CLASS : XIIth SUBJECT : MATHS

DATE : SOLUTIONS DPP NO. :4

Topic :-DIFFERENTITATION
1 (b)

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𝑦 𝑦1 𝑦2
Let 𝐷 = 𝑦3 𝑦4 𝑦5
𝑦6 𝑦7 𝑦8 |
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― 𝑝2 sin 𝑝𝑥
|
sin 𝑝𝑥 𝑝 cos 𝑝𝑥
⇒𝐷 = ― 𝑝3 cos 𝑝𝑥 𝑝4 sin 𝑝𝑥 𝑝5 cos 𝑝𝑥
― 𝑝6 sin 𝑝𝑥 ― 𝑝7 cos 𝑝𝑥 𝑝8 sin 𝑝𝑥
Taking 𝑝3 and 𝑝6common from 𝑅2 and 𝑅3row

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― 𝑝2 sin 𝑝𝑥
|
sin 𝑝𝑥 𝑝 cos 𝑝𝑥
= 𝑝9 ― cos 𝑝𝑥 𝑝 sin 𝑝𝑥 𝑝2 cos 𝑝𝑥
― sin 𝑝𝑥 ―𝑝 cos 𝑝𝑥 𝑝2 sin 𝑝𝑥

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― 𝑝2 sin 𝑝𝑥
|
sin 𝑝𝑥 𝑝 cos 𝑝𝑥
=―𝑝 ― cos 𝑝𝑥 9
𝑝 sin 𝑝𝑥 𝑝2 cos 𝑝𝑥
sin 𝑝𝑥 𝑝 cos 𝑝𝑥 ― 𝑝2 sin 𝑝𝑥
= 0 (𝑅1and 𝑅3rows are identical)
2 (d)
𝑥2 𝑥3
𝑦=1―𝑥+ ― + ...
2! 3!
―𝑥
⇒𝑦 = 𝑒
On differentiating w.r.t. 𝑥, we get
𝑑𝑦
= ― 𝑒―𝑥
𝑑𝑥
Again differentiating w.r.t. 𝑥, we get
𝑑2𝑦
= ― 𝑒―𝑥( ―1) = 𝑒―𝑥 = 𝑦
𝑑𝑥2
3 (b)
We have, 𝑓(4) = 4 and 𝑓′(4) = 1
―𝑓′(𝑥)
2 ― 𝑓(𝑥) 2 𝑓(𝑥)
∴ lim = lim [Using L′Hospital′s Rule]
𝑥→4 2 ― 𝑥 𝑥→4 ― 1
2 𝑥
2 ― 𝑓(𝑥) 𝑥 𝑓′(𝑥) 2𝑓′(4) 2
⇒ lim = lim = = =1
𝑥→4 2 ― 𝑥 𝑥→4 𝑓(𝑥) 𝑓(4) 2
4 (c)
2𝑎𝑡 2𝑎𝑡2
∵ 𝑥= and 𝑦 =
1 + 𝑡3 (1 + 𝑡3)2
∴ 2𝑎𝑦 = 𝑥2
𝑑𝑦 𝑥
⇒ =
𝑑𝑥 𝑎
5 (c)
―1
Given, 𝑦 = 𝑒𝑎 sin 𝑥
On differentiating w.r.t. 𝑥, we get
―1 1
𝑦1 = 𝑒𝑎 sin 𝑥
𝑎. 1 ― 𝑥2

⇒ 𝑦1 1 ― 𝑥2 = 𝑎𝑦
⇒ (1 ― 𝑥2)𝑦21 = 𝑎2𝑦2
Again, differentiating w.r.t. 𝑥, we get
(1 ― 𝑥2)2𝑦1𝑦2 ―2𝑥𝑦21 = 𝑎22𝑦𝑦1
⇒(1 ― 𝑥2)𝑦2 ― 𝑥𝑦1 ― 𝑎2𝑦 = 0
Using Leibnitz’s rule,
(1 ― 𝑥2)𝑦𝑛+2 + 𝑛𝐶1𝑦𝑛+1( ―2𝑥) + 𝑛𝐶2𝑦𝑛( ― 2)
―𝑥𝑦𝑛+1 ― 𝑛𝑐1𝑦𝑛 ― 𝑎2𝑦𝑛 = 0
⇒ (1 ― 𝑥2)𝑦𝑛+2 +𝑥𝑦𝑛+1( ― 2𝑛 ― 1)
+ 𝑦𝑛[ ―𝑛(𝑛 ― 1) ― 𝑛 ― 𝑎2] = 0
⇒ (1 ― 𝑥2)𝑦𝑛+2 ― (2𝑛 + 1)𝑥𝑦𝑛+1 = (𝑛2 + 𝑎2)𝑦𝑛
6 (a)
𝑥―𝑦
Since, 𝑥+𝑦 = sec―1 𝑎
(𝑥 + 𝑦)(1 ― ) ― (𝑥 ― 𝑦)(1 + )
𝑑𝑦 𝑑𝑦

