7KENDRIYA VIDYALAYA SANGTHAN, AHMEDABAD REGION
CLASS XII
STUDY MATERIAL
CONTINUITY
𝑎𝑥 + 1, 𝑖𝑓 𝑥 ≤ 3
is continuous at x=3, find a relation between a and b.
𝑏𝑥 + 3, 𝑖𝑓 𝑥 > 3
𝑎𝑥 + 1, 𝑖𝑓 𝑥 ≤ 3
Solution: As f(x)= {
is continuous at x=3
𝑏𝑥 + 3, 𝑖𝑓 𝑥 > 3
1. If f(x)= {
∴ lim+ 𝑓(𝑥)= f(3)
𝑥→3
3b+3 = 3a + 1
a-b = 2/3
𝑘(𝑥 2 − 2𝑥), 𝑥 ≤ 0
is continuous at x=0, find the value of k, if possible. Also
4𝑥 + 1,
𝑥>0
discuss continuity at x=1.
𝑘(𝑥 2 − 2𝑥), 𝑥 ≤ 0
Solution: As, f(x)={
is continuous at x=0
4𝑥 + 1,
𝑥>0
∴ lim+ 𝑓(𝑥)= f(0)
2. f(x)={
𝑥→0
1= k(0)
1=0 which is not possible for any value of k
lim 𝑓(𝑥)= f(1)
𝑥→1
5=5
for all values of k , f(x) is continuous at x = 1
𝑘𝑐𝑜𝑠𝑥
3. If
f(x)= {
𝜋−2𝑥
,
3,
𝑖𝑓 𝑥 =
Solution:
𝜋
lim𝜋 𝑓(𝑥)= f( 2 )
𝑥→
2
.lim𝜋
𝑥→
2
𝜋
2
𝜋
2( −𝑥)
2
𝑘𝑠𝑖𝑛( −𝑥)
k/2 = 3k = 6
𝑖𝑓 𝑥 ≠
=3
𝜋
2
𝜋
2
𝜋
is continuous at x=2 , find the value of k.
𝑘𝑥 + 1, 𝑖𝑓 𝑥 ≤ 𝜋
4. If, f(x)= {
is continuous at x=𝜋, find the value of k.
𝑐𝑜𝑠𝑥,
𝑖𝑓 𝑥 > 𝜋
Solution:
𝑘𝑥 + 1, 𝑖𝑓 𝑥 ≤ 𝜋
As f(x)= {
is continuous at x=𝜋
𝑐𝑜𝑠𝑥,
𝑖𝑓 𝑥 > 𝜋
∴ lim+ 𝑓(𝑥)= f(𝜋)
𝑥→𝜋
. lim+𝑐𝑜𝑠𝑥 = k 𝜋+1
𝑥→𝜋
-1 = k 𝜋+1
K = -2/ 𝜋
5, 𝑖𝑓 𝑥 ≤ 2
5. If, f(x)= {𝑎𝑥 + 𝑏, 2 < 𝑥 < 10 is continuous function. Find the values of a and b.
21, 𝑖𝑓 𝑥 ≥ 10
Solution:
5, 𝑖𝑓 𝑥 ≤ 2
As f(x)= {𝑎𝑥 + 𝑏, 2 < 𝑥 < 10is continuous function.
21, 𝑖𝑓 𝑥 ≥ 10
∴ lim+ 𝑓(𝑥)= f(2)
𝑥→2
∴ lim+ 𝑎𝑥 + 𝑏=5
𝑥→2
2a+b = 5…(1)
∴ lim− 𝑓(𝑥)= f(10)
𝑥→10
∴ lim− 𝑎𝑥 + 𝑏 = 21
𝑥→10
10a+b = 21… (2)
From equation 1 and 2,
1−𝑐𝑜𝑠𝑥
6. If,
f(x)={
𝑘𝑥 2
,
3,
a=2, b=1
𝑖𝑓 𝑥 ≠ 0
𝑖𝑓 𝑥 = 0
is continuous at x=0, find the value of k.
Solution:
1−𝑐𝑜𝑠𝑥
As, f(x)={
𝑘𝑥 2
3,
,
𝑖𝑓 𝑥 ≠ 0
𝑖𝑓 𝑥 = 0
is continuous at x=0
lim 𝑓(𝑥)= f(0)
𝑥→0
.lim
𝑥
2
𝑥2
2𝑠𝑖𝑛2
𝑥→0 4𝑘
=3
4
1
.2𝑘 = 3
K =1/6
𝑥 5 −25
7. If,
f(x)= {
𝑥−2
,
𝑖𝑓 𝑥 ≠ 2
𝑘,
𝑖𝑓 𝑥 = 2
is continuous at x= . Fnd the value of k.
Solution:
𝑥 5 −25
As f(x)= {
,
𝑥−2
𝑖𝑓 𝑥 ≠ 2
𝑘,
𝑖𝑓 𝑥 = 2
is continuous at x=
∴ lim 𝑓(𝑥)= f(2)
𝑥→2
. lim
𝑥 5 −25
=k
𝑥−2
𝑥→2
5(24) = k
K = 80
sin(𝑥−2)
8. If, f(x)= {
𝑥−2
𝑘,
,
𝑖𝑓 𝑥 ≠ 2
𝑖𝑓 𝑥 = 2
is continuous at x= . Find the value of k.
