12th Grade > Mathematics
CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option C. -> 12
:
C
f(x)=(logcotxtanx)(logtanxcotx)+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
:
C
f(x)=(logcotxtanx)(logtanxcotx)+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
Answer: Option B. -> - 1
:
B
2x+2y=2x+y
∴2x.loge2+2y.loge2dydx=2x+y.loge2(1+dydx)
∴dydx=2x+y−2x2y−2x+y
dydx|x=y=1=−1
:
B
2x+2y=2x+y
∴2x.loge2+2y.loge2dydx=2x+y.loge2(1+dydx)
∴dydx=2x+y−2x2y−2x+y
dydx|x=y=1=−1
Answer: Option C. -> 12+(g(x)−x)2
:
C
Lety=f(x)⇒x=f−1(y)
thenf(x)=x+tanx
⇒x=f−1(y)+tan(f−1(y))
⇒x=g(y)+tan(g(y))orx=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[fromEq.(i)]
12+(g(x)−x)2
:
C
Lety=f(x)⇒x=f−1(y)
thenf(x)=x+tanx
⇒x=f−1(y)+tan(f−1(y))
⇒x=g(y)+tan(g(y))orx=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[fromEq.(i)]
12+(g(x)−x)2
Answer: Option C. -> 3(d2ydx2)2
:
C
∵y=(ax+bcx+d)orcxy+dy=ax+b
Differentiating both sides w.r.t.x, then
c{xdydx+y.1}+ddydx=aorxdydx+y+(dc)dydx=(ac)
Again differentiating both sides w.r.t.x, then
orxd2ydx2+dydx+dydx+(dc)d2ydx2=0orx+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
:
C
∵y=(ax+bcx+d)orcxy+dy=ax+b
Differentiating both sides w.r.t.x, then
c{xdydx+y.1}+ddydx=aorxdydx+y+(dc)dydx=(ac)
Again differentiating both sides w.r.t.x, then
orxd2ydx2+dydx+dydx+(dc)d2ydx2=0orx+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
Answer: Option C. -> 0
:
C
f(x)=π2(aconstantfunction)
⇒ϕ′(x)=0
:
C
f(x)=π2(aconstantfunction)
⇒ϕ′(x)=0
Answer: Option C. -> In x(1+In x)−2
:
C
Since,xy=ex−y⇒ylnx=x−y∴y=x1+Inx∴dydx=lnx(1+Inx)2
:
C
Since,xy=ex−y⇒ylnx=x−y∴y=x1+Inx∴dydx=lnx(1+Inx)2
Answer: Option B. -> 1−xx
:
B
x=ey+x⇒logex=y+x
∴1x=dydx+1
⇒dydx=1−xx
:
B
x=ey+x⇒logex=y+x
∴1x=dydx+1
⇒dydx=1−xx
Answer: Option A. -> - 1
:
A
xy.yx=16
∴logexy+logeyx=loge16
⇒ylogex+xlogey=4loge2
Now, differentiating both sides w.r.t.x
yx+logexdydx+xydydx+logey.1=0
∴dydx=−(logey+yx)(logex+xy)
∴dydx|(2,2)=−(loge2+1)(loge2+1)=−1
:
A
xy.yx=16
∴logexy+logeyx=loge16
⇒ylogex+xlogey=4loge2
Now, differentiating both sides w.r.t.x
yx+logexdydx+xydydx+logey.1=0
∴dydx=−(logey+yx)(logex+xy)
∴dydx|(2,2)=−(loge2+1)(loge2+1)=−1
Answer: Option D. -> (xdydx−y)2
:
D
∵y=ln(xa+bx)x=x(lnx−ln(a+bx))
or(yx)=lnx−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[FromEq.(i)]orx3d2ydx2=(xdydx−y)2
:
D
∵y=ln(xa+bx)x=x(lnx−ln(a+bx))
or(yx)=lnx−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[FromEq.(i)]orx3d2ydx2=(xdydx−y)2
Answer: Option B. -> 78
:
B
5f(x)+3f(1x)=x+2........(i)
Replacingxby1x
∴5f(1x)+3f(x)=1x+2...(ii)
FromEq.(i).
25f(x)+15f(1x)=5x+10....(iii)
andfromEq.(ii)
9f(x)+15f(1x)=3x+6.....(iv)
SubtractingEq.(iv)from(iii),weget
∴16f(x)=5x−3x+4
∴xf(x)=5x2−3+4x16=y
∴dydx=10x+416
∴dydx|x=1=10+416
=78
:
B
5f(x)+3f(1x)=x+2........(i)
Replacingxby1x
∴5f(1x)+3f(x)=1x+2...(ii)
FromEq.(i).
25f(x)+15f(1x)=5x+10....(iii)
andfromEq.(ii)
9f(x)+15f(1x)=3x+6.....(iv)
SubtractingEq.(iv)from(iii),weget
∴16f(x)=5x−3x+4
∴xf(x)=5x2−3+4x16=y
∴dydx=10x+416
∴dydx|x=1=10+416
=78
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