MATHEMATICS
CONTENTS KEY CONCEPT .............................................................. ........................................................................... ............. Page – 2 6 EXERCISE EXERCI SE – I ............................................................................... ............................................................................... Page – 6
EXERCISE EXERCI SE – II .............................................................................Page – 8 11 EXERCISE EXERCI SE – III ............................................................................ ............................................................................ Page – 11
EXERCISE – IV ................................................................ ............................................................................. ............. Page – 12 12 14 ANSWER KEY ................................................................ ............................................................................. ............. Page Pa ge – 14
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Email:
[email protected]
KEY CONCEPTS (METHOD OF DIFFERENTIATION) 1.
DEFINITION : If x and x + h belong to the domain of a function f defined by y = f(x), then Limit f (x " h) # f (x) if it exists , is called the DERIVATIVE of f at x & is denoted by h !0 h
dy
f $(x) or 2.
dx
f (x " h) # f (x) . We have therefore , f $(x) = Limit h !0 h
The derivative of a given function f at a point x = a of its domain is defined as : Limit f (a " h) # f (a ) , provided the limit exists & is denoted by f $(a) . h !0 h
f (x) # f (a ) Note that alternatively, we can define f $(a) = Limit , provided the limit exists. x !a x#a
3.
DERIVATIVE OF f(x) FROM THE FIRST PRINCIPLE /ab INITIO METHOD:
% y Limit f (x " % x) # f (x) dy If f(x) is a derivable function then, Limit = = f (x) = $ %x !0 % x %x !0 %x dx 4.
THEOREMS ON DERIVATIVES : If u and v are derivable function of x, then, (i) (iii)
(iv) (v) 5.
6.
d dx d dx d dx
( u & v)
'
du
v
dv
dv
&v
dx
( u . v) ' u * u , /' + v .
&
dx
dx du dx
( dudx ) # u ( dvdx ) v
2
d
(ii)
dx
known as
(K u) ' K
dy dx
'
dx
, where K is any constant
PRODUCT RULE ”
“
where v 0 0 known as
If y = f(u) & u = g(x) then
du
dy du
.
du dx
QUOTIENT RULE ”
“
CHAIN RULE ”
“
DERIVATIVE OF STANDARDS FUNCTIONS : (i) D (xn) = n.xn#1 ; x 1 R, n 1 R, x > 0
(ii) D (ex) = ex
1
(iii) D (ax) = ax. ln a a > 0
(iv) D (ln x) =
(vi) D (sinx) = cosx (ix) D (secx) = secx . tanx
(vii) D (cosx) = # sinx (viii) D = tanx = sec²x (x) D (cosecx) = # cosecx . cotx
(xi) D (cotx) = # cosec²x
(xii) D (constant) = 0 where D =
(v) D (logax) =
x
1 x
logae
d dx
INVERSE FUNCTIONS AND THEIR DERIVATIVES : (a) Theorem : If the inverse functions f & g are defined by y = f(x) & x = g(y) & if
f $(x) exists & f $(x) 0 0 then g $(y) = dy dx
(b)
0 0 ,
then
dx dy
* dy ' 1 / , / + dx .
or
dy dx
.
1 f $ (x)
dx dy
'1
. This result can also be written as, if or
* dx '1 / , / dx + dy . dy
[
dx dy
0
dy dx
exists &
0]
Results : (i)
D (sin # x) ' 1
1 1# x
2
,
# 12 x 21
(ii)
D (cos # x) '
Method of differentiation
1
#1 1# x
2
,
# 12 x 21 [2]
D (tan #1 x) '
(iii)
1 1 " x2
1
x
x
2
#1
dx
= f $(u) .
dx
#1
, x
31
, x1 R
.
OR
a function of the form [f(x)]g(x) where f & g are both derivable, it will be found convinient to take the logarithm of the function first & then differentiate. This is called LOGARITHMIC DIFFERENTIATION .
In answers of dy/dx in the case of implicit functions, both x & y are present .
