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Exam: Maharashtra HSC Class 12
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Set · 5 questions
Differentiate the following w.r.t.
x
x
x
Q551
#551
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Hard
Add
log
[
4
2
x
(
x
2
+
5
2
x
3
−
4
)
3
2
]
\log\left[4^{2x}\left(\frac{x^2 + 5}{\sqrt{2x^3 - 4}}\right)^{\frac{3}{2}}\right]
lo
g
[
4
2
x
(
2
x
3
−
4
x
2
+
5
)
2
3
]
Show model answer
[QEx 1.1 Q.3 (xvi) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q552
#552
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Hard
Add
log
[
e
x
2
(
5
−
4
x
)
3
2
7
−
6
x
3
]
\log\left[\frac{e^{x^2}\left(5 - 4x\right)^{\frac{3}{2}}}{\sqrt[3]{7 - 6x}}\right]
lo
g
[
3
7
−
6
x
e
x
2
(
5
−
4
x
)
2
3
]
Show model answer
[QEx 1.1 Q.3 (xvii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q553
#553
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Hard
Add
log
(
a
cos
x
(
x
2
−
3
)
3
log
x
)
\log\left(\frac{a^{\cos x}}{\left(x^2 - 3\right)^3 \log x}\right)
lo
g
(
(
x
2
−
3
)
3
l
o
g
x
a
c
o
s
x
)
Show model answer
[QEx 1.1 Q.3 (xviii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q554
#554
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Hard
Add
y
=
(
25
)
log
5
(
sec
x
)
−
(
16
)
log
4
(
tan
x
)
y = (25)^{\log_5 (\sec x)} - (16)^{\log_4 (\tan x)}
y
=
(
25
)
l
o
g
5
(
s
e
c
x
)
−
(
16
)
l
o
g
4
(
t
a
n
x
)
Show model answer
[QEx 1.1 Q.3 (xix) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q555
#555
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
(
x
2
+
2
)
4
x
2
+
5
\frac{\left(x^2 + 2\right)^4}{\sqrt{x^2 + 5}}
x
2
+
5
(
x
2
+
2
)
4
Show model answer
[QEx 1.1 Q.3 (xx) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Set · 4 questions
A table of values of
f
f
f
,
g
g
g
,
f
′
f'
f
′
and
g
′
g'
g
′
is given |
x
x
x
|
f
(
x
)
f(x)
f
(
x
)
|
g
(
x
)
g(x)
g
(
x
)
|
f
′
(
x
)
f'(x)
f
′
(
x
)
|
g
′
(
x
)
g'(x)
g
′
(
x
)
| |---|---|---|---|---| | 2 | 1 | 6 |
−
3
-3
−
3
| 4 | | 4 | 3 | 4 | 5 |
−
6
-6
−
6
| | 6 | 5 | 2 |
−
4
-4
−
4
| 7 |
Q556
#556
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
If
r
(
x
)
=
f
[
g
(
x
)
]
r(x) = f[g(x)]
r
(
x
)
=
f
[
g
(
x
)]
find
r
′
(
2
)
r'(2)
r
′
(
2
)
.
Show model answer
[QEx 1.1 Q.4 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q557
#557
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
If
R
(
x
)
=
g
[
3
+
f
(
x
)
]
R(x) = g[3 + f(x)]
R
(
x
)
=
g
[
3
+
f
(
x
)]
find
R
′
(
4
)
R'(4)
R
′
(
4
)
.
Show model answer
[QEx 1.1 Q.4 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q558
#558
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
If
s
(
x
)
=
f
[
9
−
f
(
x
)
]
s(x) = f[9 - f(x)]
s
(
x
)
=
f
[
9
−
f
(
x
)]
find
s
′
(
4
)
s'(4)
s
′
(
4
)
.
Show model answer
[QEx 1.1 Q.4 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q559
#559
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
If
S
(
x
)
=
g
[
g
(
x
)
]
S(x) = g[g(x)]
S
(
x
)
=
g
[
g
(
x
)]
find
S
′
(
6
)
S'(6)
S
′
(
6
)
.
Show model answer
[QEx 1.1 Q.4 (iv) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q560
#560
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
Assume that
f
′
(
3
)
=
−
1
f'(3) = -1
f
′
(
3
)
=
−
1
,
g
′
(
2
)
=
5
g'(2) = 5
g
′
(
2
)
=
5
,
g
(
2
)
=
3
g(2) = 3
g
(
2
)
=
3
and
y
=
f
[
g
(
x
)
]
y = f[g(x)]
y
=
f
[
g
(
x
)]
then
[
d
y
d
x
]
x
=
2
=
?
\left[\frac{dy}{dx}\right]_{x=2} = ?
