Free Objective Test 01 Practice Test - 11th and 12th
Question 1
The general solution of the differential equation dydx=y tan x−y2sec x is
SOLUTION
Solution : C
We have dydx=y tan x−y2sec x⇒1y2dydx−1ytanx=−secx
Putting 1y=v⇒−1y2dydx=dvdx, we obtain
dvdx+tan x.v=secxwhich is linear
I.F=e∫tanxdx=elogsecx=secx
∴ The solution is
v secx=∫sec2xdx+c⇒1ysecx=tanx+c
⇒secx=y(c+tanx)
Question 2
If integrating factor of x(1−x2)dy+(2x2y−y−ax3)dx=0 is e∫Pdx, then P is equal to
SOLUTION
Solution : D
x(1−x2)dy+(2x2y−y−ax3)dx=0dydx+(2x2−1)x(1−x2)y=ax2(1−x2),∴P=2x2−1x(1−x2).
Question 3
Solution of the equation xdy=(y+xf(yx)f′(yx))dx
SOLUTION
Solution : B
We have, xdy=(y+xf(yx)f′(yx))dx
⇒dydx=yx+f(yx)f′(yx) which is homogenous
Putting y=vx⇒dydx=v+xdvdx, we obtain
v+xdvdx=v+f(v)f′(v)⇒f′(v)f(v)dv=dxx
Integrating, we get log f(v)=logx+logc
⇒logf(v)=logcx⇒f(yx)=cx
Question 4
The solution of the differential equation (1+y2)+(x−etan−1y)dydx=0, is
SOLUTION
Solution : A
dxdy+11+y2x=11+y2etan−1y
I.F=e∫11+y2dy=etan−1y
∴ Solution is x.etan−1y
=∫etan−1y.11+y2etan−1dy=12e2 tan−1y+12k⇒2x etan−1y=e2 tan−1y+k
Question 5
The orthogonal trajectories of the family of curves an−1y=xn are given by
SOLUTION
Solution : B
Differentiating, we have an−1dydx=nxn−1⇒an−1=nxn−1dxdy
Putting this value in the given equation, we have nxn−1dxdyy=xn
Replacing dydxby −dydy, we have ny=−xdxdy
⇒nydy+xdx=0⇒ny2+x2 =constant. Which is the required family of orthogonal trajectories.
Question 6
The solution of (y(1+x−1)+siny)dx+(x+logx+x cosy)dy=0 is
SOLUTION
Solution : C
The given equation can be written as y(1+x−1)dx+(x+logx)dy+sin ydx+xcos ydy=0
⇒d(y(x+logx))+d(xsiny)=0→y(x+logx)+xsiny=C
Question 7
The solution of y2−7y1+12y=0 is
SOLUTION
Solution : A
The given equation can be written as (ddx−3)(dydx−4y)=0....(i)
If dydx−4y=u then (I) reduces to dudx−3u=0
⇒duu=3dx⇒u=C1e3x. Therefore, we have dydx−4y=C1e3x which is a linear equation whose I.F.is e−4x. So ddx(ye−4x)=C1e−x
⇒ye−4x=−C1e−x+C2⇒y=C1e3x+C2e4x
Question 8
The solution lof dydx−x tan(y−x)=1
SOLUTION
Solution : A
Put y-x = z. Then dydx−1=dzdx⇒dydx=1+dzdx
Given dydx−x tan(y−x)=1⇒1+dzdx−x tan z=1⇒dzdx=x tan z⇒1tanzdz=x dx⇒∫cotz dz=∫x dx
⇒log|sin z|−logc=x22⇒log|sin zc|=x22=sinzc=ex22⇒sin(y−x)=c ex22
∴ The solution is sin(y−x)=cex22, where c is arbitrary constant.
Question 9
Solution of the differential equation : dydx=3x2y4+2xyx2−2x3y3 is
SOLUTION
Solution : B
x2dy−2x3y3dy=3x2y4dx+2xydx⇒x2dy−2xydx=3x2y4dx+2x3y3dy⇒2xydx−x2dyy2+3x2y2dx+2x3ydy=0⇒d(x2y)+d(x3y2)=0⇒x2y+x3y2=C
Question 10
The degree of the differential equation satisfying the relation √1+x2+√1+y2=λ(x√1+y2−y√1+x2) is
SOLUTION
Solution : A
On Putting x=tanA,y=tanB we get
secA+secB=λ(tanA secB−tanB secA)cosA+cosB=λ(sinA−sinB)tan(A−B2)=1λtan−1x−tan−1y=2tan−11λ
On differentiating 11+x2−11+y2dydx=0
Question 11
The order of the differential equation of all tangent lines to the parabola y=x2 is
SOLUTION
Solution : A
The parametric form of the given equation is x=t,y=t2.The equation of any tangent at t is 2xt=y+t2, Differentiating we get 2t=y1(=dydx) putting this value in the above equation, we have 2xy12=y+(y12)2⇒4xy1=4y+y21
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
SOLUTION
Solution : B
Since f"(x)=6(x-1)
⇒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
⇒f′(2)=3
3(2−1)3+c=3 [from eq(i)
⇒3+c=3⇒c=0
From Eq (i) we have
f′(x)=3(x−1)2
⇒f(x)=(x−1)3+k (Integrating)----(ii)
∴1=(2−1)3+k⇒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+(dydx)2]2/3=d2ydx2 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+(dydx)2]2/3=d2ydx2⇒[4+(dydx)2]2=[d2ydx2]3
Hence order is 2 and degree is 3.
Question 14
The order and degree of the differential equation y=xdydx+√a2(dydx)2+b2 are
1,2
2,1
1,1
2,2
SOLUTION
Solution : A
Given differential equation can be written as
y2=x2(dydx)2−2xy.dydx=a2(dydx)2+b2.
Hence it is of 1st order and 2nddegree differential equation.
Question 15
The general solution of y2 dx+(x2−xy+y2)dy=0 is
[EAMCET 2003]
tan−1(xy)+log y+c=0
2 tan−1(xy)+log x+c=0
log(y+√x2+y2)+log y+c=0
sin h−1(xy)+log y+c=0
SOLUTION
Solution : A
dxdy+x2−xy+y2y2=0dxdy+(xy)2−(xy)+1=0
Put v=x/y⇒x=vy⇒dxdy=v+ydvdyv+ydvdy+v2−v+1=0⇒dvv2+1+dyy=0⇒∫dvv2+1+∫dyy=0⇒tan−1(v)+log y+C=0⇒tan−1(x/y)+log y+c=0.