PARAMETRIC DIFFERENTIATION :
If y = f(5) & x = g(5) where 5 is a parameter , then 10.
du
#1 1 " x2
2
IMPLICIT DIFFERENTIATION : 4 (x , y) = 0 (i) In order to find dy/dx, in the case of implicit functions, we differentiate each term w.r.t. x regarding y as a functions of x & then collect terms in dy/dx together on one side to finally find dy/dx. (ii)
9.
dy
D (cot #1 x) '
(vi)
x
LOGARITHMIC DIFFERENTIATION : To find the derivative of : (i) a function which is the product or quotient of a number of functions (ii)
8.
31
, x
Note : In general if y = f(u) then 7.
1
D (sec #1 x) '
(iv)
x
#1
D (cos ec # x) '
(v)
, x 1R
dy dx
'
d y / d 5 d x / d 5
.
DERIVATIVE OF A FUNCTION W.R.T. ANOTHER FUNCTION : dy
d y / d x
f ' (x )
Let y = f(x) ; z = g(x) then d z ' d z / d x ' g' (x) . 11.
DERIVATIVES OF ORDER TWO & THREE : Let a function y = f(x) be defined on an open interval (a, b). It’s derivative, if it exists on (a, b) is a certain function f $(x) [or (dy/dx) or y ] & is called the first derivative of y w.r.t. x. If it happens that the first derivative has a derivative on (a , b) then this derivative is called the second derivative of y w. r. t. x & is denoted by f $$(x) or (d2y/dx2) or y $$. rd
Similarly, the 3 order derivative of y w. r. t. x , if it exists, is defined by
d 3y dx
' 3
* d 2 y ,, 2 // It is also dx + d x . d
denoted by f $$$(x) or y $$$. f (x)
12.
g(x)
If F(x) = l(x) m(x) n(x) , where f , g , h , l , m , n , u , v , w are differentiable functions of x then u(x)
13.
h(x)
v(x)
w(x)
f ' (x)
g' (x)
h' (x)
F $(x) = l(x)
m(x)
n(x)
u(x)
v(x)
w(x)
+
f (x)
g(x)
h(x )
l' (x)
m' (x)
n' (x)
u(x)
v(x )
w(x )
+
f (x)
g(x)
h(x)
l(x)
m(x)
n(x)
u' (x) v' (x) w' (x)
L’ HOSPITAL’S RULE : If f(x) & g(x) are functions of x such that : (i)
Limit f(x) = 0 = Limit g(x) OR Limit f(x) = 7 = Limit g(x) x !a x !a x !a x !a
(ii)
Both f(x) & g(x) are continuous at x = a
&
(iii)
Both f(x) & g(x) are differentiable at x = a
&
Method of differentiation
and
[3]
(iv)
Both f $(x) & g $(x) are continuous at x = a ,
Then
Limit f (x) = Limit f ' (x) = Limit f "( x) & so on till indeterminant form vanishes. x !a x !a x !a g(x)
14.
g' (x)
g"( x)
ANALYSIS AND GRAPHS OF SOME USEFUL FUNCTIONS : 2 tan # x 1
(i)
x
81
; * 2 x y = f(x) = sin#1 , = < # 2 tan #1 x x 31 / ; 2 + 1 " x 1 # ;# (< " 2 tan x) x 2 # 1 : HIGHLIGHTS : (a) Domain is x 1 R &
(b)
9 < <= range is ;# 2 , 2 ? : > f is continuous for all x but not diff. at x = 1 , - 1
(c)
9 2 ; 1" x dy = ;non existent dx ;# 2 :; 1 " x 2
2
(d) (ii)
for
x
for
x
for
x
21 '1 31
I in (- 1 , 1) & D in (- 7 , - 1)
Consider
@ (1 , 7)
* 1 # x 2 9 2 tan #1 x if x A 0 / = ; y = f (x) = cos , 2 1 # + 1 " x :# 2 tan x if x 2 0 -1
