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

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 x2 + y2 = 32
Solution:
The given equations are
y = x …(i)
and x2 + y2 = 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 ellipsex2a2+y2b2 and the cordinates x = 0 and x = ae, where b2 = a2 (1 – e2) 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 y2 = 4ax bounded by its locus rectum.
Solution:
The vertex of the parabola y2 = 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
dydx = a cos (x + b) ……………… (ii)
d2ydx2 = -a sin(x + b) ………………..(iii)
Eliminating a and b from equations
(i),(ii) and (iii), we get
d2ydx2 + 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.
dydx=x+12−y, (y ≠ 2)
Solution:
We have
dydx=x+12−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 – y22=x22+ x + c1
or, x2 + y2 + 2x – 4y + 2c = 0 .
or, x2 + y2 + 2x – 4y + c = 0,
Where c = 2c1
Which is the general solution of eqn. (i).

Question 6.
Find the particular solution of the differential equation
dydx=−4xy2 given that y = 1. where x = 0
Solution:
if y ≠ 0, the given differentail equation can be written as
dyy2 = -4x dx …………..(i)
Integrating both sides of equation (i), we get
∫dyy2 = -4∫x dx
−1y = -2x2 + c
y = 12x2−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 = 12x2+1

Question 7.
Show that the differential equation x – y dydx = 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
xdydx + 2y = x2(x ≠ 0).
Solution:
The given differential equation is
xdydx + 2y = x2
Dividing both sides of equation (i) by x, we get
dydx+2xy=x
Which is a linear differential equation of the type
dydx + py = Q
Where p = 2x and Q = x
So, I.F ∫2e2xdx=e2logx=elogx2=x2
Therefore, solution of the given equation is given by
y. x2 = ∫ (x)(x2) dx +c
= ∫x3 dx + c
= x24 + cx-2
Which is the general solution of the given differential equation.

Question 9.
Solve the differential equation
(tan-1 y – x)dy = (1 + y2)dx
Solution:
The given differential eqn. can be written as
dxdy+x1+y2=tan−1y1+y2
Now (i) is a linear differential eqn. of die form dxdy + P1x = Q1
Where P1 = 11+y2 and Q1 = tan−1y
Therefore, I.F = ∫e11+y2dy=etan−1y
Thus, the solution of the given differential eqn. is

substituting tan-1 y = t
So that ( 11+y2) dy = dt
I = ∫tetdt = tet – ∫1.etdt = tet – et
= et (t – 1)
I = etan-1(tan-1 y – 1)
Substituting the value of l in eqn. (ii), we get
x etan-1y = e tan-1 y(tan-1 y – 1) + c
or, x = (tan-1y – 1) + c e-tan-1y
Which is the general solution of the given differential equation.

Question 10.
Find the area enclosed by the circle x2 + y2 = a2.
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(yx)(ydx + xdy) = ysin(yx)(xdy – ydx).
Solution :
Put y = vx ⇒ dy = vdx + xdv
So the given equation becomes
x cos v(xvdx + xvdx + x2 dv) = vx sin v(yxdx + x2 dv – vxdx)
⇒ x2 cosv(2vdx+xdv) = vx2 sinvdv
⇒ 2vdx + xdv = vx tan vdv.
Dividing both sides by vac
⇒ 2dxx+dvv = tan v dv. Intergrating, we get
2logx + logv = logseev + logc
x2v = cseev xy = c sec ((xy))

Question 2.
Solve (1 + y2)dx = (tan-1 y – x)dy.
Solution:dxdy=tan−1−x1+y2⇒dxdy+x1+y2=tan−1y1+y2 ……………. (1)
which is a linear equation of the form dxdy + px = Q.

Multiplying (1) with I.F. arid integrating, we get ac
x x IF = ∫ Q x IF dy

Question 3.
Solve (x + y)2dydx = a2
Solution:

Question 4.
Show that xdy – ydx = x2+y2−−−−−−√⋅dx is a homogeneous differential eqn. and hence solve it.
Solution:
Given eqn. xdy-ydx = x2+y2−−−−−−√⋅dx .dx
or, x⋅dydx−y=x2+y2−−−−−−√

f(x) is homogeneous differential eqn
Put yn = v
y = vx
dydx=v+x⋅dvdx
From eqn (i)

Question 5.
Solve -x dydx = y. (log y – log x – 1)
Solution:
Given eqn is -x dydx = y. (log y – log x – 1
or, dydx=yx⋅(logyx−1)
put v = yx ⇒ y = vx
∴ dydx=v+x⋅dvdx
From Eqn (i)
v + x dvdx = v. (log v -1)
or x.dvdx = v. (log v – 2)
or, dvv⋅(logv−2)=dxx˙
or, ∫dvv(logv−2)=∫dxx
For L.H.S. put log v – 2 = t
1/v dv = dt
∴ From eqri. (ii),
∫dtf=∫dxx

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(yx – 2) = |x|.c
|log y – log x – 2| = c |x|

Question 6.
Solve (x2 -1) dydx+2xy = 2x2−1
Solution:

