BANSAL CLASSES

PRIVATE LIMITED
KEY CONCEPTS (FUNCTIONS)
THINGS TO REMEMBER :
1. GENERAL DEFINITION :
If to every value (Considered as real unless otherwise stated) of a variable x, which belongs to some
collection (Set) E, there corresponds one and only one finite value of the quantity y, then y is said to be
a function (Single valued) of x or a dependent variable defined on the set E ; x is the argument or
independent variable .
If to every value of x belonging to some set E there corresponds one or several values of the variable y,
then y is called a multiple valued function of x defined on E.Conventionally the word "FUNCTION” is
used only as the meaning of a single valued function, if not otherwise stated.
x f (x )  y
Pictorially : 
    , y is called the image of x & x is the pre-image of y under f.
input output

Every function from A  B satisfies the following conditions .

(i) f  Ax B (ii)  a  A  (a, f(a))  f and
(iii) (a, b)  f & (a, c)  f  b = c
2. DOMAIN, CODOMAIN & RANGE OF A FUNCTION :
Let f : A  B, then the set A is known as the domain of f & the set B is known as co-domain of f .
The set of all f images of elements of A is known as the range of f . Thus :
Domain of f = {a  a  A, (a, f(a))  f}
Range of f = {f(a)  a  A, f(a)  B}
It should be noted that range is a subset of codomain . If only the rule of function is given then the domain of
the function is the set of those real numbers, where function is defined. For a continuous function, the interval
from minimum to maximum value of a function gives the range.

3. IMPORTANT TYPES OF FUNCTIONS :

(i) POLYNOMIAL FUNCTION :
If a function f is defined by f (x) = a0 xn + a1 xn1 + a2 xn2 + ... + an1 x + an where n is a non negative integer
and a0, a1, a2, ..., an are real numbers and a0  0, then f is called a polynomial function of degree n .
NOTE : (a) A polynomial of degree one with no constant term is called an odd linear
function . i.e. f(x) = ax , a  0
(b) There are two polynomial functions , satisfying the relation ;
f(x).f(1/x) = f(x) + f(1/x). They are :
(i) f(x) = xn + 1 & (ii) f(x) = 1  xn , where n is a positive integer .
(ii) ALGEBRAIC FUNCTION :
y is an algebraic function of x, if it is a function that satisfies an algebraic equation of the form
P0 (x) yn + P1 (x) yn1 + ....... + Pn1 (x) y + Pn (x) = 0 Where n is a positive integer and
P0 (x), P1 (x) ........... are Polynomials in x.
e.g. y = x is an algebraic function, since it satisfies the equation y²  x² = 0.
Note that all polynomial functions are Algebraic but not the converse. A function that is not algebraic is
called TRANSCEDENTAL FUNCTION .
(iii) FRACTIONAL RATIONAL FUNCTION :
g(x)
A rational function is a function of the form. y = f (x) = , where
h (x )
g (x) & h (x) are polynomials & h (x)  0.

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4. DOMAINS AND RANGES OF COMMON FUNCTION :

Function Domain Range

(y = f (x) ) (i.e. values taken by x) (i.e. values taken by f (x) )

A. Algebraic Functions

(i) xn , (n  N) R = (set of real numbers) R, if n is odd

R+  {0} , if n is even
1
(ii) , (n  N) R – {0} R – {0} , if n is odd
xn
R+ , if n is even

(iii) x1 / n , (n  N) R, if n is odd R, if n is odd

R  {0} , if n is even
+
R  {0} , if n is even
+

1
(iv) 1/ n , (n  N) R – {0} , if n is odd R – {0} , if n is odd
x
R+ , if n is even R+ , if n is even

B. Trigonometric Functions
(i) sin x R [–1, + 1]
(ii) cos x R [–1, + 1]

(iii) tan x R – (2k + 1) , k I R
2

(iv) sec x , k I
R – (2k + 1) (–  , – 1 ]  [ 1 ,  )
2
(v) cosec x R – k , k  I (–  , – 1 ]  [ 1 ,  )
(vi) cot x R – k , k  I R
C. Inverse Circular Functions (Refer after Inverse is taught )