HIGHLIGHTS : (a) Domain is x 1 R & range is [0, <) (b) Continuous for all x but not diff. at x = 0
9 2 ; 1" x dy = ;non existent dx ;# 2 :; 1 " x 2
(c)
2
(d)
(iii)
for for for
30 x'0 x20 x
I in (0 , 7) & D in (- 7 , 0)
y = f (x) = tan-1
2x 1 # x2
9 2 tan #1 x ; = ; < " 2 tan #1 x ;# (< # 2 tan #1 x) :
21 x 2 #1 x31 x
HIGHLIGHTS : (a) Domain is R - {1 , -1} &
(b)
(c) (d)
< < range is * ,# , / + 2 2 f is neither continuous nor diff. at x = 1 , - 1
9 1 "2x = ; dx ;:non existent dy
2
I B x in its domain
x x
01 '1 (e)
It is bound for all x Method of differentiation
[4]
(IV)
9 # (< " 3 sin #1 x) ; y = f (x) = sin#1 (3x # 4x3) = ; 3 sin #1 x ; < # 3 sin #1 x ;:
if if if
# 1 8 x 8 # 12 # 12 8 x 8 12 1 8x81 2
HIGHLIGHTS : (a) Domain is x 1 [# 1 , 1] &
9 < <= range is ;# 2 , 2 ? : > (b)
(c) (d)
(v)
Not derivable at x '
9 3 1# x = ; ;# 3 dx ;: 1 # x dy
1 2
if x
2
1 ( # 12 ,
if x 1
2
1 2
)
(#1 , # ) ! ( 1 2
1 2
)
,1
Continuous everywhere in its domain
9 3 cos #1 x # 2 < ; y = f (x) = cos-1 (4x3 - 3x) = ;2 < # 3 cos #1 x ; 3 cos #1 x :
# 1 8 x 8 # 12 # 12 8 x 8 12 1 8x81 2
if if if
HIGHLIGHTS : (a) Domain is x 1 [- 1 , 1] & range is [0 , <] (b)
Continuous everywhere in its domain 1
but not derivable at x =
(c)
1 2
* 1 1 I in , # , / & + 2 2 . * 1 = D in ,+ 2 , 1? >
(d)
2
, #
9 3 = ; 1 # x3 dx ;# ;: 1 # x dy
1 9 ;# 1 , # 2 ./ :
!
if x
2
1 ( # 12 ,
if x 1
2
1 2
)
(#1 , # ) ! ( 1 2
1 2
)
,1
GENERAL NOTE : Concavity in each case is decided by the sign of 2 nd derivative as : 2
d y d x2
2
> 0
C
Concave upwards
D = DECREASING
;
;
d y d x2
< 0
C
Concave downwards
I = INCREASING
Method of differentiation
[5]
– Q.1
Let f , g and h are differentiable functions. If f (0) = 1 ; g (0) = 2 ; h (0) = 3 and the derivatives of their pair wise products at x = 0 are (f g)'(0) = 6 ; (g h)'(0) = 4 and (h f)'(0) = 5 then compute the value of (fgh)'(0).
Q.2
(a) If y = (cos x) lnx + (lnx)x find (b) If y = e
x2
"
x
1
ex
"e
x x
x
2
xe
"x
e
ex
dy dx
dy
. Find
" 1 " ln
x" x
dx
2
.
" 1 prove that 2y = xy' + ln y'. where ' denotes the derivative.
Q.3
If y =
Q.4
If y = ln * ,x
Q.5
Let x, y 1 R satisfying the equation tan 1x + tan 1y + tan 1(xy) =
2
2
e x Da y
+
-/ .
yx
dy
find
dx
.
–
–
–
11< 12
, then find the value of
dy dx
at x = 1. 2
Q.6
If x = cosec 5 # sin 5 ; y = cosecn 5 # sinn 5 , then show that ( x
Q.7
If a curve is represented parametrically by the equations
* +
x = sin , t "
Q.9
If x =
* < * 3< * < * 3< * 7< / + sin , t # / + sin , t " / , y = cos , t " / + cos , t # / + cos , t " / 12 . + 12 . + 12 . + 12 . + 12 . + 12 .
1 " lnt t
* d y " 4) ,, // # n 2 ( y 2 " 4) ' 0 . + dx .
7< -
d
then find the value of
Q.8
2
2
Differentiate
dt
and y =
* x y ,, # // at + y x .
3 " 2lnt t
1" x2 " 1# x2 1" x
2
#
1# x
2
t=
< 8
.
