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Exam: Maharashtra HSC Class 12
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Set · 2 questions
Differentiate the following w. r. t.
x
x
x
Q476
#476
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Hard
Add
tan
−
1
(
x
1
+
6
x
2
)
+
cot
−
1
(
1
−
10
x
2
7
x
)
\tan^{-1}\left(\frac{x}{1 + 6x^2}\right) + \cot^{-1}\left(\frac{1 - 10x^2}{7x}\right)
tan
−
1
(
1
+
6
x
2
x
)
+
cot
−
1
(
7
x
1
−
10
x
2
)
Show model answer
[QMisc II Q.4 (v) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q477
#477
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Hard
Add
tan
−
1
[
1
+
x
2
+
x
1
+
x
2
−
x
]
\tan^{-1}\left[\sqrt{\frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2} - x}}\right]
tan
−
1
[
1
+
x
2
−
x
1
+
x
2
+
x
]
Show model answer
[QMisc II Q.4 (vi) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Set · 6 questions
Q478
#478
Mathematics → Differentiation → Derivatives of Implicit Functions
·
Hard
Add
If
y
+
x
+
y
−
x
=
c
\sqrt{y + x} + \sqrt{y - x} = c
y
+
x
+
y
−
x
=
c
, then show that
d
y
d
x
=
y
x
−
y
2
x
2
−
1
\frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}
d
x
d
y
=
x
y
−
x
2
y
2
−
1
.
Show model answer
[QMisc II Q.5 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q479
#479
Mathematics → Differentiation → Derivatives of Implicit Functions
·
Hard
Add
If
x
1
−
y
2
+
y
1
−
x
2
=
1
x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1
x
1
−
y
2
+
y
1
−
x
2
=
1
, then show that
d
y
d
x
=
−
1
−
y
2
1
−
x
2
\frac{dy}{dx} = -\sqrt{\frac{1 - y^2}{1 - x^2}}
d
x
d
y
=
−
1
−
x
2
1
−
y
2
.
Show model answer
[QMisc II Q.5 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q480
#480
Mathematics → Differentiation → Derivatives of Implicit Functions
·
Hard
Add
If
x
sin
(
a
+
y
)
+
sin
a
cos
(
a
+
y
)
=
0
x\sin(a + y) + \sin a \cos(a + y) = 0
x
sin
(
a
+
y
)
+
sin
a
cos
(
a
+
y
)
=
0
, then show that
d
y
d
x
=
sin
2
(
a
+
y
)
sin
a
\frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a}
d
x
d
y
=
s
i
n
a
s
i
n
2
(
a
+
y
)
.
Show model answer
[QMisc II Q.5 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q481
#481
Mathematics → Differentiation → Derivatives of Implicit Functions
·
Hard
Add
If
sin
y
=
x
sin
(
a
+
y
)
\sin y = x \sin(a + y)
sin
y
=
x
sin
(
a
+
y
)
, then show that
d
y
d
x
=
sin
2
(
a
+
y
)
sin
a
\frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a}
d
x
d
y
=
s
i
n
a
s
i
n
2
(
a
+
y
)
.
Show model answer
[QMisc II Q.5 (iv) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q482
#482
Mathematics → Differentiation → Derivatives of Implicit Functions
·
Hard
Add
If
x
=
e
x
y
x = e^{\frac{x}{y}}
x
=
e
y
x
, then show that
d
y
d
x
=
x
−
y
x
log
x
\frac{dy}{dx} = \frac{x - y}{x \log x}
d
x
d
y
=
x
l
o
g
x
x
−
y
.
Show model answer
[QMisc II Q.5 (v) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q483
#483
Mathematics → Differentiation → Higher Order Derivatives
·
Hard
Add
If
y
=
f
(
x
)
y = f(x)
y
=
f
(
x
)
is a differentiable function then show that
d
2
x
d
y
2
=
−
(
d
y
d
x
)
−
3
⋅
d
2
y
d
x
2
\frac{d^2x}{dy^2} = -\left(\frac{dy}{dx}\right)^{-3} \cdot \frac{d^2y}{dx^2}
d
y
2
d
2
x
=
−
(
d
x
d
y
)
−
3
⋅
d
x
2
d
2
y
.
Show model answer
[QMisc II Q.5 (vi) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Set · 3 questions
Q484
#484
Mathematics → Differentiation → Differentiation of One Function with respect to Another
·
Hard
Add
Differentiate
tan
−
1
(
1
+
x
2
−
1
x
)
\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)
tan
−
1
(
x
1
+
x
2
−
1
)
w. r. t.
tan
−
1
(
2
x
1
−
x
2
1
−
2
x
2
)
\tan^{-1}\left(\frac{2x\sqrt{1 - x^2}}{1 - 2x^2}\right)
tan
−
1
(
1
−
2
x
2
2
x
1
−
x
2
)
.
Show model answer
[QMisc II Q.6 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q485
#485
Mathematics → Differentiation → Differentiation of One Function with respect to Another
·
Hard
Add
Differentiate
log
(
1
+
x
2
+
x
1
+
x
2
−
x
)
\log\left(\frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2} - x}\right)
lo
g
(
1
+
x
2
−
x
1
+
x
2
+
x
)
w. r. t.
cos
(
log
x
)
\cos(\log x)
cos
(
lo
g
x
)
.
Show model answer
[QMisc II Q.6 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q486
#486
Mathematics → Differentiation → Differentiation of One Function with respect to Another
·
Hard
Add
Differentiate
tan
−
1
(
1
+
x
2
−
1
x
)
\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)
tan
−
1
(
x
1
+
x
2
−
1
)
w. r. t.
cos
−
1
(
1
+
1
+
x
2
2
1
+
x
2
)
\cos^{-1}\left(\sqrt{\frac{1 + \sqrt{1 + x^2}}{2\sqrt{1 + x^2}}}\right)
cos
−
1
(
2
1
+
x
2
1
+
1
+
x
2
)
.