  
(i) sin–1 x [–1, + 1]  2 , 2 
 
(ii) cos–1 x [–1, + 1] [ 0, ]
  
(iii) tan–1 x R  , 
 2 2
  
(iv) cosec –1x (–  , – 1 ]  [ 1 ,  )  2 , 2  – { 0 }
 
 
(v) sec–1 x (–  , – 1 ]  [ 1 ,  ) [ 0, ] –  
2 
(vi) cot –1 x R ( 0, )

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5. EQUAL OR IDENTICAL FUNCTION :
Two functions f & g are said to be equal if :
(i) The domain of f = the domain of g.
(ii) The range of f = the range of g and
(iii) f(x) = g(x) , for every x belonging to their common domain. eg.
1 x
f(x) = & g(x) = 2 are identical functions .
x x
6. CLASSIFICATION OF FUNCTIONS :
One  One Function (Injective mapping) :
A function f : A  B is said to be a oneone function or injective mapping if different elements of A
have different f images in B . Thus for x1, x2  A & f(x1) ,
f(x2)  B , f(x1) = f(x2)  x1 = x2 or x1  x2  f(x1)  f(x2) .
Diagramatically an injective mapping can be shown as

OR

Note : (i) Any function which is entirely increasing or decreasing in whole domain, then
f(x) is oneone .
(ii) If any line parallel to xaxis cuts the graph of the function atmost at one point,
then the function is oneone .
Many–one function :
A function f : A  B is said to be a many one function if two or more elements of A have the same
f image in B . Thus f : A  B is many one if for ; x1, x2  A , f(x1) = f(x2) but x1  x2 .
Diagramatically a many one mapping can be shown as

OR

Note : (i) Any continuous function which has atleast one local maximum or local minimum, then f(x) is
manyone . In other words, if a line parallel to xaxis cuts the graph of the function atleast
at two points, then f is manyone .
(ii) If a function is oneone, it cannot be manyone and vice versa .
Onto function (Surjective mapping) :
If the function f : A  B is such that each element in B (codomain) is the f image of atleast one element
in A, then we say that f is a function of A 'onto' B . Thus f : A  B is surjective iff  b  B,  some
a  A such that f (a) = b .
Diagramatically surjective mapping can be shown as

OR

Note that : if range = codomain, then f(x) is onto.

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8. COMPOSITE OF UNIFORMLY & NON-UNIFORMLY DEFINED FUNCTIONS :
Let f : A  B & g : B  C be two functions . Then the function gof : A  C defined by
(gof) (x) = g (f(x))  x  A is called the composite of the two functions f & g .
x f (x)
Diagramatically      g (f(x)) .
Thus the image of every x  A under the function gof is the gimage of the fimage of x .
Note that gof is defined only if  x  A, f(x) is an element of the domain of g so that we can take its
g-image. Hence for the product gof of two functions f & g, the range of f must be a subset of the domain
of g.
PROPERTIES OF COMPOSITE FUNCTIONS :
(i) The composite of functions is not commutative i.e. gof  fog .
(ii) The composite of functions is associative i.e. if f, g, h are three functions such that fo (goh) &
(fog) oh are defined, then fo (goh) = (fog) oh .
(iii) The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is
defined, then gof is also a bijection.
9. HOMOGENEOUS FUNCTIONS :
A function is said to be homogeneous with respect to any set of variables when each of its terms
is of the same degree with respect to those variables .
For example 5 x2 + 3 y2  xy is homogeneous in x & y . Symbolically if ,
f (tx , ty) = tn . f (x , y) then f (x , y) is homogeneous function of degree n .
10. BOUNDED FUNCTION :
A function is said to be bounded if f(x)  M , where M is a finite quantity .
11. IMPLICIT & EXPLICIT FUNCTION :
A function defined by an equation not solved for the dependent variable is called an
IMPLICIT FUNCTION . For eg. the equation x3 + y3 = 1 defines y as an implicit function. If y has been
expressed in terms of x alone then it is called an EXPLICIT FUNCTION.
12. INVERSE OF A FUNCTION :
Let f : A  B be a oneone & onto function, then their exists a unique function
g : B  A such that f(x) = y  g(y) = x,  x  A & y  B . Then g is said to be inverse of f . Thus
g = f1 : B  A = {(f(x), x)  (x, f(x))  f} .
PROPERTIES OF INVERSE FUNCTION :
(i) The inverse of a bijection is unique .
(ii) If f : A  B is a bijection & g : B  A is the inverse of f, then fog = IB and
gof = IA , where IA & IB are identity functions on the sets A & B respectively.
Note that the graphs of f & g are the mirror images of each other in the
line y = x . As shown in the figure given below a point (x ',y ' ) corresponding to y = x2 (x >0)
changes to (y ',x ' ) corresponding to y   x , the changed form of x = y .