Solution:
sin(𝑥−2)
As f(x)= {
𝑥−2
𝑘,
,
𝑖𝑓 𝑥 ≠ 2
𝑖𝑓 𝑥 = 2
is continuous at x=
∴ lim 𝑓(𝑥)= f(2)
𝑥→2
.lim
sin(𝑥−2)
𝑥→2
𝑥−2
=k
K=1
𝑒 𝑥 −1
9. If, f(x)= { 𝑥
𝑘,
Solution:
,
𝑖𝑓 𝑥 ≠ 0
𝑖𝑓 𝑥 = 0
is continuous at x=0. Find the value of k
𝑒 𝑥 −1
As f(x)= { 𝑥
𝑘,
,
𝑖𝑓 𝑥 ≠ 0
𝑖𝑓 𝑥 = 0
is continuous at x=0
∴ lim 𝑓(𝑥)= f(0)
𝑥→0
.lim
𝑒 𝑥 −1
𝑥→0
𝑥
=k
1=k
10. Show that f(x)= |𝑥 − 3| is continuous but not differentiable at x=3
Solution:
As, f(x)= {
−𝑥 + 3 𝑖𝑓 𝑥 < 3
𝑥 − 3, 𝑖𝑓 𝑥 ≥ 3
at x=3
lim 𝑓(𝑥)= lim+ 𝑓(𝑥) =f(3)
𝑥→3−
𝑥→3
lim -(3-h)+3 = lim (3+h)-3= 0
ℎ→0
ℎ→0
0=0=0
∴ continuous at x=3
. 𝐿. 𝐻. 𝐷 = lim
ℎ→0
R.H.D = lim
ℎ→0
𝑓(3−ℎ)−𝑓(3)
−ℎ
𝑓(3+ℎ)−𝑓(3)
ℎ
=
= lim
ℎ→0
lim
−3+ℎ+3
ℎ→0
−ℎ
3+ℎ−3
=1
ℎ
As L.H.D ≠ R.H.D
∴ f(x) is not differentiable at x=3.
= -1
QUESTION BANK HIGHER ORDER THINKING SKILL
1
Solution:
2
Solution:
3
Solution:
4
Solutions:
5
Solution:
6
Solution:
7
Solution:
8
Solution:
9
Solution:
10
Solution:
DIFFERENTIABILITY
QUESTIONS ON HIGHER ORDER THINKING SKILL
1.
𝜃
𝑑𝑦
2
𝑑𝜃
If x = a(cos𝜃 + log tan ) and y =a sin 𝜃, find the value of
Soln: Y = a sin𝜃
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
= a cos𝜃
= a(-sin𝜃 +
𝜃
2
𝜃
𝑠𝑖𝑛
2
𝑐𝑜𝑠
.
1
𝜃
2
𝑐𝑜𝑠 𝑐𝑜𝑠
1
𝜃
2
. 2)
𝜋
, at 𝜃 = .
4
1−𝑠𝑖𝑛2 𝜃
)
𝑠𝑖𝑛𝜃
= a(-sin𝜃 +
1
)
𝑠𝑖𝑛𝜃
𝑑𝑦
𝑑𝑥
𝑠𝑖𝑛𝜃
𝑎 𝑐𝑜𝑠𝜃.𝑐𝑜𝑠𝜃
=
𝑎 𝑐𝑜𝑠𝜃
1
𝑑𝑦
(𝑑𝜃)
𝜃=
∗
=a(
=
𝑎 𝑐𝑜𝑠2 𝜃
𝑠𝑖𝑛𝜃
= tan𝜃
=1
𝜋
4
𝑑𝑦
2. Find,𝑑𝑥 , when a function is given by
x=
𝑠𝑖𝑛3 𝑡
√𝑐𝑜𝑠2𝑡
, y=
√𝑐𝑜𝑠2𝑡
SOLN: x=
𝑑𝑥
. 𝑑𝑡 =
𝑐𝑜𝑠3 𝑡
𝑠𝑖𝑛3 𝑡
√𝑐𝑜𝑠2𝑡
, y=
𝑐𝑜𝑠3 𝑡
√𝑐𝑜𝑠2𝑡
𝑠𝑖𝑛2 𝑡 𝑐𝑜𝑠𝑡(2𝑐𝑜𝑠2𝑡+1) 𝑑𝑦 𝑐𝑜𝑠2 𝑡 𝑠𝑖𝑛𝑡(1−2𝑐𝑜𝑠2𝑡)
(𝑐𝑜𝑠2𝑡)3/2
, 𝑑𝑡 =
(𝑐𝑜𝑠2𝑡)3/2
𝑑𝑦
.𝑑𝑥 = cot t
𝑑𝑦
−1
3. if 𝑥√1 + 𝑦 + 𝑦√1 + 𝑥=0 then prove that𝑑𝑥 = (1+𝑥)2
1. soln: 𝑥√1 + 𝑦 + 𝑦√1 + 𝑥=0
.𝑥√1 + 𝑦= -𝑦√1 + 𝑥
−𝑥
. y=1+𝑥
𝑑𝑦
−1
.𝑑𝑥 =(1+𝑥)2
4. if (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑐 2 then prove that
3
𝑑𝑦 2
) ]2
𝑑𝑥
𝑑2 𝑦
𝑑𝑥2
[1+(
soln: (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑐 2
𝑑2 𝑦
1+(y-b)
𝑑𝑥 2
𝑑𝑦
+( )2 =0
𝑑𝑥
𝑑2 𝑦
.𝑑𝑥 2 = -(1+y12)
.
3
𝑑𝑦 2
) ]2
𝑑𝑥
𝑑2 𝑦
𝑑𝑥2
[1+(
=-
(1+𝑦12 )3/2
(1+𝑦12 )
= - (1 + 𝑦12 )1/2=-(𝑐 2 )1/2= -c
5. If y = sin−1 𝑥 , prove that ( 1 - 𝑥 2 ) 𝑦2 - x 𝑦1 = 0.
is independent of a and b
Soln: If y = sin−1 𝑥 ,
𝑦1 =
1
=>
√1− 𝑥 2
2
(1 − 𝑥 2 )𝑦1 2 = 1
 ( 1 - 𝑥 ) 𝑦2 - x 𝑦1 = 0.