[
d
x
d
y
]
x
=
2
=
?
Show model answer
[QEx 1.1 Q.5 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q561
#561
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
If
h
(
x
)
=
4
f
(
x
)
+
3
g
(
x
)
h(x) = \sqrt{4f(x) + 3g(x)}
h
(
x
)
=
4
f
(
x
)
+
3
g
(
x
)
,
f
(
1
)
=
4
f(1) = 4
f
(
1
)
=
4
,
g
(
1
)
=
3
g(1) = 3
g
(
1
)
=
3
,
f
′
(
1
)
=
3
f'(1) = 3
f
′
(
1
)
=
3
,
g
′
(
1
)
=
4
g'(1) = 4
g
′
(
1
)
=
4
find
h
′
(
1
)
h'(1)
h
′
(
1
)
.
Show model answer
[QEx 1.1 Q.6 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q562
#562
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Hard
Add
Find the
x
x
x
co-ordinates of all the points on the curve
y
=
sin
2
x
−
2
sin
x
y = \sin 2x - 2 \sin x
y
=
sin
2
x
−
2
sin
x
,
0
≤
x
<
2
π
0 \leq x < 2\pi
0
≤
x
<
2
π
where
d
y
d
x
=
0
\frac{dy}{dx} = 0
d
x
d
y
=
0
.
Show model answer
[QEx 1.1 Q.7 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q563
#563
Mathematics → Differentiation → Derivatives of Composite Functions (Chain Rule)
·
Moderate
Add
Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity] "Let
f
(
x
)
=
x
2
+
5
f(x) = x^2 + 5
f
(
x
)
=
x
2
+
5
and
g
(
x
)
=
e
x
+
3
g(x) = e^x + 3
g
(
x
)
=
e
x
+
3
then
f
[
g
(
x
)
]
=
f[g(x)] =
f
[
g
(
x
)]
=
_____ and
g
[
f
(
x
)
]
=
g[f(x)] =
g
[
f
(
x
)]
=
_____. Now
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
_____ and
g
′
(
x
)
=
g'(x) =
g
′
(
x
)
=
_____. The derivative of
f
[
g
(
x
)
]
f[g(x)]
f
[
g
(
x
)]
w. r. t.
x
x
x
in terms of
f
f
f
and
g
g
g
is _____. Therefore
d
d
x
[
f
[
g
(
x
)
]
]
=
\frac{d}{dx}\left[f[g(x)]\right] =
d
x
d
[
f
[
g
(
x
)]
]
=
_____ and
[
d
d
x
[
f
[
g
(
x
)
]
]
]
x
=
0
=
\left[\frac{d}{dx}\left[f[g(x)]\right]\right]_{x=0} =
[
d
x
d
[
f
[
g
(
x
)]
]
]
x
=
0
=
_____. The derivative of
g
[
f
(
x
)
]
g[f(x)]
g
[
f
(
x
)]
w. r. t.
x
x
x
in terms of
f
f
f
and
g
g
g
is _____. Therefore
d
d
x
[
g
[
f
(
x
)
]
]
=
\frac{d}{dx}\left[g[f(x)]\right] =
d
x
d
[
g
[
f
(
x
)]
]
=
_____ and
[
d
d
x
[
g
[
f
(
x
)
]
]
]
x
=
−
1
=
\left[\frac{d}{dx}\left[g[f(x)]\right]\right]_{x=-1} =
[
d
x
d
[
g
[
f
(
x
)]
]
]
x
=
−
1
=
_____." Hint basket :
{
f
′
[
g
(
x
)
]
⋅
g
′
(
x
)
,
2
e
2
x
+
6
e
x
,
8
,
g
′
[
f
(
x
)
]
⋅
f
′
(
x
)
,
2
x
e
x
2
+
5
,
−
2
e
6
,
e
2
x
+
6
e
x
+
14
,
e
x
2
+
5
+
3
,
2
x
,
e
x
}
\left\{ f'[g(x)] \cdot g'(x),\ 2e^{2x} + 6e^x,\ 8,\ g'[f(x)] \cdot f'(x),\ 2xe^{x^2 + 5},\ -2e^6,\ e^{2x} + 6e^x + 14,\ e^{x^2 + 5} + 3,\ 2x,\ e^x \right\}
{
f
′
[
g
(
x
)]
⋅
g
′
(
x
)
,
2
e
2
x
+
6
e
x
,
8
,
g
′
[
f
(
x
)]
⋅
f
′
(
x
)
,
2
x
e
x
2
+
5
,
−
2
e
6
,
e
2
x
+
6
e
x
+
14
,
e
x
2
+
5
+
3
,
2
x
,
e
x
}
Show model answer
[QEx 1.1 Q.8 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Set · 3 questions
Find the derivative of the function
y
=
f
(
x
)
y = f(x)
y
=
f
(
x
)
using the derivative of the inverse function
x
=
f
−
1
(
y
)
x = f^{-1}(y)
x
=
f
−
1
(
y
)
in the following
Q564
#564
Mathematics → Differentiation → Derivatives of Inverse Functions
·
Easy
Add
y
=
x
+
4
3
y = \sqrt[3]{x + 4}
y
=
3
x
+
4
Show model answer
[Q1.2.3 SolvedEx.1 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q565
#565
Mathematics → Differentiation → Derivatives of Inverse Functions
·
Moderate
Add
y
=
1
+
x
y = \sqrt{1 + \sqrt{x}}