(iii) The inverse of a bijection is also a bijection .

(iv) If f & g are two bijections f : A  B , g : B  C then the inverse of gof exists and
(gof)1 = f1 o g1 .

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EXERCISE–I
Q.1 Find the domains of definitions of the following functions :
(Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)
(i) f (x) = cos2x  16  x 2 (ii) f (x) = log7 log5 log3 log2 (2x3 + 5x2  14x)

1  5x
(iii) f (x) = ln  x 2  5x  24  x  2  (iv) f (x) =
  7 x 7

 2 log10 x  1 
(v) y = log10 sin (x  3)  16  x 2 (vi) f (x) = log100 x  
 x 
1 x
(vii) f (x) =  ln x(x 2  1) (viii) f (x) = log 1 2
4x 2  1 2 x 1

1
(ix) f (x)  x 2  x  (x) f (x) = ( x 2  3x  10) . ln 2 ( x  3)
9  x2
1
cos x 
(xi) f(x) = logx (cos 2x) (xii) f (x) = 2
6  35x  6x 2

(xiii) f(x) = log1 / 3 log 4   [x] 2

5  (xiv) f(x) =
1
[x]
 log (2{x} 5) (x 2  3x  10) 
1
1 x
,

(xv) f(x) = logx sin x

  
 1 
(xvi) f(x) = log2   log1/ 2  1  + log10 log10 x  log10  4  log10 x  log10 3
 
x
 
 
sin 100  
1 1 1
(xvii) f (x) = + log1 – {x}(x2 – 3x + 10) + +
[x] 2| x| sec(sin x)
1
 7 
(xviii) f (x) = (5x  6  x 2 ) lnx + (7 x  5  2x 2 ) +  ln   x  
 2 
f
(xix) If f(x) = x 2  5 x  4 & g(x) = x + 3 , then find the domain of (x) .
g
Q.2 Find the domain & range of the following functions .
( Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)
2x
(i) y = log 5  2 (sin x  cos x)  3  (ii) y =
1 x2
(iii) f(x) =
x 2  3x  2
x2  x  6
x
(iv) f (x) = 1 | x | (v) y = 2  x  1  x

x 4 3
(vi) f (x) = log(cosec x - 1) (2  [sin x]  [sin x]2) (vii) f (x) =
x 5
Q.3 Draw graphs of the following function , where [ ] denotes the greatest integer function.
(i) f(x) = x + [x]
(ii) y = (x)[x] where x = [x] + (x) & x > 0 & x  3
(iii) y = sgn [x] (iv) sgn (x x)

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1  3 
Q.16 A function f :  ,    ,  defined as, f(x) = x2  x + 1. Then solve the equation f (x) = f 1 (x).
2  4 

Q.17 Function f & g are defined by f(x) = sin x, xR ; g(x) = tan x , xR   K  1  
 2
where K I . Find (i) periods of fog & gof. (ii) range of the function fog & gof .
Q.18 Find the period for each of the following functions :
(a) f(x)= sin4x + cos4x (b) f(x) = cosx (c) f(x)= sinx+cosx
3 2
(d) f(x)= cos x  sin x .
5 7
Q.19 Prove that the functions ; (a) f(x) = cos x (b) f(x) = sin x
(c) f(x) = x + sin x (d) f(x) = cos x2 are not periodic .
Q.20 Find out for what integral values of n the number 3 is a period of the function :
f(x) = cos nx . sin (5/n) x.
EXERCISE–II
Q.1 Let f be a oneone function with domain {x,y,z} and range {1,2,3}. It is given that exactly one of the
following statements is true and the remaining two are false .
f(x) = 1 ; f(y)  1 ; f(z)  2 . Determine f1(1)
Q.2 Solve the following problems from (a) to (e) on functional equation.
(a) The function f (x) defined on the real numbers has the property that f  f ( x ) ·1  f ( x )  = – f (x) for all
x in the domain of f. If the number 3 is in the domain and range of f, compute the value of f (3).
(b) Suppose f is a real function satisfying f (x + f (x)) = 4 f (x) and f (1) = 4. Find the value of f (21).

