - 1)^{3} + k \Rightarrow k = 0$
Hence the equation of the function is $f (x) = (x - 1)^{3}$ .

Question 13

The order and degree of the differential equation ${[4 + {(\frac{d y}{d x})}^{2}]}^{2 / 3} = \frac{d^{2} y}{d x^{2}}$ are

2,2

3,3

2,3

3,2

SOLUTION

Solution : C

Here power on the differential coefficient is fractional, therefore change it into positive integer, so
${[4 + {(\frac{d y}{d x})}^{2}]}^{2 / 3} = \frac{d^{2} y}{d x^{2}} \Rightarrow {[4 + {(\frac{d y}{d x})}^{2}]}^{2} = {[\frac{d^{2} y}{d x^{2}}]}^{3}$
Hence order is 2 and degree is 3.

Question 14

The order and degree of the differential equation $y = x \frac{d y}{d x} + \sqrt{a^{2} {(\frac{d y}{d x})}^{2} + b^{2}}$ are

1,2

2,1

1,1

2,2

SOLUTION

Solution : A

Given differential equation can be written as
$y^{2} = x^{2} {(\frac{d y}{d x})}^{2} - 2 x y . \frac{d y}{d x} = a^{2} {(\frac{d y}{d x})}^{2} + b^{2} .$
Hence it is of 1^st order and 2^nddegree differential equation.

Question 15

The general solution of $y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$ is

[EAMCET 2003]

$t a n^{- 1} (\frac{x}{y}) + l o g y + c = 0$

$2 t a n^{- 1} (\frac{x}{y}) + l o g x + c = 0$

$l o g (y + \sqrt{x^{2} + y^{2}}) + l o g y + c = 0$

$s i n h^{- 1} (\frac{x}{y}) + l o g y + c = 0$

SOLUTION

Solution : A

$\frac{d x}{d y} + \frac{x^{2} - x y + y^{2}}{y^{2}} = 0 \frac{d x}{d y} + {(\frac{x}{y})}^{2} - (\frac{x}{y}) + 1 = 0$
Put $v = x / y \Rightarrow x = v y \Rightarrow \frac{d x}{d y} = v + y \frac{d v}{d y} v + y \frac{d v}{d y} + v^{2} - v + 1 = 0 \Rightarrow \frac{d v}{v^{2} + 1} + \frac{d y}{y} = 0 \Rightarrow \int \frac{d v}{v^{2} + 1} + \int \frac{d y}{y} = 0 \Rightarrow t a n^{- 1} (v) + l o g y + C = 0 \Rightarrow t a n^{- 1} (x / y) + l o g y + c = 0.$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsObjective Test 01Objective Test 01Objective Test 01

Related Exams
Objective Test 01
Objective Test 01
Objective Test 01