𝑑2 𝑦
𝑑𝑦
6. If y= sin(log 𝑥) prove that 𝑥 2 𝑑𝑥 2 + 𝑥 𝑑𝑥 + 𝑦 = 0
Soln: We have 𝑦 = sin(log 𝑥)
Diff. w.r.t x,
𝑑𝑦
𝑥 𝑑𝑥 = cos(log 𝑥)
Again diff. w.r.t x
𝑥
𝑑2 𝑦
𝑑𝑥 2
+
𝑑𝑦
= − sin(log 𝑥) ∙
𝑑𝑥
𝑑2 𝑦
,𝑥 2 𝑑𝑥 2
1
𝑥
𝑑𝑦
+ 𝑥 𝑑𝑥 + 𝑦 = 0
1
1
7. Find dy/dx, if y =eӨ(  - )and x =e-Ө (  + )


Hence
Soln: Differentiating ‘y’ w.r.t.  we get
dy
1 
1


 e 1  2   e    
d

  


  3  2  1 
dy
 e 
      1
d
2


Differentiating ‘x’ w.r.t. 
dx
1 
1


 e  1  2   e     
d

  


8.

  3   2    1 
dx
 e 
       2
d
2


dy  e ( 3   2    1 
 , from(1)and (2)

dx  e    3   2    1 


If y = x + tanx than prove that Cos2x
Soln:
d2y
– 2y +2x = 0
dx 2
dy
d2y
 1  sec2 x  2  2sec x(sec x tan x)
dx
dx
= 2sec2 x tan x
Now cos 2 x
d2y
 2 y  2 x  cos 2 x(2sec 2 x tan x)  2( x  tan x)  2 x
2
dx
=0
9.
If y= sin−1 [
5𝑥+12√1−𝑥 2
13
soln: y = sin−1 [5𝑥
+
13
], find
𝑑𝑦
𝑑𝑥
12√1−𝑥 2
13
.
]
Y = sin−1 [sin 𝛼. 𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠𝛼. 𝑠𝑖𝑛𝜃]
Y = sin−1 (sin(𝜃 + 𝛼))
Y=𝜃+𝛼
5
Y = cos−1 𝑥 + sin−1 13
𝑑𝑦
𝑑𝑥
=
−1
√1+𝑥 2
√1+𝑥 2 −√1−𝑥 2
10. Differentiate tan−1 √1+𝑥2
+√1−𝑥 2
√1+𝑥 2 −√1−𝑥 2
Soln: y= tan−1 √1+𝑥2
+√1−𝑥 2
√1+𝑥 2 −√1−𝑥 2
y= tan−1 √1+𝑥2
= tan−1 [
𝑑𝑦
1
cos 𝜃−sin 𝜃
cos 𝜃+sin 𝜃
−1
= 2 . √1−𝑥 4 . 2𝑥 =
𝑑𝑥
𝑑𝑧
𝑑𝑥
∴
+√1−𝑥 2
=
𝑑𝑦
𝑑𝑧
−1
√1−𝑥 4
with respect to cos −1 𝑥 2 .
and z= cos −1 𝑥 2 .
, put x2=cos 2𝜃
] = tan−1 [
−1
√1−𝑥4
1−𝑡𝑎𝑛𝜃
1+𝑡𝑎𝑛𝜃
𝜋
𝜋
1
4
4
2
] = tan-1[tan( − 𝜃)]= − 𝜃= cos −1 𝑥 2
. 𝑥.
−2𝑥
. 2𝑥 = √1−𝑥 4 .
1
= 2.
QUESTIONS
1.
𝑑𝑦
if cos y = x cos(a+y) then prove that 𝑑𝑥 =
𝑐𝑜𝑠2 (𝑎+𝑦)
𝑑𝑦
sin 𝑎
𝑐𝑜𝑠2 (𝑎+𝑦)
soln : if cos y = x cos(a+y) then prove that 𝑑𝑥 =
𝑐𝑜𝑠𝑦
X= cos(𝑎+𝑦)
𝑑𝑥
.𝑑𝑦=
−siny cos(a+y)+sin(𝑎+𝑦)𝑐𝑜𝑠𝑦 𝑑𝑦
𝑐𝑜𝑠2 (𝑎+𝑦)
.𝑑𝑥 =
𝑐𝑜𝑠2 (𝑎+𝑦)
sin 𝑎
sin 𝑎
1. if y = 𝑒 𝑎𝑐𝑜𝑠
−1 𝑥
𝑑2 𝑦
𝑑𝑦
then show that (1-x2)𝑑𝑥 2 - x𝑑𝑥 - 𝑎2 𝑦 = 0
solution:
𝑑𝑦
𝑑𝑥
= 𝑒 𝑎𝑐𝑜𝑠
−𝑎
−1 𝑥
√1−𝑥 2
𝑑2 𝑦
𝑥
𝑑𝑦
√1 − 𝑥 2 𝑑𝑥 2 - √1−𝑥2 𝑑𝑥 = 𝑒 𝑎𝑐𝑜𝑠
𝑑2 𝑦
−𝑎2
−1 𝑥
√1−𝑥2
𝑑𝑦
(1-x2)𝑑𝑥 2 - x𝑑𝑥 - 𝑎2 𝑦 = 0
2. If y = 500𝑒 7𝑥 + 600𝑒 −7𝑥 , prove that
𝑑2 𝑦
𝑑𝑥 2
- 49y = 0.
