BSEB Bihar Board 12th Maths Important Questions Long Answer Type Part 5 are the best resource for students which helps in revision.

## Bihar Board 12th Maths Important Questions Long Answer Type Part 5

Application Of Integrals

Question 1.

Find the area of the region in the first quadrant enclosed by the x-axis, the liney = x and the circle x^{2} + y^{2} = 32

Solution:

The given equations are

y = x …(i)

and x^{2} + y^{2} = 32 …(ii)

Solving (i) and (ii), we find that the line and’the-circle meet at B (4,4) in, the first quadrant., Draw perpendicular BM to the x- axis.

Therefore, the required area = area of the region OBMO + area of the region BMAB. ,

Now, the area of the region OBMO

Again, the area of the region BMAB

Adding (iii) and (iv), we get

the required area = 8 + 4π – 8

= 4π

Question 2.

Find the area bounded by the ellipse\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) and the cordinates x = 0 and x = ae, where b^{2} = a^{2} (1 – e^{2}) and e < 1.

Solution:

The required area of the region BOB’RFSB us enclosed by the ellipse and the line x = 0 and x = ae

Now, the area of the region BOBRFSB

Question 3.

Find the area of the parabola y^{2} = 4ax bounded by its locus rectum.

Solution:

The vertex of the parabola y^{2} = 4ax is at origin (0,0).

The equation of the locus rectum ∠SL’ is x = a. Also, parabola is symmetrical about the x-axis.

The required area of the region OLL’ = 2 (area of the region OLSO).

Question 4.

Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.

Solution:

We have

y = a sin {x + b) ………….(i)

Differentiating both sides of the eqn. (i)

with respect to x, we get

\(\frac{d y}{d x}\) = a cos (x + b) ……………… (ii)

\(\frac{d^{2} y}{d x^{2}}\) = -a sin(x + b) ………………..(iii)

Eliminating a and b from equations

(i),(ii) and (iii), we get

\(\frac{d^{2} y}{d x^{2}}\) + y = 0

Which is free form the arbitrary constants a and b hence this the required differential equation.

Question 5.

Find the general solution of the differential equation.

\(\frac{d y}{d x}=\frac{x+1}{2-y}\), (y ≠ 2)

Solution:

We have

\(\frac{d y}{d x}=\frac{x+1}{2-y}\)

Separating the variables in equation (i), we get

(2-y)dy = (x + 1)dx

Integrating both sides of equation (ii), we get

∫(2-y)dy = ∫(x + 1)dx

2y – \(\frac{y^{2}}{2}=\frac{x^{2}}{2}\)+ x + c_{1}

or, x^{2} + y^{2} + 2x – 4y + 2c = 0 .

or, x^{2} + y^{2} + 2x – 4y + c = 0,

Where c = 2c_{1}

Which is the general solution of eqn. (i).

Question 6.

Find the particular solution of the differential equation

\(\frac{d y}{d x}=-4 x y^{2}\) given that y = 1. where x = 0

Solution:

if y ≠ 0, the given differentail equation can be written as

\(\frac{d y}{y^{2}}\) = -4x dx …………..(i)

Integrating both sides of equation (i), we get

\(\int \frac{d y}{y^{2}}\) = -4∫x dx

\(-\frac{1}{y}\) = -2x^{2} + c

y = \(\frac{1}{2 x^{2}-c}\) ………….(ii)

Substituting y,= 1 and x = 0 in equation (ii), we get c = -1

Now substituting the value of c in equation (ii), we get the particular

solution of the given differential equation as y = \(\frac{1}{2 x^{2}+1}\)

Question 7.

Show that the differential equation x – y \(\frac{d y}{d x}\) = x + 2y is homogeneous and solve It.

Solution:

The given differential equation can be expressed as dy x + 2y

Therefore, f(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.

Question 8.

Find the general solution of the differential equation

x\(\frac{d y}{d x}\) + 2y = x^{2}(x ≠ 0).

Solution:

The given differential equation is

x\(\frac{d y}{d x}\) + 2y = x^{2}

Dividing both sides of equation (i) by x, we get

\(\frac{d y}{d x}+\frac{2}{x} y=x\)

Which is a linear differential equation of the type

\(\frac{d y}{d x}\) + py = Q

Where p = \(\frac{2}{x}\) and Q = x

So, I.F \(\int_{e}^{2} \frac{2}{x} d x=e^{2 \log x}=e^{\log ^{x^{2}}}=x^{2}\)

Therefore, solution of the given equation is given by

y. x^{2} = ∫ (x)(x^{2}) dx +c

= ∫x^{3} dx + c

= \(\frac{x^{2}}{4}\) + cx^{-2}

Which is the general solution of the given differential equation.

