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11th And 12th > Mathematics

METHODS OF DIFFERENTIATION MCQs

Methods Of Differentiation

Total Questions : 55 | Page 4 of 6 pages
Question 31.


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


  1.     0
  2.     1
  3.     11x2
  4.     11+x2
 Discuss Question
Answer: Option B. -> 1
:
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
Question 32.


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


  1.     0
  2.     1
  3.     - 1 
  4.     None of these
 Discuss Question
Answer: Option A. -> 0
:
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
Question 33.


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


  1.     2
  2.     cos a
  3.     sin a
  4.     None of these
 Discuss Question
Answer: Option C. -> sin a
:
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
Question 34.


If (x+y)+(yx)=a,then d2ydx2 equals


  1.     2a
  2.     2a2
  3.     2a2
  4.     None of these
 Discuss Question
Answer: Option C. -> 2a2
:
C
Given,
(x+y)+(yx)=a..........(i)
x+yyx=2xa.......(ii)
Adding Eqs.(i) and (ii), then
2x+y=a+2xa
Squaring,4x+4y=a2+4x2a2+4x
4+4dydx=0+8xa2+4
0+4d2ydx2=8a2
d2ydx2=2a2
Question 35.


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


  1.     2
  2.     - 2
  3.     1
  4.     - 1
 Discuss Question
Answer: Option D. -> - 1
:
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
Question 36.


If xcos y+ycos x=5. Then


  1.     at x = 0, y = 0, y’ = 0
  2.     at x = 0, y = 1, y’ = 0
  3.     at x = y = 1, y’ = – 1
  4.     at x = 1, y = 0, y’ = 1
 Discuss Question
Answer: Option C. -> at x = y = 1, y’ = – 1
:
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
Question 37.


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


  1.     f(x,y)=yx
  2.     f(x,y)=y2x2
  3.     f(x,y)=2y2x2
  4.     f(x,y)=x2y2
 Discuss Question
Answer: Option D. -> f(x,y)=x2y2
:
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
Question 38.


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


  1.     (3ba3)cosec4θ cot4θ
  2.     (3ba3)cosec4θ cotθ
  3.     (3ba3)cosec4θ cotθ
  4.     None of the above
 Discuss Question
Answer: Option C. -> (3ba3)cosec4θ cotθ
:
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
Question 39.


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


  1.     12y1
  2.     y2x2y32xy1
  3.     (2y1)
  4.     None of these
 Discuss Question
Answer: Option B. -> y2x2y32xy1
:
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
Question 40.


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


  1.     - 2
  2.     - 1
  3.     0
  4.     1
 Discuss Question
Answer: Option D. -> 1
:
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
 

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