Soln: y = 500𝑒 7𝑥 + 600𝑒 −7𝑥 ,
𝑑𝑦
𝑑𝑥
= 7 ( 500𝑒 7𝑥 - 600𝑒 −7𝑥 )
𝑑2 𝑦
= 49 (500𝑒 7𝑥 + 600𝑒 −7𝑥 ) = 49 y
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑦
3. If y = 3 cos ( log x ) + 4 sin ( log x ) , show that 𝑥 2 𝑑𝑥 2 + x𝑑𝑥 + y = 0.
Soln:
𝑑𝑦
y = 3 cos ( log x ) + 4 sin ( log x ) ,
1
1
= -3 sin log x .𝑥+ 4cos log x . 𝑥
𝑑𝑥
𝑑𝑦
x 𝑑𝑥 = -3 sin log x + 4 cos log x
x
𝑑2 𝑦
𝑑𝑥 2
+
𝑑𝑦
𝑑𝑥
= - ( 3 cos ( log x ) + 4 sin ( log x ) ) = -y
𝑑2 𝑦
𝑑𝑦
𝑥 2 𝑑𝑥 2 + x𝑑𝑥 + y = 0.
Hence ,
4. If y = tan−1 𝑥 , prove that ( 1 + 𝑥 2 ) 𝑦2 + 2x 𝑦1 = 0.
Soln :y = tan−1 𝑥 ,
1
(1 + 𝑥 2 ) 𝑦1 = 1
𝑦1 =
1+ 𝑥 2
=>
(1 + 𝑥 2 ) 𝑦2 + 2x𝑦1 = 0
5. Find
𝑑2 𝑦
𝑑𝑥 2
=>
, if x = a cost , y = b sin t.
Soln: x = a cost , y = b sin t.
𝑑𝑥
𝑑𝑡
𝑑𝑦
= −𝑎 sin 𝑡
,
𝑑𝑦
𝑑𝑡
= 𝑏 cos 𝑡
𝑏
= - 𝑎 cot 𝑡
𝑑𝑥
𝑑2 𝑦
𝑑𝑥 2
=
𝑏
𝑎
𝑐𝑜𝑠𝑒𝑐 2 𝑡 .
𝟐
1
−𝑎 sin 𝑡
𝒅𝒚
6. If y = (log x)cos x + 𝒙𝒙𝟐 +𝟏
, find
−𝟏
𝒅𝒙
Soln: y=u+v
𝑏
= -𝑎2 𝑐𝑜𝑠𝑒𝑐 3 𝑡
log u = cosx log(logx)
1 𝑑𝑢
𝑢 𝑑𝑥
𝑑𝑢
1
1
= 𝑐𝑜𝑠𝑥. log 𝑥 . 𝑥 + log(𝑙𝑜𝑔𝑥) (− sin 𝑥)
1
1
= (log 𝑥)cos 𝑥 (𝑐𝑜𝑠𝑥. log 𝑥 . 𝑥 − log(𝑙𝑜𝑔𝑥) (sin 𝑥))
𝑑𝑥
𝑑𝑣
=
𝑑𝑥
𝑑𝑦
(𝑥 2 −1 ).2𝑥−(𝑥 2 +1).2𝑥
(𝑥 2 −1)2
− 4𝑥
= (𝑥 2 −1)2
1
1
− 4𝑥
= (log 𝑥)cos 𝑥 (𝑐𝑜𝑠𝑥. log 𝑥 . 𝑥 − log(𝑙𝑜𝑔𝑥) (sin 𝑥)) + (𝑥 2 −1)2
𝑑𝑥
7. If xy= 𝒆𝒙−𝒚 , prove that
𝒅𝒚
𝒅𝒙
=
𝐥𝐨𝐠 𝒙
(𝟏+𝐥𝐨𝐠 𝒙)𝟐
Soln : Y log x = x-y
Y(1+log x)=x
Y=
𝑑𝑦
𝑑𝑥
8.
=
(1+log 𝑥).1−𝑥.
1
𝑥
(1+𝑙𝑜𝑔𝑥)2
𝐥𝐨𝐠 𝒙
= (𝟏+𝐥𝐨𝐠
𝒙)𝟐
𝝅
𝒅𝟐 𝒚
𝟐
𝒅𝒕𝟐
If x= a(cos t + t sin t) and y= a(sin t – t cos t) , 0<t< , find
Sol :
𝒅𝒚
𝒅𝒙
𝒅𝟐 𝒙
𝒅𝒕𝟐
𝒅𝟐 𝒚
𝒅𝒕𝟐
𝒅𝟐 𝒚
𝒅𝒙𝟐
𝒅𝒙
𝒅𝒕
,
𝒅𝟐 𝒙
𝒅𝒕𝟐
and
= 𝒂(−𝒔𝒊𝒏 𝒕 + 𝒔𝒊𝒏 𝒕 + 𝒕 𝒄𝒐𝒔 𝒕) = 𝒂𝒕 𝒄𝒐𝒔 𝒕
= 𝒂(𝒄𝒐𝒔𝒕 − 𝒄𝒐𝒔 𝒕 + 𝒕. 𝒔𝒊𝒏𝒕) = 𝒂𝒕 𝒔𝒊𝒏𝒕
𝒅𝒕
𝒅𝒚
9.
𝒙
𝟏+𝒍𝒐𝒈 𝒙
= 𝒕𝒂𝒏 𝒕
= 𝒂(−𝒕 𝒔𝒊𝒏𝒕 + 𝒄𝒐𝒔 𝒕)
= 𝒂(𝒕 𝒄𝒐𝒔 𝒕 + 𝒔𝒊𝒏 𝒕)
= sec2t .
𝒅𝒕
𝒅𝒙
If y = 𝐭𝐚𝐧−𝟏 [
= sec2t .
𝟏
𝒂𝒕 𝒄𝒐𝒔 𝒕
=
√𝟏+𝐬𝐢𝐧𝟐 𝒙+ √𝟏−𝐬𝐢𝐧𝟐 𝒙
𝟏
𝒂𝒕𝒄𝒐𝒔𝟑 𝒕
] , find
.