Question 9.

Solve the differential equation

(tan^{-1} y – x)dy = (1 + y^{2})dx

Solution:

The given differential eqn. can be written as

\(\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}}\)

Now (i) is a linear differential eqn. of die form \(\frac{d x}{d y}\) + P_{1}x = Q_{1}

Where P_{1} = \(\frac{1}{1+y^{2}}\) and Q_{1} = \(\frac{\tan ^{-1} y}{1+y^{2}}\)

Therefore, I.F = \(\int_{e} \frac{1}{1+y^{2} d y}=e^{\tan ^{-1} y}\)

Thus, the solution of the given differential eqn. is

substituting tan^{-1} y = t

So that ( \(\frac{1}{1+y^{2}}\)) dy = dt

I = ∫te^{t}dt = te^{t} – ∫1.e^{t}dt = te^{t} – e^{t}

= e^{t} (t – 1)

I = e^{tan-1}(tan^{-1} y – 1)

Substituting the value of l in eqn. (ii), we get

x etan^{-1}y = e tan^{-1} y(tan^{-1} y – 1) + c

or, x = (tan^{-1}y – 1) + c e^{-tan-1y}

Which is the general solution of the given differential equation.

Question 10.

Find the area enclosed by the circle x^{2} + y^{2} = a^{2}.

Solution:

The whole area enclosed by the given circle = 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x-0 and(x = a)

As the region AOBA lies in the first quadrant y is taken as positive.

Integrating, we get the whole area enclosed by die given circle.

Differential Equation

Question 1.

Solve x cos(\(\frac{{y}}{{x}}\))(ydx + xdy) = ysin(\(\frac{{y}}{{x}}\))(xdy – ydx).

Solution :

Put y = vx ⇒ dy = vdx + xdv

So the given equation becomes

x cos v(xvdx + xvdx + x^{2} dv) = vx sin v(yxdx + x^{2} dv – vxdx)

⇒ x^{2} cosv(2vdx+xdv) = vx^{2} sinvdv

⇒ 2vdx + xdv = vx tan vdv.

Dividing both sides by vac

⇒ \(\frac{2 d x}{x}+\frac{d v}{v}\) = tan v dv. Intergrating, we get

2logx + logv = logseev + logc

x^{2}v = cseev xy = c sec (\(\left(\frac{x}{y}\right)\))

Question 2.

Solve (1 + y^{2})dx = (tan^{-1} y – x)dy.

Solution:\(\frac{d x}{d y}=\frac{\tan ^{-1}-x}{1+y^{2}} \Rightarrow \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}}\) ……………. (1)

which is a linear equation of the form \(\frac{d x}{d y}\) + px = Q.

Multiplying (1) with I.F. arid integrating, we get ac

x x IF = ∫ Q x IF dy

Question 3.

Solve (x + y)2\(\frac{d y}{d x}\) = a^{2}

Solution:

Question 4.

Show that xdy – ydx = \(\sqrt{x^{2}+y^{2}} \cdot d x\) is a homogeneous differential eqn. and hence solve it.

Solution:

Given eqn. xdy-ydx = \(\sqrt{x^{2}+y^{2}} \cdot d x\) .dx

or, \(x \cdot \frac{d y}{d x}-y=\sqrt{x^{2}+y^{2}}\)

……………..(i)

f(x) is homogeneous differential eqn

Put \(\frac{y}{n}\) = v

y = vx

\(\frac{d y}{d x}=v+x \cdot \frac{d v}{d x}\)

From eqn (i)

Question 5.

Solve -x \(\frac{d y}{d x}\) = y. (log y – log x – 1)

Solution:

Given eqn is -x \(\frac{d y}{d x}\) = y. (log y – log x – 1

or, \(\frac{d y}{d x}=\frac{y}{x} \cdot\left(\log \frac{y}{x}-1\right)\)

put v = \(\frac{y}{x}\) ⇒ y = vx

∴ \(\frac{d y}{d x}=v+x \cdot \frac{d v}{d x}\)

From Eqn (i)

v + x \(\frac{d v}{d x}\) = v. (log v -1)

or x.\(\frac{d v}{d x}\) = v. (log v – 2)

or, \(\frac{d v}{v \cdot(\log v-2)}=\frac{d x}{\dot{x}}\)

or, \(\int \frac{d v}{v(\log v-2)}=\int \frac{d x}{x}\)

For L.H.S. put log v – 2 = t

1/v dv = dt

∴ From eqri. (ii),

\(\int \frac{d t}{f}=\int \frac{d x}{x}\)

or, log|t| = log|x| + log|c|

or, log|t| = log |x|; c

or, |t| = |x| .c

or, |log v — 2| = |x| c

or, |log(\(\frac{y}{x}\) – 2) = |x|.c

|log y – log x – 2| = c |x|

Question 6.

