Free Objective Test 02 Practice Test - 11th and 12th
Question 1
If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to
SOLUTION
Solution : C
Let y=f(x)⇒x=f−1(y)
then f(x)=x+tanx
⇒ x=f−1(y)+tan(f−1(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+(x−g(x))2 [from Eq.(i)]
12+(g(x)−x)2
Question 2
If x=secθ−cosθ,y=sec10θ−cos10θ and (x2+4)(dydx)2=k(y2+4), then k is equal to
SOLUTION
Solution : D
∵x2+4=(secθ−cosθ)2+4=(secθ+cosθ)2 ⋯(i)Similarly,y2+4=(sec10θ+cos10θ)2 ⋯(ii)Now,dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ)and dydθ=10sec9θsecθtanθ−10cos9θ(−sinθ)=10tanθ(sec10θ−cos10θ)⇒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
Question 3
If y=ln(xa+bx)x,then x3d2ydx2 is equal to
SOLUTION
Solution : D
∵y=ln(xa+bx)x=x(ln x−ln(a+bx))
or (yx)=ln x−ln(a+bx)
Differentiating both sides w.r.t.x,then
xdydx−y.1x2=1x−ba+bx=ax(a+bx) ⋯(i)
or (xdydx−y)=(axa+bx)
Again taking logarithm on both sides, then
ln(xdydx−y)=ln (ax)−ln(a+bx)
Differetiating both sides w.r.t.x, then
xd2ydx2+dydx−dydx(xdydx−y)=1x−ba+bx=ax(a+bx)=(xdydx−y)x2 [From Eq.(i)]or x3d2ydx2=(xdydx−y)2
Question 4
If y=(ax+bcx+d) , then 2dydx.d3ydx3 is equal to
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.d2ydx2−dydx.d3ydx3)(d2ydx2)2+0=0∴2dydx.d3ydx3=3(d2ydx2)2
Question 5
If xy=ex−y then dydx=
SOLUTION
Solution : C
Since, xy=ex−y⇒y ln x=x−y∴y=x1+ln x∴dydx=ln x(1+ln x)2
Question 6
If x=ey+ey+ey+ey+⋯∞
SOLUTION
Solution : B
x=ey+x⇒loge x=y+x
∴1x=dydx+1
⇒dydx=1−xx
Question 7
If xy.yx=16,then dydx at (2,2) is
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
Question 8
If 5f(x) + 3f (1x) = x + 2 and y = x f(x), then (dydx)x=1 is equal to
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)=5x−3x+4
∴xf(x)=5x2−3+4x16=y
∴dydx=10x+416
∴dydx|x=1=10+416
=78
Question 9
If 2x+2y=2x+y then the value of dydx at x=y=1 is
SOLUTION
Solution : B
2x+2y=2x+y
∴2x.loge 2+2y.loge 2dydx=2x+y.loge 2(1+dydx)
∴dydx=2x+y−2x2y−2x+y
dydx|x=y=1=−1
Question 10
If f(x) = sin−1(sin x)+cos−1(sin x) and ϕ(x)=f(f(f(x))),then ϕ′(x) is equal to
SOLUTION
Solution : C
f(x)=π2(a constant function)
⇒ϕ′(x)=0
Question 11
If f(x)=(logcotxtan x)(logtanxcot x)−1+tan−1(x√(4−x2)) then f'(0) is equal to
SOLUTION
Solution : C
f(x)=(logcotxtan x)(logtan xcot x)+tan−1(x√(4−x2))
f(x)=1+tan−1(x√(4−x2))
Now Put x=2sinθ, we get
f(x)=1+tan−1(2sinθ√4−4sin2θ)⇒f(x)=1+tan−1(2sinθ2cosθ)⇒f(x)=1+tan−1(tanθ)⇒f(x)=1+θ⇒f(x)=1+sin−1(x2)⇒f′(x)=0+1√1−(x2)2⋅12⇒f′(0)=12
Question 12
If x2+y2=t−1t and x4+y4=t2+1t2,then x3ydydx equals
SOLUTION
Solution : C
x2+y2=t−1t
On squaring both the sides
=x4+y4+2x2y2 = t2+1t2−2
Given that -
x4+y4=t2+1t2
∴x2y2=−1
⇒x2.2ydydx+y2.2x=0
⇒x3ydydx=−x2y2=1
Question 13
If y=√(1+cos 2 θ1−cos2 θ),dydθ at θ=3π4 is
SOLUTION
Solution : B
y=√(1+cos 2 θ1−cos 2 θ)
=|cotθ|=−cot θ(θ=3π4)
∴dydθ=cosec2θ
dydθ|θ=3π/4=(√2)2=2
Question 14
If y=√sin x+√sin x+√sin x+.....∞,then dydx is equal to
SOLUTION
Solution : B
y=√sin x+y⇒y2−y=sin x
∴(2y−1)dydx=cos x
Question 15
If √(x2+y2)=a.etan−1(y/x) a>0,then y"(0) is equal to
aeπ/2
SOLUTION
Solution : C
√(x2+y2)=a.etan−1(y/x)
⇒12√x2+y2(2x+2yy′) =a.etan−1(y/x)×11+y2x2×xy′−yx2.....(i)
⇒x+yy′√x2+y2=√(x2+y2)×xy′−yx2+y2
[from Eq.(i)]
∴x+yy′=xy′−y⇒y′=x+yx−y
∴y"=2(xy′−y)(x−y)2
y"(0)=2(0−y(0)){0−y(0)}2=−2aeπ/2=−2ae−π/2