(c) Let 'f' be a function defined from R+  R+ . If [ f (xy)]2 = x  f ( y) 2 for all positive numbers x and y and
f (2) = 6, find the value of f (50).
(d) Let f (x) be a function with two properties
(i) for any two real number x and y, f (x + y) = x + f (y) and
(ii) f (0) = 2.
Find the value of f (100).
(e) Let f be a function such that f (3) = 1 and f (3x) = x + f (3x – 3) for all x. Then find the value of f (300).

Q.3(a) A function f is defined for all positive integers and satisfies f(1) = 2005 and f(1)+ f(2)+ ... + f(n) = n2f(n)
for all n > 1. Find the value of f(2004).
(b) If a, b are positive real numbers such that a – b = 2, then find the smallest value of the constant L for
which x 2  ax  x 2  bx < L for all x > 0.
(c) Let f (x) = x2 + kx ; k is a real number. The set of values of k for which the equation f (x) = 0 and
f  f ( x )  = 0 have same real solution set.
(d) If f (2x + 1) = 4x2 + 14x, then find the sum of the roots of the equation f (x) = 0.
ax  b 5
Q.4 Let f (x) = for real a, b and c with a  0. If the vertical asymptote of y = f (x) is x = – and the
4x  c 4
3
vertical asymptote of y = f –1 (x) is x = , find the value(s) that b can take on.
4

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1
Q.18 Find the set of real x for which the function f(x) = is not defined, where [x]
   
x  1  12  x  11
denotes the greatest integer function.
Q.19 A is a point on the circumference of a circle. Chords AB and AC divide the area of the circle into three
equal parts . If the angle BAC is the root of the equation, f (x) = 0 then find f (x) .
Q.20 If for all real values of u & v, 2 f(u) cos v = f(u + v) + f(u  v), prove that, for all real values of x
(i) f(x) + f( x) = 2a cos x (ii) f( x) + f( x) = 0
(iii) f( x) + f(x) =  2b sin x . Deduce that f(x) = a cos x  b sin x, a, b are arbitrary constants.

EXERCISE–III
Q.1 If the functions f , g , h are defined from the set of real numbers R to R such that ;
0, if x  0

f (x)= x2  1, g (x) = x 2  1 , h (x) =  ; then find the composite function ho(fog) & determine
 x , if x  0
whether the function (fog) is invertible & the function h is the identity function. [REE '97, 6]

 
2
Q.2(a) If g (f(x)) = sin x & f (g(x)) = sin x , then :
(A) f(x) = sin2 x , g(x) = x (B) f(x) = sin x , g(x) = x
(C) f(x) = x2 , g(x) = sin x (D) f & g cannot be determined
(b) If f(x) = 3x  5, then f1(x)
1 x 5
(A) is given by (B) is given by
3x  5 3
(C) does not exist because f is not oneone (D) does not exist because f is not onto
[JEE'98, 2 + 2]
Q.3 If the functions f & g are defined from the set of real numbers R to R such that f(x) = ex,
g(x) = 3x  2, then find functions fog & gof. Also find the domains of functions (fog)1 & (gof)1.
[ REE '98, 6 ]
Q.4 If the function f : [1, )  [1, ) is defined by f(x) = 2x (x  1), then f1(x) is : [ JEE '99, 2 ]
x (x  1)
 1
(A)  
 2
(B)
1
2
1  1  4 log2 x  (C)
1
2