. Show that y
w. r. t.
dy dx
2
dy ' 2x * , / "1 . + dx .
1# x4 . g ( x ),
Q.10
9 Let g(x) be a polynomial, of degree one & f (x) be defined by f(x) = ; 1 / x 1 " x ;: * , / , Find the continuous function f(x) satisfying f $(1) = f(#1) + 2 " x .
Method of differentiation
x80
. x
30
[6]
Q.11
1
If y = x +
, prove that
1
x" x"
dy dx
1
' 2#
1
1
1
x" 1
Let f (x) = x +
Q.13
Find the derivative with respect to x of the function:
(a)
Let
+
2x " 2x " 2x "
#1 *
, 2x +
d (sin
1# x
#1
2
-/ -/ . .
x)
1 x"....................
......... 7 . Compute the value of f (100) · f ' (100).
Q.12
d * , sin
1
x"
x"...............
.
x
= p, where x 1
* # 1 , , + 2
/ and 2 .
1
2 * * - #1 , 1 " x # 1 / / , d tan , , // x + . . = q, +
d (tan # x ) 1
where x 1 R – {0}. Find the value of 2 {p + q} + pq. [Note: where {k} denotes fractional part of k.] (b)
Let f (x) = sin
* , +
#1 ,
/ , find the value of / 2 4 x " 8x " 13 . 2x " 2
d (tan
Let f(x) = cos 1 (4x3 – 3x). If f '(0) = p and f ' ,
Q.15
Suppose f (x) = tan sin #1 ( 2 x )
–
x)
1 2
.
(a) (b)
Find the domain and range of f . Express f (x) as an algebaric function of x.
(c)
Find f ' (1 4) .
(a)
Let f (x) = x 2 # 4x # 3, x > 2 and let g be the inverse of f. Find the value of g $ where f (x) = 2. Let f : R!R be defined as f (x) = x3 + 3x2 + 6x – 5 + 4e2x and g(x) = f 1(x), then find g ' ( – 1).
(b)
(c)
–
Suppose f
1
–
is the inverse function of a differentiable function f and let G(x) =
If f(3) = 2 and f '(3) = (d)
1 #
f 1 ( x )
.
1
, find G ' (2). 9 Let f : R ! R be a differentiable bijective function. Suppose g is the inverse function of f such that G(x) = x2 g(x). If f(2) = 1 and f '(2) =
Q.17
#1
when x =
* # 4 - = q, then find the value of (2p – 3q). / + 5 .
Q.14
Q.16
d (tan f ( x ))
If y = tan
1
#1 2
" tan #1
1 2
" tan #1
x " x "1 x " 3x " 3 x Find dy/dx , expressing your answer in 2 terms.
1 2
, then find G'(1).
1 2
" 5x " 7
" tan #1
Method of differentiation
1 x
2
" 7 x " 13
+...... to n terms.
[7]
u
1
* +
- * 1 - prove that 2 dy + 1 = 0. / @ , ,1/ dx 2 . + 2 .
1
Q.18
If y = tan #1
Q.19
If y = cot #1
Q.20
If y = tan
Q.21
(a) (b)
Q.22
Prove that the second order derivative of a single valued function parametrically represented by x = 4(t)
1 # u2
& x = sec #1
1 " sin x
" 1 " sin x #
1 # sin x
#1
, find
1 # sin x
*
x
1
–
2 u2
, u 1 , 0,
#1 + sin ,, 2 tan
1" 1# x2
+
dy dx
* < - @ * < ,< - . / , / + 2 . + 2 .
if x 1 , 0,
1# x
/ , then find 1" x /
dy dx
for x 1 ( – 1, 1).
If y = y(x) and it follows the relation exy + y cos x = 2, then find (i) y ' (0) and (ii) y '' (0). A twice differentiable function f (x) is defined for all real numbers and satisfies the following conditions f (0) = 2; f ' (0) = – 5 and f '' (0) = 3. The function g (x) is defined by g (x) = e ax + f (x) B x 1 R, where 'a' is any constant. If g ' (0) + g '' (0) = 0. Find the value(s) of 'a'.
and y = E(t),F < t < G where 4(t) and E(t) are differentiable functions and 4'(t) 0 0 is given by
d2y dx 2
d 2 y - * d 2 x - * dy * dx - * , / ,, 2 // # ,, 2 // , / + dt . + dt . + dt . + dt .
