Free Objective Test 02 Practice Test - 11th and 12th

Question 1

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

[MP PET 1995; Pb. CET 2002]

tan x

secx

-sec x

cot x

SOLUTION

Solution : B

The differential equation
is $\frac{d y}{d x} - y t a n x = - y^{2} s e c x I . F . = e^{- \int t a n x d x}$
This is Bernoulli's equation i.e. reducible to
linear equation.
Dividing the equation by $y^{2},$ we get
$\frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x . . . . . . . . . . . . (i)$
Put $\frac{1}{y} = y \Rightarrow - \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d Y}{d x}$
Equation (i) reduces to $- \frac{d y}{d x} = - y t a n x = - s e c x \Rightarrow \frac{d Y}{d x} + Y t a n x = s e c x,$ Which is a linear equation
Hence $I . F . = e^{- \int t a n x d x} = s e c x .$

Question 2

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 $- \frac{d x}{d y}$ 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 3

The order of the differential equation whose general solution is given by $y = C_{1} e^{2 x + C_{2}} + C_{3} e^{x} + C_{4} s i n (x + C_{5})$ is

A. 5

B. 4

C. 3

D. 2

SOLUTION

Solution : B

$y = C_{1} e^{2 x + C_{2}} + C_{3} e^{x} + C_{4} s i n (x + C_{5}) = C_{1} . e^{C_{2}} e^{2 x} + C_{3} e^{x} + C_{4} (s i n x c o s C_{5} + c o s x s i n C_{5}) = A e^{2 x} + C_{3} e^{x} + B s i n x + D c o s x$
Here, $A = C_{1} e^{C_{2}}, B = C_{4} c o s C_{5}, D = C_{4} s i n C_{5}$
(Since equation consists of four arbitrary constants)
$∴$ order of differential equation = 4.

Question 4

The degree and order of the differential equation of the family of all parabolas whose axis is x–axis, are respectively

A. 1,2

B. 3,2

C. 2,3

D. 2,1

SOLUTION

Solution : A

Equation of family of parabolas with x-axis as axis is $y^{2} = 4 a (x + α)$ where $a, α$ are two arbitrary constants. So differential equation is of order 2 and degree 1.

Question 5

If $y_{1} (x)$ is a solution of the differential equation $\frac{d y}{d x} - f (x) y = 0$ , then a solution of the differential equation $\frac{d y}{d x} + f (x) y = r (x)$

\frac{1}{y_{1} (x)} \int r (x) y_{1} (x) d x

y_{1} (x) \int \frac{r (x)}{y_{1} (x)} d x

\int r (x) y_{1} (x) d x

N o n e o f t h e s e

SOLUTION

Solution : A

$\frac{d y}{d x} - f (x) . y = 0 \frac{d y}{y} = f (x) d x$
ln $y = \int f (x) d x$
$y_{1} (x) = e^{\int f (x) d x}$ Then for given equation I.F = $e^{\int f (x) d x}$
Hence Solution $y . y_{1} (x) = \int r (x) . y_{1} (x) d x$
$y = \frac{1}{y_{1} (x)} \int r (x) . y_{1} (x) d x$

Question 6

If $x \frac{d y}{d x} = y (l o g y - l o g x + 1)$ , then the solution of the equation is

y l o g (\frac{x}{y}) = c x

x l o g (\frac{y}{x}) = c y

l o g (\frac{y}{x}) = c x

l o g (\frac{x}{y}) = c x

SOLUTION

Solution : C

$\frac{d y}{d x} = \frac{y}{x} (l o g \frac{y}{x} + 1)$
Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$∴ v + x \frac{d v}{d x} = v l o g v + v \Rightarrow \frac{1}{v l o g v} d v = \frac{1}{x} d x \Rightarrow \int \frac{1}{v l o g v} d v = \int \frac{1}{x} d x \Rightarrow l o g (l o g v) = l o g x + l o g c$
$\Rightarrow l o g \frac{y}{x} = c x$