y
=
1
+
x
Show model answer
[Q1.2.3 SolvedEx.1 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q566
#566
Mathematics → Differentiation → Derivatives of Inverse Functions
·
Easy
Add
y
=
log
x
y = \log x
y
=
lo
g
x
Show model answer
[Q1.2.3 SolvedEx.1 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q567
#567
Mathematics → Differentiation → Derivatives of Inverse Functions
·
Moderate
Add
Find the derivative of the inverse of function
y
=
2
x
3
−
6
x
y = 2x^3 - 6x
y
=
2
x
3
−
6
x
and calculate its value at
x
=
−
2
x = -2
x
=
−
2
.
Show model answer
[Q1.2.3 SolvedEx.2 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Q568
#568
Mathematics → Differentiation → Derivatives of Inverse Functions
·
Moderate
Add
Let
f
f
f
and
g
g
g
be the inverse functions of each other. The following table lists a few values of
f
f
f
,
g
g
g
and
f
′
f'
f
′
|
x
x
x
|
f
(
x
)
f(x)
f
(
x
)
|
g
(
x
)
g(x)
g
(
x
)
|
f
′
(
x
)
f'(x)
f
′
(
x
)
| |---|---|---|---| |
−
4
-4
−
4
|
2
2
2
|
1
1
1
|
1
3
\frac{1}{3}
3
1
| |
1
1
1
|
−
4
-4
−
4
|
−
2
-2
−
2
|
4
4
4
| find
g
′
(
−
4
)
g'(-4)
g
′
(
−
4
)
.
Show model answer
[Q1.2.3 SolvedEx.3 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q569
#569
Mathematics → Differentiation → Derivatives of Inverse Functions
·
Moderate
Add
Let
f
(
x
)
=
x
5
+
2
x
−
3
f(x) = x^5 + 2x - 3
f
(
x
)
=
x
5
+
2
x
−
3
. Find
(
f
−
1
)
′
(
−
3
)
\left(f^{-1}\right)'(-3)
(
f
−
1
)
′
(
−
3
)
.
Show model answer
[Q1.2.3 SolvedEx.4 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q570
#570
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Moderate
Add
Using derivative prove that
sin
−
1
x
+
cos
−
1
x
=
π
2
\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}
sin
−
1
x
+
cos
−
1
x
=
2
π
.
Show model answer
[Q1.2.6 SolvedEx.1 · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
Report
Set · 5 questions
Differentiate the following w. r. t. x.
Q571
#571
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Easy
Add
sin
−
1
(
x
3
)
\sin^{-1}(x^3)
sin
−
1
(
x
3
)
Show model answer
[Q1.2.6 SolvedEx.2 (i) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q572
#572
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Moderate
Add
cos
−
1
(
2
x
2
−
x
)
\cos^{-1}(2x^2-x)
cos
−
1
(
2
x
2
−
x
)
Show model answer
[Q1.2.6 SolvedEx.2 (ii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q573
#573
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Easy
Add
sin
−
1
(
2
x
)
\sin^{-1}(2^x)
sin
−
1
(
2
x
)
Show model answer
[Q1.2.6 SolvedEx.2 (iii) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q574
#574
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Easy
Add
cot
−
1
(
1
x
2
)
\cot^{-1}\left(\frac{1}{x^2}\right)
cot
−
1
(
x
2
1
)
Show model answer
[Q1.2.6 SolvedEx.2 (iv) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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Q575
#575
Mathematics → Differentiation → Derivatives of Inverse Trigonometric Functions
·
Moderate
Add
cos
−
1
(
1
+
x
2
)
\cos^{-1}\left(\sqrt{\frac{1+x}{2}}\right)
cos
−
1
(
2
1
+
x
)
Show model answer
[Q1.2.6 SolvedEx.2 (v) · Maharashtra State Board (Class 12) — Differentiation (Balbharati textbook)]
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