'
* dx , / + dt .
3
3
.
2
d y
on the ellipse 3x2 + 4y2 = 12.
Q.23
Find the value of the expression y
Q.24
If f : R ! R is a function such that f (x) = x3 + x2 f $(1) + xf $$(2) + f $$$(3) for all x 1 R , then prove that f (2) = f (1) # f (0).
Q.25
Let P(x) be a polynomial of degree 4 such that P(1) = P(3) = P(5) = P'(7) = 0. If the real number x 0 1, 3, 5 is such that P(x) = 0 can be expressed as x = p/q where 'p' and 'q' are relatively prime, then find ( p + q).
dx
2
–
Q.1
Let f and g be two real-valued differentiable functions on R. If f '(x) = g(x) and g'(x) = f(x) B x 1 R and f(3) = 5, f '(3) = 4 then find the value of ( f 2 (< ) # g 2 (< ) ) . y
arc sin 2
"y 'e 2
x
2
" y2
" y2 ) , x > 0. ' 3 (x # y) 2( x 2
Q.2
If
Q.3
Find a polynomial function f (x) such that f (2x) = f ' (x) f " (x).
x
1
Q.4
If 2x '
y5
"y
#
. Prove that
d2y
1 5 then
(x
2
# 1)
2
d y dx
"x 2
dx
2
dy
' ky , then find the value of 'k'. dx Method of differentiation
[8]
2
d y
Q.5
Let y = x sin kx. Find the possible value of k for which the differential equation 2 + y = 2k cos kx dx holds true for all x 1 R.
Q.6
The function f : R ! R satisfies f (x2) · f ''(x) = f '(x) · f '(x2) for all real x. Given that f (1) = 1 and f '''(1) = 8, compute the value of f '(1) + f ''(1).
Q.7
2 x * d y dy Show that the substitution z = ln , tan / changes the equation cot x " " 4 y cos ec2 x ' 0 to 2 + 2 . dx dx 2 2
(d y/dz ) + 4 y = 0.
Q.8
If the dependent variable y is changed to 'z' by the substitution y = tan z then the differential equation d2y dx
2
' 1"
2(1 " y) * dy 1" y
2
2
, / + dx .
2
2
is changed to
d z dx
2
= cos
2
* dz z " k , / , then find the value of k . + dx .
Q.9
H1 " (dy dx ) I Show that R =
Q.10
Let f (x) =
Q.11
Suppose f and g are two functions such that f , g : R ! R,
2 3 / 2
2
d y dx
sin x x
can be reduced to the form R 2/3 =
2
1
(d 2 y
dx 2
" 2 / 3
)
1
(d 2 x dy2 )2 / 3
.
if x 0 0 and f (0) = 1. Define the function f ' (x) for all x and find f '' (0) if it exist.
f (x) = ln * ,1 " 1 " x 2 -/
+
g (x) = ln * , x " 1 " x 2 -/
and
.
+
'
.
* * 1 - / // " g' (x ) at x = 1. x + + .
then find the value of x eg(x) ,, f ,
Q.12
Let f (x) = (a) (b)
9 ;:
xe
x
x"x
x 2
#x
3
x
80 30
then prove that
f is continuous and differentiable for all x. f ' is continuous and differentiable for all x.
Q.13 f : [0, 1] ! R is defined as f (x) = (a)
Q.14
f is differentiable in [0, 1]
9 ; ;:
* 1 / + x 2 .
x 3 (1 # x ) sin ,
if 0 2 x 8 1
, then prove that if x ' 0
0
(b)
f is bounded in [0, 1] (c)
* x " y + k / =
Let f(x) be a derivable function at x = 0 & f ,
f (x)
" f (y) k
f ' is bounded in [0, 1]
(k 1 R , k 0 0, 2). Show that
f (x) is either a zero or an odd linear function.
