11th And 12th > Mathematics
METHODS OF DIFFERENTIATION MCQs
Methods Of Differentiation
Total Questions : 55
| Page 6 of 6 pages
Answer: Option C. ->
0
:
C
f(x)=π2(a constant function)
⇒ϕ′(x)=0
:
C
f(x)=π2(a constant function)
⇒ϕ′(x)=0
Answer: Option C. ->
1
:
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
:
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
Answer: Option B. ->
cos x2y−1
:
B
y=√sin x+y⇒y2−y=sin x
∴(2y−1)dydx=cos x
:
B
y=√sin x+y⇒y2−y=sin x
∴(2y−1)dydx=cos x
Answer: Option B. ->
2
:
B
y=√(1+cos 2 θ1−cos 2 θ)
=|cotθ|=−cot θ(θ=3π4)
∴dydθ=cosec2θ
dydθ|θ=3π/4=(√2)2=2
:
B
y=√(1+cos 2 θ1−cos 2 θ)
=|cotθ|=−cot θ(θ=3π4)
∴dydθ=cosec2θ
dydθ|θ=3π/4=(√2)2=2
Answer: Option C. ->
−2ae−π/2
:
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
:
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
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