Question 7

The solution of $\frac{d y}{d x} + \frac{x}{y} = x^{2}$ is

\frac{1}{y} = c x - x l o g x

\frac{1}{x} = c y - y l o g y

\frac{1}{x} = c x - x l o g y

\frac{1}{y} = c x - y l o g x

SOLUTION

Solution : B

$\frac{d x}{d y} + \frac{x}{y} = x^{2} \Rightarrow x^{- 2} \frac{d x}{d y} + \frac{x^{- 1}}{y} = 1$
Put $x^{- 1} = t . T h e n - x^{- 2} \frac{d x}{d y} = \frac{d t}{d y} \Rightarrow x^{- 2} \frac{d x}{d y} = - \frac{d t}{d y} ∴ - \frac{d t}{d y} + \frac{t}{y} = 1 \Rightarrow \frac{d t}{d y} - \frac{1}{y} t = - 1$
It is linear in t. Here $P = - \frac{1}{y} . Q = - 1$
$I . F = e^{\int (\frac{1}{y}) d y} = e^{- l o g y} = \frac{1}{y}$
$∴$ Solution is $t (\frac{1}{y}) = \int (- 1) \frac{1}{y} d y = - l o g y + c \Rightarrow t = - y l o g y + c y \Rightarrow x^{- 1} c y - y l o g y .$

Question 8

The solution of the differential equation $(x^{2} s i n^{3} y - y^{2} c o s x) d x + (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = 0$ is

x^{3} s i n^{3} y = 3 y^{2} s i n x + C

x^{3} s i n^{3} y + 3 y^{2} s i n x = C

x^{2} s i n^{3} y + y^{3} s i n x = C

2 x^{2} s i n y + y^{2} s i n x = C

SOLUTION

Solution : A

$(x^{2} s i n^{3} y - y^{2} c o s x) d x + (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = 0 \frac{d y}{d x} = \frac{y^{2} c o s x - x^{2} s i n^{3} y}{x^{3} c o s y s i n^{2} - 2 y s i n x} (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = (y^{2} c o s x - x^{2} s i n^{3} y) d x = 0 (\frac{x^{3}}{3} d s i n^{3} y - s i n d y^{2}) - s i n^{3} y d (\frac{x^{3}}{3}) + y^{2} d s i n x = 0$
$\frac{x^{3}}{3} d s i n^{2} y + s i n^{3} y d (\frac{x^{3}}{3}) - (s i n d y^{2} + y^{2} d s i n x)$
$d (\frac{x^{3}}{3} s i n^{3} y) - d (y^{2} s i n x) = 0 \frac{x^{3}}{3} s i n^{3} y - y^{2} s i n x = c$

Question 9

The solution of $\frac{d y}{d x} = \frac{y}{x} + t a n \frac{y}{x}$ is

y s i n (\frac{y}{x}) = c x

y s i n (\frac{y}{x}) = c y

s i n (\frac{x}{y}) = c x

s i n (\frac{y}{x}) = c x

SOLUTION

Solution : D

Put $y = v x . T h e n \frac{d y}{d x} = v + x \frac{d v}{d x}$
Given equation is $\frac{d y}{d x} = \frac{v}{x} + t a n \frac{y}{x} \Rightarrow v + x t a n v \Rightarrow c o t v d v = \frac{d x}{x} \Rightarrow l o g s i n v = l o g x + l o g c$
$\Rightarrow s i n v = c x \Rightarrow s i n (\frac{y}{x} = c x)$

Question 10

The curves satisfying the differential equation $(1 - x^{2}) y^{1} + x y = a x$ are

A. Ellipse and hyperbola

B. Ellipse and parabola

C. Ellipse and straight line

D. Circle and parabola

SOLUTION

Solution : A

The given equation is linear DE and can be written as
$\frac{d y}{d x} + \frac{x}{1 - x^{2}} y = \frac{a x}{1 - x^{2}}$
Its integrating factor is $e^{\int \frac{x}{1 - x^{2}} d x} = e^{- (\frac{1}{2}) ln (1 - x^{2})} = \frac{1}{\sqrt{1 - x^{2}}} [- 1 < x < 1]$ and $if x^{2} > 1$ then $I . F . = \frac{1}{\sqrt{x^{2} - 1}}$
$\frac{d}{d x} (y \frac{1}{\sqrt{1 - x^{2}}}) = \frac{a x}{(1 - x^{2})^{\frac{3}{2}}} = - \frac{1}{2} \frac{- 2 a x}{(1 - x^{2})^{\frac{3}{2}}}$
$\Rightarrow y \frac{1}{\sqrt{1 - x^{2}}} = \frac{a}{\sqrt{1 - x^{2}}} + C \Rightarrow y = a + C \sqrt{1 - x^{2}}$
$\Rightarrow (y - a)^{2} = C^{2} (1 - x^{2}) \Rightarrow (y - a)^{2} + C^{2} x^{2} = C^{2}$
Thus, if $- 1 < x < 1$ the given equation represents an ellipse.
If $x^{2} > 1$ then the solution is of the form $- (y - a)^{2} + C^{2} x^{2} = C^{2}$ which represents a hyperbola.