Method of differentiation
[9]
Q.15
Let
f ( x " y) # f ( x )
=
2
f ( y) # a
+ xy for all real x and y. If f (x) is differentiable and f $(0) exists
2
for all real permissible values of 'a' and is equal to Column-I
Q.16
3
(A)
ln (1 " x ) · sin
(B)
9 ;:
ln (1 " x ) · sin
9 ;:
ln , 1 "
(D)
g (x) =
u (x) =
x
2
, if x
* +
0,
v (x) = Lim t !0
( x#a )
4
If f ( x) = ( x#b)
4
1 x
0,
2x
<
30
if x 8 0 , if x
30
if x 8 0
sin x /, if x 3 0 2 . if x 8 0
tan
#1 * 2
, 2/ + t
(P)
continuous everywhere but not differenti able at x ' 0
(Q)
differentiable at x ' 0 but derivative is discontinuous at x ' 0
(R)
differentiable and has continuous derivative
(S)
continuous and differentiable at x ' 0
( x#a )
3
1
4
( x#a )
2
1
( x#b)
3
1 then f $ (x) = J . ( x#b) 4 4 1 ( x#c)
( x#b)
2
1 . Find the value of
( x#c)
2
1
( x#c)3
( x#a )
2
sin( x" x )
2
cos( x#x )
cos( x"x )
Q.19
1
0,
( x#c) 4
Q.18
Column-II
9 f (x) = ; :
(C)
Q.17
2 5a # 1 # a . Prove that f (x) is positive for all real x.
If f(x) = sin( x#x ) sin 2 x
2
J.
#cos( x"x 2 ) 2 sin( x#x ) then find f '(x).
2
sin2 x 2
0
a "x
b"x
c" x
Let f (x) = " " x
m" x
n " x . Show that f
p"x
q"x
r"x
$$ (x) = 0
and that f (x) = f (0) + k x where k denotes
the sum of all the co-factors of the elements in f (0).
Q.20
If Y = sX and Z = tX, where all the letters denotes the functions of x and suffixes denotes the differentiation w.r.t. x then prove that X X1 X2
Y Y1 Y2
Z s Z1 = X3 1 s2 Z2
t1 t2
Method of differentiation
[10]
– Evalute the following limits using L’Hospital ’s Rule or otherwise :
Q.1
Lim
Q.3
Lim
Q.5
Lim
Q.7
x !0
x !0
x " ln * , x
2 9 1 1# x = # 2 ? ; #1 x sin x x > :
1
;: x 2
#
1 sin
2
x !0
? x>
x! <
1 x
the differentiability of f(x) at x = 0.
Q.9
If Lim x !0
" x3 2 3 4 2x .ln (1 " x ) # 2x " x a sin x # bx " cx
Q.10
Evaluate: Lim
Q.11
(a)
x
6000
x !0"
)
#
2 e2x # 1
, x 0 0 is continuous at x = 0 & examine
2
exists & is finite, find the values of a, b, c & the limit.
# (sin x ) 6000
1 # cos x · cos 2x · cos 3x........cos nx x2
x !0
(where n 1 N).
5
Find the value of Lim
(b)
(
Limit cot x # ln 2 x
2 6000 x · (sin x )
x !0
If Lim
(b)
if x 0 0 . Check the continuity and derivability of f(x) at x = 0. if x ' 0
x
Q.8
3
1 # sin x " ln (sin x )
2
Find the value of f (0) so that the function f (x)=
x x #1
" 1 # x -/ .
sin x # (sin x )sin x
Lim
Q.6
2
x·tan x
M| x | , Let f (x ) ' L K 1,
x
x Q.4(a) xLim !0 "
1 " sin x # cos x " l n (1 # x )
x !0
+
Lim
Q.2
2
3
has the value equal to 253, find the value of n
3
1 # cos x cos 2x cos 3x
x !0
x
2
.
n
1 # cos 3x · cos 9 x · cos 27 x......... cos 3 x If Lim = 310, find the value of n. 1 1 1 1 x !0 1 # cos x · cos x · cos x......... cos n x 3 9 27 3
(c)
Q.12
n
Q.13
( !7
If Lim a n n
n
" b ) ln n
–
Let a1 > a2 > a3 ............ an > 1; p1 > p2 > p3......... > p n > 0 ; such that p 1 + p2 + p3 + ...... + p n = 1
(
Also F (x) = p a x 1 1 (a) Lim F( x ) x !0"
Q.14
has the value equal to e 3, find the value of (4b + 3a).