Solve (x^{2} -1) \(\frac{d y}{d x}\)+2xy = \(\frac{2}{x^{2}-1}\)

Solution:

Question 7.

Solve \(\frac{d y}{d x}\) – 3y cot x = sin x given that y = 2 when x = \(\frac{\pi}{2}\)

Solution:

\(\frac{d y}{d x}\) – 3y cot x = sin x

or, 2 = -2 + c

c = 4

From eqn. (ii)

y = -2sin^{2}x + 4sin^{3} x

Question 8.

Solve \(\frac{d y}{d x}+\frac{1}{x}=\frac{e^{y}}{x}\)

Solution:

⇒ iog|t| = log |x| + logc

⇒ log | 1 – e^{-y} | = log |x| .c)

∴ 1 – e^{-y} = c |x|

Question 9.

Solve (x + 2y^{3} )\(\frac{d y}{d x}\) = y

Solution:

Given (x + 2y^{3} )\(\frac{d y}{d x}\) = y

or, x + 2y^{3} = y\(\frac{d x}{d y}\)

or, y. \(\frac{d x}{d y}\) – x = 2y^{3}

or, \(\frac{d x}{d y}+\left(\frac{-1}{y}\right)\). x = 2y^{2} …………. (i)

From eqn (i)

Question 10.

\(\frac{d y}{d x}\) + ycot x = 2y^{2} cos x

Solution:

\(\frac{1}{y^{2}} \frac{d y}{d x}+\frac{1}{y} \cot x\) = 2cos x …………………..(1)

put \(\frac{1}{y}\) ⇒ \(\frac{1}{y^{2}} \frac{d y}{d x}=\frac{-d z}{d x}\)

∴ (1) ⇒ \(\frac{-d z}{d x}\) + zcotx = 2cos x

\(\frac{d z}{d x}\) z cot x = -2 cos x

∴ I.F = e^{-∫cotxdx} = e^{-logsinx} = cosec x

Multiplying (2) by IF and integrating, we have

zx cosec x = -∫ cosx . cosec xdx = -2 ∫ cot x dx

⇒ z cosec x = -2 log sin x + k ⇒ \(\frac{cosecx}{y}\) = -2logsin x + k

Question 11.

\(\frac{d y}{d x}\) + y sec x = tan x

Solution:

I. F = e^{∫secx} dx = e^{∫log(sec+tanx)} = sec x + tan x

Multiplying the given eqn. with I.F.and integrating we get.

∴ y x (sec x + tan x) =∫ tanx(secx + tanx)dx

= ∫ tanx secx dx + ∫ (sec^{2}x – 1)dx

⇒ y(secx + tan x) = sec x + tan x – x + k.

Question 12.

Evaluate ∫sec^{3} xdx

Solution:

I = ∫sec^{3} xdx = ∫sec^{2} x.sec xdx

⇒ I = secx∫sec^{2}xdx – ∫ [\(\frac{d}{d x}\) (secx)∫sec^{2}xdx]dx

= secx tanx – ∫secx – tanx tanxdx

= sec x tanx- ∫sec x(sec^{2} x – 1)dx

= sec x tan x – ∫sec^{3} xdx + ∫secdx

= secx tanx – I + log(secx + tanx) + k

⇒ 2I = sec x . tan x + log (secx + tan x) + k

⇒ I = \(\frac { 1 }{ 2 }\) secx tanx + \(\frac { 1 }{ 2 }\)log(secx + tanx) + c .

Question 13.

Evaluate ∫cos\(\sqrt{x} d x\)

Solution:

Let \(\sqrt{x} d x\) = z

⇒ x = z^{2} ⇒ dx = 2zdz

∴ I = ∫cos z. 2zdz = 2 ∫zcoszdz

= 2z∫coszdz – 2∫\(\frac{d}{d z}\)(z)∫ coszdz]dz

= 2zsinz- 2∫sinzdz

= 2 z sinz + 2 cos z + c

⇒ I = \(2 \sqrt{x} \sin \sqrt{x}+2 \cos \sqrt{x}\) + c

For each of the differential equations in exercises from 14 to 17, find a particular solution satisfying the given condition.