⇒ 𝑑𝑥
2
𝑑𝑥
=0
(𝑥 + 𝑦)
𝑑𝑦 𝑑𝑦 𝑦
⇒𝑥 + 𝑦 ― 𝑥 + 𝑦 ― (𝑥 + 𝑦 + 𝑥 ― 𝑦) = 0⇒ =
𝑑𝑥 𝑑𝑥 𝑥
7 (c)
𝑥
Given, 𝑦 = 𝑥 + 𝑥2 + 𝑥3 +...⇒𝑦 = 1 ― 𝑥
𝑦
⇒𝑥= = 𝑦 ― 𝑦2 + 𝑦3 ― ...
1+𝑦
On differentiating w.r.t. 𝑦, we get
𝑑𝑥
= 1 ― 2𝑦 + 3𝑦2 ― ...
𝑑𝑦
8 (b)
1 ― sin 2𝑥 cos 𝑥 ― sin 𝑥
Let 𝑦 = 1 + sin 2𝑥
= cos 𝑥 + sin 𝑥
1 ― tan 𝑥 𝜋
=
1 + tan 𝑥 (
= tan ― 𝑥
4
)
 𝑑𝑦
= ― sec2 ( ― 𝑥)
𝜋
⇒ 𝑑𝑥 4

9 (b)
Let 𝑢 = log10 𝑥 and 𝑣 = 𝑥2
𝑑𝑢 log10 𝑒 𝑑𝑣
∴ = and = 2𝑥
𝑑𝑥 𝑥 𝑑𝑥
𝑑𝑢 𝑑𝑢 /𝑑𝑥 log10 𝑒
∴ 𝑑𝑣
= 𝑑𝑣 /𝑑𝑥 = 𝑥
/2𝑥
log10 𝑒
=
2𝑥2
10 (d)
∵ 𝑦 = 𝑥 In (𝑎 +𝑥 𝑏𝑥) = 𝑥(In 𝑥 ― In(𝑎 + 𝑏𝑥))
𝑦
or ( ) = In 𝑥 ― In (𝑎 + 𝑏𝑥)
𝑥
On differentiating both sides w.r.t. 𝑥, we get
𝑑𝑦
𝑥
𝑑𝑥
𝑥2
― 𝑦.1
=
(1
―
𝑏
=
𝑎
)
𝑥 𝑎 + 𝑏𝑥 𝑥(𝑎 + 𝑏𝑥)
…(i)

𝑎𝑥
or (𝑥 ― 𝑦) = 𝑎 + 𝑏𝑥
𝑑𝑦
𝑑𝑥
On taking log on both sides, we get
𝑑𝑦
𝑑𝑥 (
In 𝑥 )
― 𝑦 = In (𝑎𝑥) ― In (𝑎 + 𝑏𝑥)

On differentiating both sides w.r.t. 𝑥, we get

𝑑2𝑦 𝑑𝑦 𝑑𝑦
𝑥 2+ ―
𝑑𝑥 𝑑𝑥 𝑑𝑥 1 𝑏 𝑎
= ― =
𝑑𝑦
𝑥
𝑑𝑥
―𝑦 ( )
𝑥 𝑎 + 𝑏𝑥 𝑥(𝑎 + 𝑏𝑥)