" p 2 a 2x " ....... " p n a nx ) . Compute 1x
(b) Lim F( x )
(c) Lim F( x ) x! #7
x !7
Let x1, x1, x1 , x2, x3, x4, ............., x 8 be 10 real zeroes of the polynomial P(x) = x 10 + ax2 + bx + c where a, b, c 1 R. If the value of Q(x1) =
p q
where p and q are coprime to each other,,
Q(x) = (x – x2) (x – x3)....... (x – x8) and x1 =
1 2
, then find the value of p + q.
Method of differentiation
[11]
Q.15
Column-I contains function defined on R and Column-II contains their properties. Match them. Column-I Column-II
(A)
* 1 " tan < , / 2 n / Lim , < / n !7 , , 1 " sin / 3n . +
(B)
Lim
x !0 "
n
equal
1
equals
1
(P)
e
(Q)
e2
(R)
e
(S)
e< /6
(1 " cosec x ) ln (sin x ) 1x
* 2 #1 Lim , cos x / equals x ! 0 + < .
(C)
2/ <
–
– Q.1 (a)
If
ln (x + y) = 2xy, then y ' (0) = (B) – 1
(A) 1
(b)
M #1 * x " c b sin , /, N 2 + . NN 1 f (x) = L N 2ax / 2 #1 Ne , NK x
(C) 2
(D) 0
[ JEE 2004 (Scr.)]
1
# 2x20 2
at x ' 0 02x2
.
1 2
If f (x) is differentiable at x = 0 and | c | < 1/2 then find the value of 'a' and prove that 64b 2 = 4 – c2. [JEE 2004, 4 out of 60] Q.2 (a)
If y = y(x) and it follows the relation x cos y + y cos x = <, then y"(0) (A) 1 (B) – 1 (C) < (D) – <
(b)
If P(x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that P(1) = 1, P(0) = 0 and P'(x) > 0 B x 1 [0, 1], then (A) S = 4 (B) S = {(1 – a)x2 + ax, 0 < a < 2 (C) (1 – a)x2 + ax, a 1 (0, 7) (D) S = {(1 – a)x2 + ax, 0 < a < 1
(c)
If f (x) is a continuous and differentiable function and f (1 n ) = 0, B n A 1 and n 1 I, then (A) f (x) = 0, x 1 (0, 1] (C) f '(x) = 0 = f ''(x), x 1 (0, 1]
Q.3
1 / x
For x > 0, Lim (sin x ) x !0
(A) 0
" (1 x )sin x
(B) – 1
(B) f (0) = 0, f ' (0) = 0 (D) f (0) = 0 and f ' (0) need not to be zero [JEE 2005 (Scr.)] is (C) 1 Method of differentiation
(D) 2
[JEE 2006, 3] [12]
Q.4
d 2x dy
2
equals
* d 2 y (A) ,, 2 // + dx .
#1
#1
* d 2 y - * dy - #2 / (C) ,, 2 // , dx dx + . + .
* d 2 y - * dy - #3 / (B) – ,, 2 // , dx dx + . + .
* d 2 y -* dy - #3 / (D) – ,, 2 //, + dx .+ dx . [JEE 2007, 3]
Q.5 (a)
Let g (x) = ln f (x) where f (x) is a twice differentiable positive function on (0, f (x + 1) = x f (x). Then for N = 1, 2, 3
* +
g' ' , N "
1 -
* 1 / # g ' ' , / = 2 . + 2 .