Question 7.
Solve dydx – 3y cot x = sin x given that y = 2 when x = π2
Solution:
dydx – 3y cot x = sin x


or, 2 = -2 + c
c = 4
From eqn. (ii)
y = -2sin2x + 4sin3 x

Question 8.
Solve dydx+1x=eyx
Solution:

⇒ iog|t| = log |x| + logc
⇒ log | 1 – e-y | = log |x| .c)
∴ 1 – e-y = c |x|

Question 9.
Solve (x + 2y3 )dydx = y
Solution:
Given (x + 2y3 )dydx = y
or, x + 2y3 = ydxdy
or, y. dxdy – x = 2y3
or, dxdy+(−1y). x = 2y2 …………. (i)
From eqn (i)

Question 10.
dydx + ycot x = 2y2 cos x
Solution:
1y2dydx+1ycotx = 2cos x …………………..(1)
put 1y ⇒ 1y2dydx=−dzdx
∴ (1) ⇒ −dzdx + zcotx = 2cos x
dzdx 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 ⇒ cosecxy = -2logsin x + k

Question 11.
dydx + 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 + ∫ (sec2x – 1)dx
⇒ y(secx + tan x) = sec x + tan x – x + k.

Question 12.
Evaluate ∫sec3 xdx
Solution:
I = ∫sec3 xdx = ∫sec2 x.sec xdx
⇒ I = secx∫sec2xdx – ∫ [ddx (secx)∫sec2xdx]dx
= secx tanx – ∫secx – tanx tanxdx
= sec x tanx- ∫sec x(sec2 x – 1)dx
= sec x tan x – ∫sec3 xdx + ∫secdx
= secx tanx – I + log(secx + tanx) + k
⇒ 2I = sec x . tan x + log (secx + tan x) + k
⇒ I = 12 secx tanx + 12log(secx + tanx) + c .

Question 13.
Evaluate ∫cosx−−√dx
Solution:
Let x−−√dx = z
⇒ x = z2 ⇒ dx = 2zdz
∴ I = ∫cos z. 2zdz = 2 ∫zcoszdz
= 2z∫coszdz – 2∫ddz(z)∫ coszdz]dz
= 2zsinz- 2∫sinzdz
= 2 z sinz + 2 cos z + c
⇒ I = 2x−−√sinx−−√+2cosx−−√ + c

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

Question 14.
(x3 + x2 + x + x + 1)dydx = 2x2 + x; y = 1 when x = 0
Solution:
(x3 + x2 + x + x + 1)dydx = 2x2 + x;
or, (x3 + x2 + x + 1)dy = (2x2 + x )dx

2x2 + x = A(x2 + 1) + (Bx + C)(x + 1)
= A(x2 +1) + B(x2 + x) + C( x + 1)
Put x = -1, 2 – 1 = A( 1 + 1) ⇒ A = 12
Comparing the cofficients of x2 and x
2 = A + B and 1 = B + C
B = 2 – A = 2 – 12=32
C = 1, B = 1 – 32=−12

[Note : for integrating ∫2xx2+1dx, put x2 + 1 = t 1
We have y = 1 when x = 0
Putting these values
1 = 12logl + 34log1 – 12 tan-10 + c
= 0 + c is c = 1
Thus the solution is
y = 12 log(x + 1) + 34 (x2 + 1) – 12 tan-1 x + 1
= 1422log(x + 1) + 34log(x2 + 1) – 12 tan-1 x + z
= 14[Iog(x + 1)2 + log(x2 + 1)3] – tan-1 x + 1
= 14 log(x + 1)2(x2+1)2 – 12tan-1 x + 1
This is the required solution.

Question 15.
x(x2 – 1)dy/dx = 1
Solution:
x(x2 – 1)dy/dx = 1
or x(x2 – 1)dy = dx
∴ dy=dxx(x2−1)=dxx(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 = 12
Put x = -1, 1 = C (-1) (-2) ∴ C = 12
∴ y = ∫dxx+12∫1x−1dx+12∫dxx+1
= – log x + 12 log; (x -1) + 12 log(x +1) + log C
Multing by 2
2y = -2 log x + log (x – 1) + log (x + 1) + 2 log C.
= log x2−1x2 + 2 log C.
When x = 2, y = 0
∴ 0 = log 4−14 + 2 log C.
2 log C = -log34
Solution is

Question 16.
cos dydx = a; (a ∈ R), y = 1 when x = 0
Solution:
cos dydx = a ∴ dydx = cos-1a
or dy = (cos-1a)dx
Integrating
∫ dy = (cos-1a)∫dx or, y = (cos-1a)x + C
we have y = 2 when x = 0
2 = C
∴ Solution y = x(cos-1a) + 2 or cos-1 a = y−2x
a = cos y−2x is the reqd. solution.

Question 17.
Solution:
dydx = ytan x; y = 1 when x = 0
Solution:
dydx = ytan x
or, dyy = tan xdx
Integrating we get
∫dyy=∫tanxdx
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.