𝑑𝑦
𝑥
𝑑𝑥
―𝑦 ( )
= [from Eq.(i)]
𝑥2
𝑑2𝑦 2
or 𝑥3𝑑𝑥2 = (𝑥 ― 𝑦)
𝑑𝑦
𝑑𝑥
12 (a)
On differentiating given equation w.r.t. 𝑥, we get
𝑑𝑦 𝑑𝑦 𝑑𝑦
4𝑥 ― 3𝑥 ― 3𝑦 + 2𝑦 +1+2 ―0=0
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑦 3𝑦 ― 4𝑥 ― 1
⇒ =
𝑑𝑥 2𝑦 ― 3𝑥 + 2
13 (c)
Given, 𝑦 = sin 𝑥 + 𝑦 ⇒ 𝑦2 = sin 𝑥 + 𝑦
𝑑𝑦 𝑑𝑦
⇒2𝑦 = cos 𝑥 +
𝑑𝑥 𝑑𝑥
 𝑑𝑦
⇒(2𝑦 ― 1) = cos 𝑥
𝑑𝑥
14 (b)
We have,
2𝑥 + 2𝑦 = 2𝑥+𝑦
Differentiating with respect to 𝑥, we get
𝑑𝑦 𝑑𝑦
2𝑥 log 2 + 2𝑦 log 2
𝑑𝑥 (
= 2𝑥+𝑦 log 2 1 +
𝑑𝑥)
𝑑𝑦 𝑑𝑦
⇒2𝑥 + 2𝑦
𝑑𝑥 (
= 2𝑥+𝑦 1 +
𝑑𝑥 )
𝑑𝑦 2 ― 2𝑥 𝑥+𝑦 𝑑𝑦 2―4
⇒
𝑑𝑥 2
= 𝑥+𝑦 ⇒ ( ) =
― 2𝑦 𝑑𝑥 (1, 0) 4 ― 2
= ―1

15 (a)
Given, 𝑦 = sec( tan―1 𝑥)
⇒ 𝑦 = sec( sec―1 1 + 𝑥2) = 1 + 𝑥2
On differentiating w.r.t. 𝑥,we get
𝑑𝑦 1 𝑥
= (2𝑥) =
𝑑𝑥 2 1 + 𝑥 2
1 + 𝑥2
16 (b)
Given, 𝑓(𝑥) = (𝑥 ― 7)2(𝑥 ― 2)7
⇒ 𝑓(𝜃) = (𝜃 ― 7)2(𝜃 ― 2)7
On differentiating w.r.t. 𝜃, we get
⇒ 𝑓′(𝜃) = 2(𝜃 ― 7)(𝜃 ― 2)7 + 7(𝜃 ― 2)6(𝜃 ― 7)2
put 𝑓′(𝜃) = 0
⇒ (𝜃 ― 7)(𝜃 ― 2)6[2(𝜃 ― 2) + 7(𝜃 ― 7)] = 0
53
⇒ 9𝜃= 53⇒𝜃 = 9
17 (b)
We have,
1 1
𝑥2 + 𝑦2 = 𝑡 ― 𝑡 and 𝑥4 + 𝑦4 = 𝑡2 + 𝑡2
2 1
⇒(𝑥2 + 𝑦2) = 𝑡2 + ―2
𝑡2
2
⇒(𝑥2 + 𝑦2) = 𝑥4 + 𝑦4 ― 2
⇒2𝑥2𝑦2 = ―2
⇒𝑥2𝑦2 = ―1
1 𝑑𝑦 2 𝑑𝑦
⇒𝑦2 = ― 2 ⇒2𝑦 = 3 ⇒𝑥3𝑦 =1
𝑥 𝑑𝑥 𝑥 𝑑𝑥
18 (b)
Let 𝑓(𝑥) = |𝑥 ― 1| +|𝑥 ― 5|

{
―2𝑥 + 6, 𝑥<1
⇒𝑓(𝑥) = 4 1≤𝑥<5
2𝑥 ― 6 𝑥≥5
 {
𝑑 ―2, 𝑥<1
∴ (𝑓(𝑥)) = 0 1<𝑥<5
𝑑𝑥 2 𝑥>5
Hence, ( (𝑓(𝑥)))𝑥=3 = 0
𝑑
𝑑𝑥
19 (b)
π
sin―1 𝑥 + sin―1 𝑦 =
2
―1
π ―1
⇒ sin 𝑥 = ― sin 𝑦
2
⇒sin―1 𝑥 = cos―1 𝑦
⇒𝑦 = 1 ― 𝑥2
On differentiating w.r.t. 𝑥, we get
𝑑𝑦 1 𝑥
= ( ―2𝑥) = ―
𝑑𝑥 2 1 ― 𝑥2 𝑦
20 (c)

𝑓[𝑓(𝑓(𝑥))] = 𝑓 𝑓[ (1 ―1 𝑥)] = 𝑓(1 ― 1 1 )

1―𝑥

[ ∵ 𝑓(𝑥) = 1 ―1 𝑥]
1―𝑥 1
= 𝑓( ) =
1―𝑥
―𝑥 1+( )
𝑥
⇒ 𝑓[𝑓(𝑓(𝑥))] = 𝑥
∴ The derivative of composite function is equal to 1.
 ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. B D B C C A C B B D

Q. 11 12 13 14 15 16 17 18 19 20
A. D A C B A B B B B C