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

If y = tan1(1+sinx1sinx),π2<x<π, then dydx equals

A. 12
B. 1
C. 12
D. 1

#### SOLUTION

Solution : A

y=tan1 (1cos(π2+x)1+cos(π2+x))=tan1tan(π4+x2))(i)
Now,       π2<x<π
π4<x2<π2
or       π2<π4+x2<3π4
=tan{π(π4+x2)}
From Eq.(i),
y=tan1tan{π(π4+x2)}
=π(π4+x2)
=3π4x2
(principal value of tan1 x inπ2 to π2)
dydx=12

If f(x) = |x|, then f’(x), where x 0 is equal to

A. 1
B. 0
C. 1
D. |x|x

#### SOLUTION

Solution : D

f(x)={x,x0x,x<0f(x)=1,x>0,i.e,|x|x,x>01,x>0,i.e,|x|x,x>0=|x|x,x0

If (1x2n)+(1y2n)=a(xnyn), then (1x2n1y2n)dydx is equal to

A. xn1yn1
B. yn1xn1
C. xy
D. 1

#### SOLUTION

Solution : A

Put xn=sin θ and yn=sin ϕthen,   (cosθ+cosϕ)=a(sinθsinϕ)2 cos(θ+ϕ2)cos(θϕ2)=2a cos(θ+ϕ2)sin(θϕ2)cot(θϕ2)=a(θϕ2)=cot1aθϕ=2cot1aor           sin1xnsin1yn=2 cot1aDifferentiating both sides,we havenxn1(1x2n)nyn1(1y2n)dydx=0           (1x2n1y2n)dydx=xn1yn1

If d2xdy2(dydx)3+d2ydx2=k, then k is equal to

A. 0
B. 1
C. 2
D. None of these

#### SOLUTION

Solution : A

dydx=(dxdy)1d2ydx2=(dxdy)2{ddx(dxdy)}d2ydx2=(dxdy)2{ddy(dxdy)dydx}d2ydx2=(dydx)2{d2xdy2.dydx}d2ydx2=(dydx)3d2xdy2d2ydx2+(dydx)3d2xdy2=0

If y=(1+x)(1+x2)(1+x4)(1+x2n) then dydx at x = 0 is

A. 0
B. -1
C. 1
D. None of these

#### SOLUTION

Solution : C

Since,  y=(1+x)(1+x2)(1+x4)(1+x2n),(1x)y=(1x2)(1+x2)(1+x4)(1+x2n)=(1x4)(1+x4)(1+x2n+1)y=1x2n+1(1x)dydx=(1x)(2n+1.x2n)(1x2n+1)(1)(1x)2dydxx=0=(1)(0)(1)(1)(1)2=1

The derivative of sin1(2x1+x2) with respect to tan1(2x1x2) is

A. 0
B. 1
C. 11x2
D. 11+x2

#### SOLUTION

Solution : B

Let u=sin1(2x1+x2)=2 tan1x
dudx=21+x2
and v=tan1(2x1+x2)=2 tan1x
dvdx=21+x2
dudv=(dudx)(dvdx)=1

Let f(x)=loge{u(x)v(x)},u(2)=4,v(2)=2,u(2)=2,v(2)=1, then f(2) is equal to

A. 0
B. 1
C. - 1
D. None of these

#### SOLUTION

Solution : A

f(x)=loge{u(x)v(x)}
=loge u(x)logev(x)
f(x)=u(x)u(x)v(x)v(x)
f(2)=u(2)u(2)v(2)v(2)
=4221
=22=0

If sin y = x sin (a + y) and dydx=A1+x22xcos a, then the value of A is

A. 2
B. cos a
C. sin a
D. None of these

#### SOLUTION

Solution : C

x=sin ysin(a+y).....(i)
dxdy=sin(a)sin2(a+y)
dydx=sin2(a+y)sin(a)=A1+x22x cos a
Put x=0,y=0,
then A=sin a

If (x+y)+(yx)=a,then d2ydx2 equals

A. 2a
B. 2a2
C. 2a2
D. None of these

#### SOLUTION

Solution : C

Given,
(x+y)+(yx)=a..........(i)
x+yyx=2xa.......(ii)
2x+y=a+2xa
Squaring,4x+4y=a2+4x2a2+4x
4+4dydx=0+8xa2+4
0+4d2ydx2=8a2
d2ydx2=2a2

If sin (x+y)=loge(x+y),thendydx is equal to

A. 2
B. - 2
C. 1
D. - 1

#### SOLUTION

Solution : D

sin(x+y)=loge(x+y),
cos(x+y)(1+dydx)=1(x+y)(1+dydx)
(1+dydx)(cos(x+y)1x+y)=0
cos(x+y)1(x+y) 0
1+dydx=0
dydx=1

If xcos y+ycos x=5. Then

A. at x = 0, y = 0, y’ = 0
B. at x = 0, y = 1, y’ = 0
C. at x = y = 1, y’ = – 1
D. at x = 1, y = 0, y’ = 1

#### SOLUTION

Solution : C

xcos y+ycos x=5
ecos y logex+ecos x logey=5
ecos y loge x{cosyxloge x(sin y)dydx}+ecosx logey{cos xydydxsin x logey}=0
Putx=y=1, (cos 10)+(cos 1dydx0)=0
dydx=1
or y=1

If (1x6)+(1y6)=a(x3y3) and dydx=f(x,y)(1y61x6),then

A. f(x,y)=yx
B. f(x,y)=y2x2
C. f(x,y)=2y2x2
D. f(x,y)=x2y2

#### SOLUTION

Solution : D

Put x3=sin θ,y3=sin ϕ,
then cosθ+cosϕ=a(sin θsinϕ)
2 cos(θ+ϕ2) cos(θϕ2) =2a cos (θ+ϕ2) sin (θϕ2)
cot(θϕ2) = a
(θϕ2)=cot1a
sin1x3sin1y3=2 cot1 a
3x2(1x6)3y2(1y6)dydx=0
dydx=x2y2(1y61x6)
f(x,y)=x2y2

If x = a cos θ,y=b sin θ,then d3ydx3 is equal to

A. (3ba3)cosec4θ cot4θ
B. (3ba3)cosec4θ cotθ
C. (3ba3)cosec4θ cotθ
D. None of the above

#### SOLUTION

Solution : C

x=acosθdxdθ=a sin θ and y=b sin θ dydθ=b cos θdydx=ba cot θd2ydx2=ba cosec2θ dθdx=ba2cosec3θd3ydx3=3ba2 cosec2θ(cosec θ cot θ)dθdx=3ba2 cosec3θ cot θ (1a sin θ)=3ba3 cosec4θ cot θ

If y=x+y+x+y+...., then dydx is equal to

A. 12y1
B. y2x2y32xy1
C. (2y1)
D. None of these

#### SOLUTION

Solution : B

y=x+y+y
(y2x)=2y
or (y2x)2=2y
Differentiating both sides w.r.t. x, then
2(y2x)(2ydydx1)=2dydx
dydx=(y2x)2y32xy1

The derivative of cos1(x1xx1+x) at x=1 is

A. - 2
B. - 1
C. 0
D. 1

#### SOLUTION

Solution : D

We have, y=cos1(x1xx1+x)y=cos1(1x21+x2)
Now, Put x=tanθ
We get y=cos1(1tan2θ1+tan2θ)y=cos1(cos2θ)y=2θy=2tan1xdydx=21+x2dydx|x=1=21+(1)2=2