1  1  4 log2 x  (D) not defined

Q.5 The domain of definition of the function, y (x) given by the equation, 2x + 2y = 2 is :
(A) 0 < x  1 (B) 0  x  1 (C)  < x  0 (D)  < x < 1
[ JEE 2000 Screening), 1 out of 35 ]
Q.6 Given x = {1, 2, 3, 4}, find all oneone, onto mappings, f : X  X such that,
f (1) = 1 , f (2)  2 and f (4)  4 . [ REE 2000, 3 out of 100 ]
 1 , x  0

Q.7(a) Let g (x) = 1 + x  [ x ] & f (x) =  0 , x  0 . Then for all x , f (g (x)) is equal to
1 , x0

(A) x (B) 1 (C) f (x) (D) g (x)
1
(b) If f : [1 , )  [2 , ) is given by , f (x) = x + , then f 1 (x) equals
x
x  x2  4 x x  x2  4
(A) (B) 2 (C) (D) 1  x2  4
2 1 x 2

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Q.12 For the following questions, choose the correct answer from the codes (A),(B),(C) & (D) defined as
follows.
(A) Statement I is true, Statement II is also true;Statement II is correct explanation of Statement I.
(B) Statement I is true, Statement II is also true;Statement II is NOT correct explanation of Statement I.
(C) Statement I is true, Statement II is false.
(D) Statement I is false, Statement II is true.

(a) Let F (x) be an indefinite integral of sin2 x.

Statement-I: The function F(x) satisfies F  x     F  x  for all real x.
because
Statement-II: sin 2  x     sin 2 x for all real x.

(b) Let f (x) = 2 + cos x for all real x.

Statement-I: For each real t, there exists a point c in t , t    such that f'(c) = 0.
because
Statement-II: f  t   f  t  2  for each real t.

x2
(c) Statement-I: The curve y   x  1 is symmetric with respect to the line x = 1.
2
because
Statement-II: A parabola is symmetric about its axis. [IIT JEE 2007;3+3+3]

Q.13 Match the Column

Match the condition/expression in Column I with statement in Column II.
x2  6x  5
Let f  x   2 .
x  5x  6
Column I Column II
(A) If -1 < x < 1, then f (x) satisfies (P) 0 <f (x) < 1
(B) If 1 < x < 2, then f (x) satisfies (Q) f (x) < 0
(C) If 3 < x < 5, then f (x) satisfies (R) f (x) > 0
(D) If x > 5, then f (x) satisfies (S) f (x) < 1 [IIT JEE 2007;6]

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xy
P5 tan1 x + tan1 y = tan1 where x > 0 , y > 0 & xy < 1
1  xy

xy
=  + tan1 where x > 0 , y > 0 & xy > 1
1  xy

xy
tan1 x  tan1y = tan1 where x > 0 , y > 0
1  xy

sin1 x + sin1 y = sin1 x 1  y  y 1  x  where x  0 , y  0 & (x2 + y2)  1

2 2
P6 (i)


Note that : x2 + y2  1  0  sin1 x + sin1 y 
2

sin1 x + sin1 y =   sin1 x 1  y  y 1  x  where x  0 , y  0 & x2 + y2 > 1

2 2
(ii)


Note that : x2 + y2 >1  < sin1 x + sin1 y < 
2

(iii) 
sin–1x – sin–1y = sin 1 x 1  y 2  y 1  x 2  where x > 0 , y > 0

(iv) 
cos1 x + cos1 y = cos1 x y  1 x 2 1 y 2  where x  0 , y  0

 x  y  z  xyz 
P7 If tan1 x + tan1 y + tan1 z = tan1 1  x y  y z  z x  if, x > 0, y > 0, z > 0 & xy + yz + zx < 1
 
Note : (i) 1 1 1
If tan x + tan y + tan z =  then x + y + z = xyz

(ii) If tan1 x + tan1 y + tan1 z = then xy + yz + zx = 1
2

1 1  x = tan1 2 x
2
2x
P8 2 tan1 x = sin1 = cos
1  x2 1  x2 1  x2
Note very carefully that :
 2 tan 1 x if x 1
2x  1  x2  2 tan 1 x
if x  0
sin1 =   2 tan 1 x if x1 cos1 =  1
1 x 1  x2  2 tan x if x  0
2