𝒅𝒚
𝒅𝒙
√𝟏+𝒔𝒊𝒏𝟐𝒙 − √𝟏−𝐬𝐢𝐧 𝟐𝒙
-1 (𝒔𝒊𝒏 𝒙+𝒄𝒐𝒔 𝒙)+ (𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙)
Soln: Y= tan [ (𝒔𝒊𝒏𝒙+
𝒄𝒐𝒔 𝒙)−(𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙)
.
]
= tan-1[𝒕𝒂𝒏 𝒙] = x
𝒅𝒚
𝒅𝒙
= 𝟏.
DIFFERENTIAL AND APPROXIMATIONS
QUS:1.
Using Differential ,find the approximate value of √25.3.
𝒅𝟐 𝒚
𝒅𝒙𝟐
.
SOL
QUS:2.
1
Using Differential ,find the value of (81)4
SOL:
QUS:3
1
Using Differential ,find the approximate value of (0.999)10
SOL:
QUS:4
SOL:
QUS:5
?
SOL:
QUS:6
SOL:
QUS:7
1
Using Differential ,find the value of (81)4
SOL:
QUS:8
SOL:
QUS:9
SOL:
QUS:10
SOL:
INCREASING DECREASING FUNCTIONS
Qn.No
1
Ans
2
Ans
3
Ans
4
Ans
5
Ans
f(x) = sin 𝑥 + cos 𝑥 , 0 ≤ x ≤ 2𝜋
f ’(x) = cos x – sin x ,
f ’(x) = 0 , gives cos x = sin x ,
𝜋 5𝜋
x= ,
4
4
Sign of f ’(x)
+
Nature
increasing
)
-
Decreasing
( , 2𝜋)
+
increasing
Interval
𝜋
[0 , )
4
𝜋 5𝜋
( ,
4 4
5𝜋
4
6
ANS
7
Ans
8
Ans
9
Find the intervals in which the function f given by f(x) = -2𝑥 3 - 9 𝑥 2 - 12𝑥 +1
is strictly decreasing .
Ans
f(x) = -2𝑥 3 - 9 𝑥 2 - 12𝑥 +1
f ’(x) = -6𝑥 2 - 18𝑥 - 12 = -6(𝑥 2 +3x + 2 ) = -6( x +1 ) ( x+2)
Critical points are -1, -2.
Interval
x +1
x+2 -6( x+1 )
Nature
(x+2)
Decreasing
x -2
+
+
increasing
-2 x -1
+
+
Decreasing
x -1
10
Find the intervals in which the function f given by f(x) = 4𝑥 3 – 6 𝑥 2 72𝑥 + 30 is strictly increasing
f(x) = 4𝑥 3 – 6 𝑥 2 - 72𝑥 + 30
f ’(x) = 12 𝑥 2 - 12𝑥 - 72 = 12(𝑥 2 –x – 6 ) = 12( x-3 ) ( x+2)
Critical points are -2,3.
Interval
x-3
x+2
12( x-3 ) ( x+2) Nature
+
increasing
x -2
+
Decreasing
-2 x 3
+
+
+
increasing
x 3
***********************************************************************************
MAXIMA AND MINIMA
1.
Find the point on the curve y2 = 4x which is nearest to the point (2, -8).
Solution: Let P(h, k)be the nearest point on the curve y 2 = 4x to the point (2, −8). Then
𝐷 = √(ℎ − 2)2 + (𝑘 + 8)2
𝐷2 = 𝑍(𝑠𝑎𝑦) = (ℎ − 2)2 + (𝑘 + 8)2 ……….(1)
and k 2 = 4h
………….(2)
k2
2
𝑍 = ( 4 − 2) + (𝑘 + 8)2
By (1) and (2), we get
𝑑𝑍
For critical point 𝑑𝑘 = 0
k2
𝑘 ( − 2) + 2(𝑘 + 8) = 0
4
k 3 + 64 = 0
k = −4
𝑑2 𝑍
Now, 𝑑𝑘 2 )
𝑘=−4
=
3k2
4
)
𝑘=−4
= 12 > 0
So, at 𝑘 = −4, Z has minimum value.
By eq. (2), ℎ = 4
So the nearest point on the curve y 2 = 4x to the point (2, −8) is (4, −4).
2.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a
square from each corner and folding up the flaps to form the box. What should be the side
of the square to be cut off so that the volume of the box is the maximum possible.
Solution:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑥 𝑉(𝑥) = 𝑥 (18 – 2𝑥)2 = 324 𝑥 + 4𝑥 3 – 72𝑥 2
2
𝑉’ (𝑥) = 324 + 12 𝑥 – 144 𝑥
𝐹𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉’ (𝑥) = 0
x
18-x
18-x
or, 324 + 12 𝑥 − 144 𝑥 = 0
2
or, 𝑥 = 9 𝑜𝑟 𝑥 = 3
𝑥 = 9 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 , 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑥 = 3.
𝑉’’ (𝑥 ) = 24 𝑥 – 144.
𝑉’’ (𝑥))𝑥 = 3 = −72 < 0.
Therefore volume of the box is maximum for 𝑥 = 3.
3.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so
that its depth is 2 m and volume is 8𝑚3 . If building of the tank costs Rs 70 per square meters for
the base and Rs 45 per sqarw meter for sides. What is the cost of least expensive tank?