Question 14.

(x^{3} + x^{2} + x + x + 1)\(\frac{d y}{d x}\) = 2x^{2} + x; y = 1 when x = 0

Solution:

(x^{3} + x^{2} + x + x + 1)\(\frac{d y}{d x}\) = 2x^{2} + x;

or, (x^{3} + x^{2} + x + 1)dy = (2x^{2} + x )dx

2x^{2} + x = A(x^{2} + 1) + (Bx + C)(x + 1)

= A(x^{2} +1) + B(x^{2} + x) + C( x + 1)

Put x = -1, 2 – 1 = A( 1 + 1) ⇒ A = \(\frac{1}{2}\)

Comparing the cofficients of x^{2} and x

2 = A + B and 1 = B + C

B = 2 – A = 2 – \(\frac{1}{2}=\frac{3}{2}\)

C = 1, B = 1 – \(\frac{3}{2}=-\frac{1}{2}\)

[Note : for integrating \(\int \frac{2 x}{x^{2}+1} d x\), put x^{2} + 1 = t 1

We have y = 1 when x = 0

Putting these values

1 = \(\frac{1}{2}\)logl + \(\frac{3}{4}\)log1 – \(\frac{1}{2}\) tan^{-1}0 + c

= 0 + c is c = 1

Thus the solution is

y = \(\frac{1}{2}\) log(x + 1) + \(\frac{3}{4}\) (x^{2} + 1) – \(\frac{1}{2}\) tan^{-1} x + 1

= \(\frac{1}{4}2\)2log(x + 1) + \(\frac{3}{4}\)log(x^{2} + 1) – \(\frac{1}{2}\) tan^{-1} x + z

= \(\frac{1}{4}\)[Iog(x + 1)^{2} + log(x^{2} + 1)^{3}] – tan^{-1} x + 1

= \(\frac{1}{4}\) log(x + 1)^{2}(x^{2}+1)^{2} – \(\frac{1}{2}\)tan^{-1} x + 1

This is the required solution.

Question 15.

x(x^{2} – 1)dy/dx = 1

Solution:

x(x^{2} – 1)dy/dx = 1

or x(x^{2} – 1)dy = dx

∴ \(d y=\frac{d x}{x\left(x^{2}-1\right)}=\frac{d x}{x(x-1)(x+1)}\)

Integrating, we get

1= A (x – 1) (x + 1) + Bx (x + I) + Cx (x-1)

Put x-0,1 = A (-1)(+1) ∴ A = -1

Put x = 1, 1 = B = 1.2 ∴ B = \(\frac{1}{2}\)

Put x = -1, 1 = C (-1) (-2) ∴ C = \(\frac{1}{2}\)

∴ y = \(\int \frac{d x}{x}+\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{d x}{x+1}\)

= – log x + \(\frac{1}{2}\) log; (x -1) + \(\frac{1}{2}\) log(x +1) + log C

Multing by 2

2y = -2 log x + log (x – 1) + log (x + 1) + 2 log C.

= log \(\frac{x^{2}-1}{x^{2}}\) + 2 log C.

When x = 2, y = 0

∴ 0 = log \(\frac{4-1}{4}\) + 2 log C.

2 log C = -log\(\frac{3}{4}\)

Solution is

Question 16.

cos \(\frac{d y}{d x}\) = a; (a ∈ R), y = 1 when x = 0

Solution:

cos \(\frac{d y}{d x}\) = a ∴ \(\frac{d y}{d x}\) = cos^{-1}a

or dy = (cos^{-1}a)dx

Integrating

∫ dy = (cos^{-1}a)∫dx or, y = (cos^{-1}a)x + C

we have y = 2 when x = 0

2 = C

∴ Solution y = x(cos^{-1}a) + 2 or cos^{-1} a = \(\frac{y-2}{x}\)

a = cos \(\frac{y-2}{x}\) is the reqd. solution.

Question 17.

Solution:

\(\frac{d y}{d x}\) = ytan x; y = 1 when x = 0

Solution:

\(\frac{d y}{d x}\) = ytan x

or, \(\frac{d y}{y}\) = tan xdx

Integrating we get

\(\int \frac{d y}{y}=\int \tan x d x\)

log y = – log cos x + log C

or, log y + log cos x = log C

or, lyg y cosx = log C

∴ y cos x = C

Putting y = 1, x = 0 => C = 1

∴ y cos x = 1

or, y = sec x is the solution.