 
    2 tan x
1
 if x  1

 2tan 1 x if x 1
2x 
tan1 =   2tan 1 x if x  1
 
1  x2
  2tan 1 x if x 1

REMEMBER THAT :
3
(i) sin1 x + sin1 y + sin1 z =  x = y = z = 1
2
(ii) cos1 x + cos1 y + cos1 z = 3  x = y = z = 1
(iii) tan1 1+ tan1 2 + tan1 3 =  and tan1 1 + tan1 12 + tan1 13 = 2

Bansal Classes Functions & Trig.-- IV [18]

 BANSAL CLASSES
PRIVATE LIMITED
  
7. (a) y = sin 1 (sin x) , x  R , y   ,  , 7.(b) y = sin (sin 1 x) ,
 2 2
Periodic with period 2  = x
x  [ 1 , 1] , y  [ 1 , 1] , y is aperiodic

8. (a) y = cos 1(cos x), x  R, y[0, ], periodic with period 2  8. (b) y = cos (cos 1 x) ,
= x = x
x  [ 1 , 1] , y  [ 1 , 1], y is aperiodic

9. (a) y = tan (tan 1 x) , x  R , y  R , y is aperiodic 9. (b) y = tan 1 (tan x) ,

=x = x
      
x  R  (2 n  1) 2 n  I  , y    2 , 2  ,
 
periodic with period 

10. (a) y = cot 1 (cot x) , 10. (b) y = cot (cot 1 x) ,

= x = x
x  R  {n } , y  (0 , ) , periodic with  x  R , y  R , y is aperiodic

Bansal Classes F u n cti on s & Tr i g.- - IV [20]

 BANSAL CLASSES
PRIVATE LIMITED
Q.4 Find the domain of definition the following functions.
( Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)
2x 1  x2
(i) f(x) = arc cos (ii) cos (sin x)  sin 1
1 x 2x
 x  3
(iii) f (x) = sin 1    log10 ( 4  x )
 2 
1  sin x
(iv) f(x) =  cos 1 (1  {x}) , where {x} is the fractional part of x .
log 5 (1  4x 2 )
3  2x
(v) f (x) = 3  x  cos 1   
  log6 2 x  3  sin log 2 x
1
 5 
 3 
(vi) f (x) = log10 (1  log7 (x2  5 x + 13)) + cos1  
9 x 
 2  sin 2 

sin 1  x2 
(vii) f(x) = e
x 
 tan 1   1  n
2 
 x  [x] 
2 sin x  1 1  
(viii) f(x) = sin(cos x) + ln ( 2 cos2 x + 3 cos x + 1) + ecos  

 2 2 sin x 

Q.5 Find the domain and range of the following functions .

(Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)

(i) f (x) = cot1(2x  x²) (ii) f (x) = sec1 (log3 tan x + logtan x 3)
 2 x2  1   
(iii) f(x) = cos1 
 x2  1 
 
(iv) f (x) = tan 1  log 4 5x 2  8x  4 


  5

Q.6 Find the solution set of the equation, 3 cos1 x = sin1  1  x 2 (4 x 2  1)  .

 

Q.7 Prove that:


(a) sin–1 cos (sin1 x) + cos–1 sin (cos–1 x) = , | x |  1
2
(b) 2 tan1 (cosec tan1x  tan cot1x) = tan1x (x  0)
 2mn   2pq   2MN 
(c) tan1   + tan1  2  = tan1  2  where M = mp  nq, N = np + mq,
m  n 
2 2
p  q 
2
 M  N2 

n q N
1 ; 1 and 1
m p M
(d) tan (tan1 x + tan1 y + tan1 z) = cot (cot1 x + cot1 y + cot1 z)

x 1  1 
Q.8 Find the simplest value of, arc cos x + arc cos   3  3x2  , x   , 1
 2 2  2 

2 2
x y x 2.xy y
Q.9 If cos1 + cos1 =  then prove that 2  cos   2  sin 2  .
a b a ab b

Bansal Classes Functions & Trig.-- IV [22]

 BANSAL CLASSES
PRIVATE LIMITED
EXERCISE–II
 1 a  1 a 2b
Q.1 Prove that: (a) tan   cos 1  + tan   cos 1  =
4 2 b 4 2 b a

cos x  cos y  x y  ab x  b  a cos x 

(b) cos1 = 2 tan1  tan . tan  (c) 2 tan1  a  b . tan 2  = cos1  
1  cos x cos y 2 2    a  b cos x 

 1  x2  1  x2 
Q.2 If y = tan1   prove that x² = sin 2y..
 1  x2  1  x2 
 

Q.3 If u = cot1 cos2  tan1 cos2 then prove that sin u = tan2 .