Solution: Let the length and the breadth be 𝑥 𝑎𝑛𝑑 𝑦. Then
Volume of the tank = 8𝑚3 = 2𝑥𝑦 ...............(i)
Let C be the cost of making the tank
𝐶 = 70𝑥𝑦 + 45 × 2(2𝑥 + 2𝑦)
𝐶 = 70𝑥𝑦 + 180(𝑥 + 𝑦)
From equation (i)
4
4
𝐶 = 70𝑥. + 180(𝑥 + )
𝑥
𝑥
4
𝐶 = 280 + 180 (𝑥 + 𝑥) .................(ii)
𝑑𝐶
𝑑𝑥
4
= 180 (1 − 𝑥 2 )
𝑑𝐶
For maxima and minima, 𝑑𝑥 = 0
4
⇒ 180 (1 − 𝑥 2 ) = 0
⇒ 𝑥 =  2 as 𝑥 ≠ −2, 𝑥 = 2
𝑑2 𝐶
8
Now 𝐷𝑥 2 = 180 × 𝑥 3
𝑑2 𝐶
(𝐷𝑥 2 )
𝑥=2
= 180 > 0 ⇒
𝐶 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 𝑥 = 2
By equation (ii)
𝐶𝑥=2 = 𝑅𝑠. 1000
4.
Find the maximum area of an
𝑥2
𝑦2
isosceles triangle inscribed in the ellipse 𝑎2 + 𝑏2 = 1 with its vertex at one end of the major axis.
Solution:
1
1
Area of ∆𝐴𝐵𝐶 = 2 × 𝐴𝐵 × 𝐷𝐶 = 2 × 2𝑏 𝑠𝑖𝑛 𝜃(𝑎 − 𝑎 𝑐𝑜𝑠𝜃)
= 𝑎𝑏𝑠𝑖𝑛𝜃(1 − 𝑐𝑜𝑠𝜃)-----(i)
𝑑𝐴
𝑑𝜃
= 𝑎𝑏(sin2 𝜃 + cos 𝜃 − cos2 𝜃)
𝑑𝐴
For maxima and minima 𝑑𝜃 = 0
𝑎𝑏(sin2 𝜃 + cos 𝜃 − cos2 𝜃) = 0
𝑐𝑜𝑠𝜃 − 𝑐𝑜𝑠2𝜃 = 0
𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑠2𝜃
2𝜃 = 2𝑛𝜋 ± 𝜃
𝜃
𝜃 = 𝑛𝜋 ± 2----(ii)
As 𝜃 ∈ (0, 𝜋) by equation (ii)
𝜃=𝜋−
𝜃=
𝜃
2
2𝜋
3
2
𝑑 𝐴
= 𝑎𝑏(2𝑠𝑖𝑛2𝜃 − 𝑠𝑖𝑛𝜃)
𝑑𝜃 2
𝑑2 𝐴
[𝑑𝜃2 ]
𝜃=
2𝜋
3
< 0 ⇒ 𝐴 is maximum.
By equation (i)
𝐴max =
3√3
4
𝑎𝑏 sq.unit
5.
Show that the volume of the greatest cylinder which can be inscribed in a cone of height
4
ℎ and semi-vertical angle ∝ is 27 𝜋ℎ3 𝑡𝑎𝑛2 ∝
Solution: Let 𝑥 𝑎𝑛𝑑 𝑦 be the radius and height of the inscribed cylinder
Then semi vertical angle =∝
Volume (V) of the cylinder = 𝜋𝑥 2 𝑦
= 𝜋𝑡𝑎𝑛2 ∝ (ℎ2 𝑦 − 2ℎ𝑦 2 + 𝑦 3 )
𝑑𝑉
=0
𝑑𝑦
ℎ
(𝑦 − ℎ)(3𝑦 − ℎ) = 0
𝑦 = ℎ, 3
And
𝑑2 𝑉
𝑑𝑦 2
=0
When 𝑦 =
𝑑2 𝑣
ℎ
3
[𝑑𝑦 2 ] = 2𝜋𝑡𝑎𝑛2 ∝ ℎ < 0
ℎ
𝑉is maximum when 𝑦 = 3
ℎ3
2
ℎ3
Maximum value = 𝜋𝑡𝑎𝑛2 ∝ [ 3 − 9 ℎ3 + 27]
4
= 27 𝜋ℎ3 𝑡𝑎𝑛2 ∝
6.
Show that the semi vertical angle of a right circular cone of given surface area and
1
maximum volume is 𝑠𝑖𝑛−1 3.
Solution: 𝐺𝑖𝑣𝑒𝑛: 𝜋𝑟(𝑙 + 𝑟) = 𝑆
𝑉 =
𝑉
2
1 2
𝜋𝑟 ℎ
3
𝜋2 4 2
𝑟 2𝑠2
2𝜋𝑠𝑟 4
2
=
𝑟 (𝑙 − 𝑟 ) =
−
9
9
9
𝑑𝑉
2𝑟𝑠 2
8𝜋𝑠𝑟 3
2𝑉
=
−
𝑑𝑟
9
9
𝑑𝑉
𝑟
1
= 0 𝑎𝑛𝑑 𝑢𝑠𝑖𝑛𝑔 𝑠 = 4𝜋𝑟 2 𝑤𝑒 𝑔𝑒𝑡 𝑙 = 3𝑟 𝑜𝑟 =
= 𝑠𝑖𝑛 𝛼
𝑑𝑟
𝑙
3
1
𝛼 = 𝑠𝑖𝑛−1
3
𝑑2𝑉
= −𝑣𝑒
𝑑𝑟 2
7.
Show that the volume of the largest cone that can be
inscribed in a sphere of radius R is 8/27 of the volume of the
Solution:
sphere.