Show model answer
[QMisc II Q.6 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Set · 5 questions
Q487
#487
Mathematics → Differentiation → Higher Order Derivatives
·
Hard
Add
If
y
2
=
a
2
cos
2
x
+
b
2
sin
2
x
y^2 = a^2\cos^2 x + b^2\sin^2 x
y
2
=
a
2
cos
2
x
+
b
2
sin
2
x
, show that
y
+
d
2
y
d
x
2
=
a
2
b
2
y
3
y + \frac{d^2y}{dx^2} = \frac{a^2 b^2}{y^3}
y
+
d
x
2
d
2
y
=
y
3
a
2
b
2
.
Show model answer
[QMisc II Q.7 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q488
#488
Mathematics → Differentiation → Higher Order Derivatives
·
Hard
Add
If
log
y
=
log
(
sin
x
)
−
x
2
\log y = \log(\sin x) - x^2
lo
g
y
=
lo
g
(
sin
x
)
−
x
2
, show that
d
2
y
d
x
2
+
4
x
d
y
d
x
+
(
4
x
2
+
3
)
y
=
0
\frac{d^2y}{dx^2} + 4x\frac{dy}{dx} + (4x^2 + 3)y = 0
d
x
2
d
2
y
+
4
x
d
x
d
y
+
(
4
x
2
+
3
)
y
=
0
.
Show model answer
[QMisc II Q.7 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q489
#489
Mathematics → Differentiation → Higher Order Derivatives
·
Hard
Add
If
x
=
a
cos
θ
x = a\cos\theta
x
=
a
cos
θ
,
y
=
b
sin
θ
y = b\sin\theta
y
=
b
sin
θ
, show that
a
2
[
y
d
2
y
d
x
2
+
(
d
y
d
x
)
2
]
+
b
2
=
0
a^2\left[y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2\right] + b^2 = 0
a
2
[
y
d
x
2
d
2
y
+
(
d
x
d
y
)
2
]
+
b
2
=
0
.
Show model answer
[QMisc II Q.7 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q490
#490
Mathematics → Differentiation → Higher Order Derivatives
·
Moderate
Add
If
y
=
A
cos
(
log
x
)
+
B
sin
(
log
x
)
y = A\cos(\log x) + B\sin(\log x)
y
=
A
cos
(
lo
g
x
)
+
B
sin
(
lo
g
x
)
, show that
x
2
y
2
+
x
y
1
+
y
=
0
x^2 y_2 + x y_1 + y = 0
x
2
y
2
+
x
y
1
+
y
=
0
.
Show model answer
[QMisc II Q.7 (iv) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q491
#491
Mathematics → Differentiation → Higher Order Derivatives
·
Moderate
Add
If
y
=
A
e
m
x
+
B
e
n
x
y = Ae^{mx} + Be^{nx}
y
=
A
e
m
x
+
B
e
n
x
, show that
y
2
−
(
m
+
n
)
y
1
+
(
m
n
)
y
=
0
y_2 - (m + n)y_1 + (mn)y = 0
y
2
−
(
m
+
n
)
y
1
+
(
mn
)
y
=
0
.
Show model answer
[QMisc II Q.7 (v) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Set · 6 questions
Differentiate the following
w
.
r
.
t
.
x
w.\ r.\ t.\ x
w
.
r
.
t
.
x
.
Q492
#492
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Easy
Add
y
=
x
2
+
5
y = \sqrt{x^2 + 5}
y
=
x
2
+
5
Show model answer
[Q1.1.3 SolvedEx.1 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q493
#493
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Easy
Add
y
=
sin
(
log
x
)
y = \sin(\log x)
y
=
sin
(
lo
g
x
)
Show model answer
[Q1.1.3 SolvedEx.1 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q494
#494
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Easy
Add
y
=
e
tan
x
y = e^{\tan x}
y
=
e
t
a
n
x
Show model answer
[Q1.1.3 SolvedEx.1 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q495
#495
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Easy
Add
log
(
x
5
+
4
)
\log(x^5 + 4)
lo
g
(
x
5
+
4
)
Show model answer
[Q1.1.3 SolvedEx.1 (iv) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q496
#496
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Easy
Add
5
3
cos
x
−
2
5^{3\cos x - 2}
5
3
c
o
s
x
−
2
Show model answer
[Q1.1.3 SolvedEx.1 (v) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q497
#497
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
y
=
3
(
2
x
2
−
7
)
5
y = \frac{3}{(2x^2 - 7)^5}
y
=
(
2
x
2
−
7
)
5
3
Show model answer
[Q1.1.3 SolvedEx.1 (vi) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Set · 3 questions
Differentiate the following
w
.
r
.
t
.
x
w.\ r.\ t.\ x
w
.
r
.
t
.
x
.
Q498
#498
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
y
=
sin
x
3
y = \sqrt{\sin x^3}
y
=
sin
x
3
Show model answer
[Q1.1.3 SolvedEx.2 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q499
#499
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
y
=
cot
2
(
x
3
)
y = \cot^2(x^3)
y
=
cot
2
(
x
3
)
Show model answer
[Q1.1.3 SolvedEx.2 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q500
#500
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
y
=
log
[
cos
(
x
5
)
]
y = \log\left[\cos(x^5)\right]
y
=
lo
g
[
cos
(
x
5
)
]
Show model answer
[Q1.1.3 SolvedEx.2 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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