Question 11

The solution of $\frac{d y}{d x} + 1 = e^{x + y}$ is

e^{- (x + y)} + x + c = 0

e^{- (x + y)} - x + c = 0

e^{x + y} + x + c = 0

e^{x + y} - x + c = 0

SOLUTION

Solution : A

$x + y = z \Rightarrow 1 + \frac{d y}{d x} = \frac{d z}{d x} \frac{d y}{d x} + 1 = e^{x + y} \Rightarrow \frac{d y}{d x} = e^{- z} d z = d x \Rightarrow \int e^{- z} d z = \int d x \Rightarrow - e^{- z} = x + c \Rightarrow e^{- (x + y)} + x + c = 0.$

Question 12

Solution to the differential equation $\frac{x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + . . . . .}{1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + . . . . .} = \frac{d x - d y}{d x + d y}$ is

2 y e^{2 x} = C . e^{2 x} + 1

2 y e^{2 x} = C . e^{2 x} - 1

y e^{2 x} = C . e^{2 x} + 2

2 x e^{2 y} = C . e^{x} - 1

SOLUTION

Solution : B

Applying C and D, we get
$\frac{d y}{d x} = \frac{e^{- x}}{e^{x}} = e^{- 2 x} \Rightarrow 2 y = - e^{- 2 x} + C$
or $2 y e^{2 x} = C . e^{2 x} - 1$ .

Question 13

A curve is such that the mid point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line y = x. If the curve passes through (1, 0), then the curve is

2 y = x^{2} - x

y = x^{2} - x

y = x - x^{2}

y = 2 (x - x^{2})

SOLUTION

Solution : C

The point on y-axis is $(0, y - x \frac{d y}{d x})$
According to given condition,
$\frac{x}{2} = y - \frac{x}{2} \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = 2 \frac{y}{x} - 1$
Putting $\frac{y}{x} = v$ we get
$x \frac{d v}{d x} = v - 1 \Rightarrow l n ∣ ∣ \frac{y}{x} - 1 ∣ ∣ = l n | x | + c \Rightarrow 1 - \frac{y}{x} = x$ (as f(1)=0).

Question 14

The family of curves passing through $(0, 0)$ and satisfying the differential equation $\frac{y^{''}}{y^{'}} = 1$ $(where y^{'} = \frac{d y}{d x} and y^{''} = \frac{d^{2} y}{d x^{2}})$ is

y = k

y = k x

y = k (e^{x} + 1)

y = k (e^{x} - 1)

SOLUTION

Solution : D

$\frac{d p}{d x} = p (where p = \frac{d y}{d x})$
$ln p = x + c \Rightarrow p = e^{x + c}$
$\frac{d y}{d x} = k e^{x} \Rightarrow y = k e^{x} + λ$
Satisfying $(0, 0)$ , So $λ = - k$
$y = k (e^{x} - 1)$

Question 15

S1: The differential equation of parabolas having their vertices at the origin and foci on the x-axis is an equation whose variables are separable
S2: The differential equation of the straight lines which are at a fixed distance p from the origin is an equation of degree 2
S3: The differential equation of all conics whose both axes coincide with the axes of coordinates is an equation of order 2

A. TTT

B. TFT

C. FFT

D. TTF

SOLUTION

Solution : A

$S_{1}$ -Equation of parabola is $y^{2} = \pm 4 a x$
$2 y \frac{d y}{d x} = \pm r a$
D.E of parabola $\Rightarrow y^{2} = 2 y x \frac{d y}{d x}$
$2 \frac{d y}{y} = \frac{d x}{x}$
Which is variable seperable
$S_{2}$ -Equation of line which is fixed distance. P from origin can be equation of tangent to circle
$s^{2} + y^{2} = p^{2}$
Line is $y = m x + p \sqrt{1 + m^{2}} (m = \frac{d y}{d x})$
${(y - x \frac{d y}{d x})}^{2} = P (1 + {(\frac{d y}{d x})}^{2})$
So, degree is 2
$S_{3}$ -Equation of conic whose both axis co-incide with co-ordinate axis is $a x^{2} + b y^{2} = 1$
As there are two constants, so order of D.E is 2

Free Objective Test 02 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

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