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒 = (1/3) 𝜋𝑟 2 ℎ = (1/3) 𝜋 (𝑅 2 – 𝑥 2 ) ( 𝑅 + 𝑥) − − − − − − (1)
3𝑉/ 𝜋 = 𝑅 3 – 𝑥 3 + 𝑅 2 𝑥 – 𝑥 2 𝑅
𝑍(𝑥) = 𝑅 3 – 𝑥 3 + 𝑅 2 𝑥 – 𝑥 2 𝑅
𝑍’ (𝑥) = −3𝑥 2 + 𝑅 2 – 2𝑥𝑅
𝐹𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑜𝑙𝑢𝑚𝑒 𝑍’ (𝑥) = 0
R
−3𝑥 2 + 𝑅 2 – 2𝑥𝑅 = 0
x
𝑥 = 𝑅/3 𝑜𝑟 𝑥 = − 𝑅
𝑥 = −𝑅 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 .
r
𝑍’’ (𝑥) = −6𝑥 − 2𝑅 < 0
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚. 𝐼𝑛 (1) 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒 =
1
𝜋×
3
8𝑅 2
9
×
4𝑅
3
8
= 27 ×
4𝜋𝑅 3
3
=
8
27
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒.
8.
A wire of length 28 cm is cut into two pieces. One of the pieces is turned in the form of a
square and the other into a circle. Find the length of each piece so that the sum of the areas of the
two is minimum.
SOLUTION:
9.
SOLUTION;
10.
SOLUTION;
11.
SOLUTION;
12.
SOLUTION;
RATE OF CHANGE QUANTITIES
Q1
The total revenue received from the sale of x souveniers in connection with PEACE DAY
is given by R (x) = 3 x2 +40 x+ 10. Find the marginal revenue when 100 souveniers were sold.
Solution- we have R (x) = 3 x2 +40 x+ 10
𝑅 ′ (𝑥) = 6x +40
𝑅 ′ (100) = 6x100 +40 =640
Q 2. For the curve y = 5x – 2x3,if x increases at the rate of 2 units / s,then how fast is the slope of
the curve changing when x = 3 ?
𝑑𝑥
Solution: - We have 𝑑𝑡 =2 units / s
𝑑𝑦
y = 5x – 2x3, Slope S =𝑑𝑥 =5- 6 x2
𝑑𝑆
= -12 x.
𝑑𝑡
𝑑𝑆
𝑑𝑥
𝑑𝑡
( at x =3 )= -12×3×2 =-72 units/ s
𝑑𝑡
Q 3. Radius of a variable circle is changing at the rate of 5 cm/ s.What is the radius of the circle
at atime when its area is changing at the rate of 100 cm2/s ?
Solution: - The area A of a circle with radius r = A =𝜋𝑟2
𝑑𝐴
𝑑𝑡
𝑑𝑟
=2πr 𝑑𝑡 or 100 =2πr×5
r=
10
𝜋
cm
Q 4. The side of a square is changing at the rate of 5cm/sec. How fast is its area increasing when
the length of its side is 8cm.
Solution: - Let side of square at any time be x cm.
𝑑𝑥/𝑑𝑡=5cm/sec
Now, area of square ,A=x2
∴ 𝑑𝐴/𝑑𝑡 =80𝑐𝑚2/ Ans
Q5 . The radius of an air bubble is increasing at the rate of 12 cm/s. At what rate is the volume of
the bubble increasing when the radius is 1cm?
Solution : Let r be the radius of bubble and v be the volume of bubble at any time t.
4
Then 𝑑𝑟/𝑑𝑡 =12𝑐𝑚 𝑎𝑛𝑑 𝑟=1𝑐𝑚, V =3 πr3
𝑑𝑉
By solving , we get 𝑑𝑣/𝑑𝑡= 𝑑𝑟 ×
𝑑𝑟
𝑑𝑡
= 48π 𝑐𝑚3/s 8.
Q 6. A man 2m high, walks at a uniform speed of 6m/min away from a lamp post 5m high. Find
the rate at which the length of his shadow increases
Solution : Let AB is the Lamp post.
At any time t, let the man be at a distance of x m away from the lamp post and
Y m be length of his shadow
Then AB= 5m and CD= 2m
𝑑𝑥
𝑑𝑡
=6m/min
By solving we get
𝑑𝑦
𝑑𝑡
=4m/min
Q.7 Sand is pouring from a pipe at the rate of 12cm3/sec. The falling sand forms a cone on the
ground in such a way that the height of the cone is always one –sixth of the radius of the base.
How fast is the height of the sand cone increasing when the height is 4cm?
Solution: Let r be the radius h be the height and V be the volume of sand cone at any time t.
Then
𝑑𝑣
𝑑𝑡
1
=12𝑐𝑚3 /s and h= 6 r
By solving we get
𝑑ℎ
=
1
𝑑𝑡 48𝜋
cm/sec
Q. 8 The radius of a circle is increasing uniformly at the rate of 3cm/s. find the rate at which the
area of the circle is increasing when the radius is 10cm.
Solution: - Let r be the radius and A be the area of the circle at any time t.
𝑑𝑟
Now 𝑑𝑡 =3cm/s and r=10cm
Now A=π𝑟 2
By solving it we get
𝑑𝐴
𝑑𝑡
=60𝜋𝑐𝑚2 /s
Q. 9 Find the point on the curve 𝑦 2 =3x for which the abscissa and ordinate change at the same
rate.
Solution: - Given 𝑦 2 =3x and
𝑑𝑦
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
𝑑𝑦
𝑑𝑥
=3
𝑑𝑡
𝑑𝑡
3
𝑦=
2
2𝑦
3
3
3
So, 𝑥 = 4 and point is (4 , 2)
Q. 10 An edge of a variable cube is increasing at the rate of 3cm/s. How fast is the volume of the
cube increasing when the edge is 10 cm long.
