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

If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to

A. 11+(g(x)x)2
B. 11(g(x)x)2
C. 12+(g(x)x)2
D. 12(g(x)x)2

#### SOLUTION

Solution : C

Let y=f(x)x=f1(y)
then             f(x)=x+tanx
x=f1(y)+tan(f1(y))
x=g(y)+tan(g(y)) or x=g(x)+tan(g(x))          (i)
Differentiating both sides, then we get
1=g(x)+sec2g(x).g(x)
g(x)=11+sec2(g(x))=11+1+tan2(g(x))
=12+(xg(x))2                 [from Eq.(i)]
12+(g(x)x)2

If x=secθcosθ,y=sec10θcos10θ and (x2+4)(dydx)2=k(y2+4), then k is equal to

A. 1100
B. 1
C. 10
D. 100

#### SOLUTION

Solution : D

x2+4=(secθcosθ)2+4=(secθ+cosθ)2         (i)Similarly,y2+4=(sec10θ+cos10θ)2            (ii)Now,dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ)and dydθ=10sec9θsecθtanθ10cos9θ(sinθ)=10tanθ(sec10θcos10θ)dydx=(dydθ)(dxdθ)=10tanθ(sec10θ+cos10θ)tanθ(secθ+cosθ)(dydx)2=100(sec10θ+cos10θ)(secθ+cosθ)2=100(y2+4)(x2+4)or      (x2+4)(dydx)2=100(y2+4)
On comparing with the expression given we get k = 100

If y=ln(xa+bx)x,then x3d2ydx2 is equal to

A. (dydx+x)2
B. (dydxy)2
C. (xdydx+y)2
D. (xdydxy)2

#### SOLUTION

Solution : D

y=ln(xa+bx)x=x(ln xln(a+bx))
or (yx)=ln xln(a+bx)
Differentiating both sides w.r.t.x,then
xdydxy.1x2=1xba+bx=ax(a+bx)      (i)
or (xdydxy)=(axa+bx)
Again taking logarithm on both sides, then
ln(xdydxy)=ln (ax)ln(a+bx)
Differetiating both sides w.r.t.x, then
xd2ydx2+dydxdydx(xdydxy)=1xba+bx=ax(a+bx)=(xdydxy)x2           [From Eq.(i)]or      x3d2ydx2=(xdydxy)2

If y=(ax+bcx+d) , then 2dydx.d3ydx3 is equal to

A. (d2ydx2)2
B. 3d2ydx2
C. 3(d2ydx2)2
D. 3d2xdy2

#### SOLUTION

Solution : C

y=(ax+bcx+d)or c xy+dy=ax+b
Differentiating both sides w.r.t.x, then
c{xdydx+y.1}+ddydx=aor xdydx+y+(dc)dydx=(ac)
Again differentiating both sides w.r.t.x, then
or xd2ydx2+dydx+dydx+(dc)d2ydx2=0or x+2dydx(d2ydx2)+dc=0
Again differentiating both sides w.r.t.x, then
1+2(d2ydx2.d2ydx2dydx.d3ydx3)(d2ydx2)2+0=02dydx.d3ydx3=3(d2ydx2)2

If xy=exy then dydx=

A. (1+ln x)1
B. (1+ln x)2
C. ln x(1+ln x)2
D. None of these

#### SOLUTION

Solution : C

Since, xy=exyy ln x=xyy=x1+ln xdydx=ln x(1+ln x)2

If x=ey+ey+ey+ey+

A. 1x
B. 1xx
C. x1+x
D. None of these

#### SOLUTION

Solution : B

x=ey+xloge x=y+x
1x=dydx+1
dydx=1xx

If xy.yx=16,then dydx at (2,2) is

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

#### SOLUTION

Solution : A

xy.yx=16
logexy+loge yx=loge 16
y loge x+x loge y=4 loge 2
Now, differentiating both sides w.r.t.x
yx+loge x dydx+xydydx+loge y.1=0
dydx=(loge y+yx)(loge x+xy)
dydx|(2,2)=(loge 2+1)(loge 2+1)=1

If 5f(x) + 3f (1x) = x + 2 and y = x f(x), then (dydx)x=1 is equal to

A. 14
B. 78
C. 1
D. None of these

#### SOLUTION

Solution : B

5f(x)+3f(1x)=x+2........(i)
Replacing x by1x
5f(1x)+3f(x)=1x+2...(ii)
From Eq.(i).
25f(x)+15f(1x)=5x+10....(iii)
and from Eq.(ii)
9f(x)+15f(1x)=3x+6.....(iv)
Subtracting Eq.(iv) from (iii), we get
16f(x)=5x3x+4
xf(x)=5x23+4x16=y
dydx=10x+416
dydx|x=1=10+416
=78

If 2x+2y=2x+y then the value of dydx at x=y=1 is

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

#### SOLUTION

Solution : B

2x+2y=2x+y
2x.loge 2+2y.loge 2dydx=2x+y.loge 2(1+dydx)
dydx=2x+y2x2y2x+y
dydx|x=y=1=1

If f(x) = sin1(sin x)+cos1(sin x) and ϕ(x)=f(f(f(x))),then ϕ(x) is equal to

A. 1
B. sin x
C. 0
D. None of these

#### SOLUTION

Solution : C

f(x)=π2(a constant function)
ϕ(x)=0

If f(x)=(logcotxtan x)(logtanxcot x)1+tan1(x(4x2)) then f'(0) is equal to

A. 2
B. 2
C. 12
D. 0

#### SOLUTION

Solution : C

f(x)=(logcotxtan x)(logtan xcot x)+tan1(x(4x2))
f(x)=1+tan1(x(4x2))
Now Put x=2sinθ, we get
f(x)=1+tan1(2sinθ44sin2θ)f(x)=1+tan1(2sinθ2cosθ)f(x)=1+tan1(tanθ)f(x)=1+θf(x)=1+sin1(x2)f(x)=0+11(x2)212f(0)=12

If x2+y2=t1t and x4+y4=t2+1t2,then x3ydydx equals

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

#### SOLUTION

Solution : C

x2+y2=t1t
On squaring both the sides
=x4+y4+2x2y2 = t2+1t22
Given that -
x4+y4=t2+1t2
x2y2=1
x2.2ydydx+y2.2x=0
x3ydydx=x2y2=1

If y=(1+cos 2 θ1cos2 θ),dydθ at θ=3π4 is

A. 2
B. 2
C. ±2
D. None of these

#### SOLUTION

Solution : B

y=(1+cos 2 θ1cos 2 θ)
=|cotθ|=cot θ(θ=3π4)
dydθ=cosec2θ
dydθ|θ=3π/4=(2)2=2

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

A. 2y1cos x
B. cos x2y1
C. 2x1cos x
D. cos x2x1

#### SOLUTION

Solution : B

y=sin x+yy2y=sin x
(2y1)dydx=cos x

If (x2+y2)=a.etan1(y/x) a>0,then y"(0) is equal to

A. 2aeπ/2
B.
aeπ/2
C. 2aeπ/2
D. Does not exist

#### SOLUTION

Solution : C

(x2+y2)=a.etan1(y/x)
12x2+y2(2x+2yy) =a.etan1(y/x)×11+y2x2×xyyx2.....(i)
x+yyx2+y2=(x2+y2)×xyyx2+y2
[from Eq.(i)]
x+yy=xyyy=x+yxy
y"=2(xyy)(xy)2
y"(0)=2(0y(0)){0y(0)}2=2aeπ/2=2aeπ/2