M 1 1 Q 1 # " " " " 4 1 ..... L (A) 2P 9 25 # ( 2 N 1 ) K O (C)
(b)
7) such that
M
1
K
9
(B) 4L1 "
M 1 1 Q 1 # 4L1 " " " ..... " 2P ( 2 N " 1) O K 9 25
"
M
1
K
9
(D) 4L1 "
1 25
"
" ..... "
1 25
Q 2P ( 2 N # 1) O
" ..... "
1
Q 2P ( 2 N " 1) O 1
Let f and g be real valued functions defined on interval ( – 1, 1) such that g''(x) is continuous, g(0) 0 0, g'(0) = 0, g''(0) 0 0, and f (x) = g (x) sin x. STATEMENT-1 : Lim [g(x) cot x – g(0) cosec x] = f ''(0) x !0
and STATEMENT-2 : f '(0) = g(0) (A) Statement-1 is True, Statement-2 is True ; statement-2 is a correct explanation for statement-1 (B) Statement-1 is True, Statement-2 is True ; statement-2 is NOT a correct explanation for statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True [JEE 2008, 3 + 3] x
Q.6
If the function f ( x ) ' x
Q.7
Let f (5) = sin ,, tan
* +
3
" e 2 and g(x) = f 1(x), then the value of g$(1) is
#< - -/ / / , where 4 cos 25 . .
#1* sin 5
, +
–
< 5 <
< 4
. Then the value of
[JEE 2009, 4] d d(tan 5)
(f (5) ) , is [JEE 2011, 4]
Method of differentiation
[13]
ANSWER KEY EXERCISE – I Q.1
16
Q.2
(a)
x 9 1n(cos x) = # tan x 1nx? " (1nx ) Dy = (cosx)lnx ; x : >
dy
(b)
Q.4
Q.9
Q.12
y x
.
dx
= e
xe
x
.x
ex
x 9 e x x = x x e e#1 xe ee ex 9 1 " " " " "e x1nx =? e 1 nx e x x [ 1 e 1 nx ] x e ; ? ; :x > :x >
" x "n x . " n y " 1 "n x ( 1 # x # y "n a)
x "n x
1" 1# x
9 1 = ; 1nx " 1n(1nx)? ; : >
Q.5
–
* ,1 " , +
3 -
/ 2 ./
2 1 3 # 9; " ln =? x 3 :6 2>
4
Q.10
x6
100
Q.13
f (x) =
Q.7
8
if x 8 0
1 / x
* 1 " x , / + 2 " x .
if x
(a) 2; (b) 1
30
Q.14
21
2x * # 1 1 16 3 , / , ( – 7 , 7) ; (b) f (x) = Q.15 (a) , ; (c) 2 + 2 2 . 9 1 # 4x Q.16
(a) 1/6, (b)
1 14
(c) – 1; (d) 6
Q.17
1 1 "( x " n ) 2
#
1
1
Q.19
1" x2
2
or #
1 2
1 # 2x Q.20
Q.23
2 1# x
2
#9 4
Q.21
(a) (i) y ' (0) = – 1 ; (ii) y '' (0) = 2 ; (b) a = 1, – 2
Q.25
100
EXERCISE – II Q.1
Q.6
9
6
Q.3
Q.8
Q.16
4x 3
Q.4
9
k=2
Q.11
zero
Q.18
2(1 + 2x) . cos 2(x + x2)
Q.10
25
9 f ' (x) = ; ;:
Q.5 x cos x # sin x x
2
if x 0 0 ; f '' (0) = if x ' 0
0
(A) R, S; (B) Q, S; (C) P ; (D) R, S
k = 1, – 1 or 0
Q.17
Method of differentiation
–
1 3
3
[14]
EXERCISE – III Q.1
5
Q.2
6
Q.3
6
#
1
Q.4
3
2
Q.8
Continuous but not derivable at x = 0
Q.9
a = 6, b = 6, c = 0;
Q.12
7
Q.13
(a) 1 (b) does not exist
Q.5
#
1 2
f (0) = 1 ; differentiable at x = 0, f$ (0+) = # (1/3) ; f $(0#) = # (1/3)
Q.6
Q.14 31
Q.7
1
3
Q.10
40 p
p
1000
Q.11
(a) 11, (b) 22, (c) 4
p
(a) a1 1 · a 2 2 .....a n n ; (b) a1 ; (c) an
Q.15 (A) S; (B) P ; (C) R
EXERCISE – IV Q.1 (a) A; (b) a = 1 Q.3
C
Q.4
Q.2 (a) C; (b) B; (c) B D
Q.5
(a) A, (b) A
Q.6
2
Q.7
Method of differentiation
1
[15]