Solution: - Let side of the cube is x. Then
Volume 𝑉 = 𝑥 3
Now
𝑑𝑉
)
𝑑𝑉
𝑑𝑡
𝑑𝑡 𝑥=10
𝑑𝑥
= 3𝑥 2 𝑑𝑡
= 3(10)2 . 3 = 900 𝑐𝑚3 /𝑠
as
𝑑𝑥
𝑑𝑡
= 3 𝑐𝑚/𝑠
TANGENTS AND NORMALS
1. Find the equations of the tangent to the curve y =𝑥 2 + 4𝑥 + 1 at the point whose xcoordinate is 3.
2. Find the equations of the normal to the curve y =𝑥 2 + 4𝑥 + 1 at the point whose xcoordinate is 3.
3. Find the slope of the normal to the curve y = 𝑥 2 - 3𝑥 + 2 at the point whose x-coordinate
is 3.
𝜋
4. Find the equations of the tangent to the curve x = cos 𝑡 , y = sin 𝑡at t = 4 .
5. Find the equation of the normal at the point (1,1) on the curve 2y + 𝑥 2 = 3.
6. Find the equation of tangent to the curve y = √3𝑥 − 2 , which is parallel to the line
4x -2y + 5 =0.
7. Find the equations of the tangent to the parabola 𝑦 2 = 4𝑎𝑥 at the point ( a𝑡 2 , 2𝑎𝑡 )
𝑥2
8. Find the equations of the tangent to the parabola𝑎2 -
𝑦2
𝑏2
= 1 at the point (h,k .
9. Prove that the curves x = 𝑦 2 and xy = k cut at right angles if 8𝑘 2 = 1.
10. On which point the line y = x + 1 is a tangent to the curve 𝑦 2 = 4𝑥
ROLL’S THEOREM AND MEAN VALUE THEOREM
1.
2.
3.
4.
Examine if the Roll’s theorem is applicable for the function f(x)=[x], if x ∈ [5,9].
5.
Examine if the Roll’s theorem is applicable for the function f(x)=x2-1 if x ∈ [1,2].
SOLUTION:TANGENTS AND NORMALS
1. curve y = 𝑥 2 + 4𝑥 + 1
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 2x + 4
at x = 3 is 10. So m = 10.
𝑥 = 3 , y = 22
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
y - 22 = 10 ( x – 3 ) i. e. 10x –y = 8.
2. curve y = 𝑥 2 + 4𝑥 + 1
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 2x + 4
at x = 3 is 10.
1
So slope of normal is -10.
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑖𝑠
1
y - 22 = - 10( x – 3 ) i. e. x + 10y = 223
3. y = 𝑥 2 - 3𝑥 + 2
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 2x -3.
at x = 3 is 3.
1
So slope of normal is -3.
4. x = cos 𝑡 , y = sin 𝑡
𝑑𝑥
𝑑𝑡
𝑑𝑦
𝑑𝑥
= − sin 𝑡
at t =
𝜋
4
𝑑𝑦
,
𝑑𝑡
𝑑𝑦
= cos 𝑡
=>𝑑𝑥 = - cot t
= -1
𝜋
x = cos 4 =
1
,
√2
𝜋
y = sin 4 =
1
√2
Hence, 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
y-
1
√2
= -1( x -
1
√2
)
=>
x + y = √2 .
5. Curve 2y + 𝑥 2 = 3
𝑑𝑦
𝑑𝑦
2𝑑𝑥 +2x = 0
=>𝑑𝑥 = -x
𝑑𝑦
=>𝑑𝑥 𝑎𝑡 (1,1) = −1
So slope of normal is 1.
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑖𝑠
=>
x–y=0
y – 1 = 1( x – 1 )
6. curve y = √3𝑥 − 2 ,
𝑑𝑦
=
𝑑𝑥
3
2√3𝑥−2
Slope of the line
4x -2y + 5 =0 is 2.
Given
3
2√3𝑥−2
41
=2
=> x = 48and y =
3
4
Hence, 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
3
41
y - 4 = 2 ( x - 48 )
=>
48x -24 y -23 = 0
7. parabola 𝑦 2 = 4𝑎𝑥
𝑑𝑦
𝑑𝑦
2y𝑑𝑥 = 4a
=>𝑑𝑥 =
2𝑎
𝑑𝑦
=>𝑑𝑥 at the point ( a𝑡 2 , 2𝑎𝑡 ) =
𝑦
1
𝑡
Hence, 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
1
y – 2at = ( x - a𝑡 2 )
8.
𝑥2
𝑎2
-
𝑦2
𝑏2
𝑑𝑦
= 1 =>𝑑𝑥 =
x – ty + a𝑡 2 = 0
=>
𝑡
𝑏2 𝑥
𝑎2 𝑦
𝑏2 ℎ
𝑑𝑦
at the point ( h,k = 𝑎2 𝑘
𝑑𝑥
Hence, 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑖𝑠
y–k=
9. x = 𝑦 2
𝑏2 ℎ
𝑎2 𝑘
ℎ𝑥
(x–h)
𝑑𝑦
=>𝑑𝑥 =
=>𝑎2 -
𝑘𝑦
𝑏2
=1
1
2𝑦
𝑑𝑦
and xy = k
=>𝑑𝑥 =
1
−𝑦
𝑥
−𝑦
cut at right angles if 2𝑦x 𝑥 = -1
so
1
x=2
1 21
xy = k, =>𝑥 2 𝑦 2 = 𝑘 2 =>(2)
Since
2
1
, 𝑦2 = 2
= 𝑘2
 8𝑘 2 = 1
𝑑𝑦
10. 𝑦 2 = 4𝑥 =>2y𝑑𝑥 = 4
𝑑𝑦
2
=>𝑑𝑥 = 𝑦
Slope of the line y = x + 1 is 1.

2
𝑦
=1
=>
y = 2. Then x = 1.
 So required point is ( 1, 2).
SOLUTIONS OF ROLL’S AND M.V.THEOREM
1
.
2
.
3
.
4
.
Hence Roll’s theorem not satisfied.
5
.
Hence Roll’s theorem not satisfied.