1  x  1  x2 
Q.4 If  = 2 arc tan   &  = arc sin   for 0 < x < 1 , then prove that + =, what the
1  x 
2
1  x
value of  +  will be if x > 1.

 1
Q.5 If x  1,  then express the function f (x) = sin–1 (3x – 4x3) + cos–1 (4x3 – 3x) in the form of
 2
–1
a cos x + b , where a and b are rational numbers.

Q.6 Find the sum of the series:

1 2 1 n  n1
(a) sin1 + sin1 + ..... + sin1 + ...... 
2 6 n (n  1)
1 2 2 n 1
(b) tan1 + tan1 + ..... + tan1 1  22n 1 + ..... 
3 9
(c) cot17 + cot113 + cot121 + cot131 + ...... to n terms.
1 1 1 1
(d) tan1 + tan1 2 + tan1 2 + tan1 2 to n terms.
x x1 2
x  3x  3 x  5x  7 x  7 x  13
1 1 1 1
(e) tan1 + tan1 + tan1 + tan1 + ..... 
2 8 18 32

Q.7 Solve the following

(a) cot1x + cot1 (n²  x + 1) = cot1 (n  1)
x x
(b) sec1  sec1 = sec1b  sec1a a  1; b  1, a  b.
a b
x  1 2x  1 23
(c) tan1 + tan1 2 x  1 = tan1
x1 36

3  1 1    3  1 1  
cosec2  tan sec2  tan as an integral polynomial in  & .
 
Q.8 Express  +
2 2  2 2

Q.9 Find the integral values of K for which the system of equations ;
 2 K 2
arc cos x  (arc sin y) 
 4 possesses solutions & find those solutions.
4
(arc sin y) 2 . (arc cos x)  
 16

Bansal Classes Functions & Trig.-- IV [24]

 BANSAL CLASSES
PRIVATE LIMITED
EXERCISE–III

Q.1 The number of real solutions of tan1 x (x  1) + sin1 x 2  x  1 = is :
2
(A) zero (B) one (C) two (D) infinite [JEE '99, 2 (out of 200)]

Q.2 Using the principal values, express the following as a single angle :
 1  1 142
3 tan1   + 2 tan1   + sin1 . [ REE '99, 6 ]
  2  5 65 5

ax bx
Q.3 Solve, sin1 + sin1 = sin1x, where a2 + b2 = c2, c  0. [REE 2000(Mains), 3 out of 100]
c c

Q.4 Solve the equation:

cos1  6x  cos 3 1
3x 2  
2
[ REE 2001 (Mains), 3 out of 100]

 x2 x 3  
 ........ + cos–1  x 2  x  x  ........ = for 0 < | x | <
4 6
Q.5 If sin–1  x   2 then x equals to
 2 4   2 4  2
[JEE 2001(screening)]
(A) 1/2 (B) 1 (C) – 1/2 (D) – 1

x2  1
Q.6 Prove that cos tan–1 sin cot –1 x = [JEE 2002 (mains) 5]
x2  2


Q.7 Domain of f (x) = sin 1 (2x )  is
6
 1 1  1 3  1 1  1 1
(A)   ,  (B)  ,  (C)  ,  (D)  , 
 2 2  4 4  4 4  4 2
[JEE 2003 (Screening) 3]

Q.8  
If sin cot 1 ( x  1)  cos(tan 1 x ) , then x =
1 1 9
(A) – (B) (C) 0 (D)
2 2 4
[JEE 2004 (Screening)]
MATCH THE COLUMN
Q.9 Let (x, y) be such that

sin 1  ax   cos 1  y   cos 1  bxy  
2
Match the statements in Column I with statement in Column II and indicate your answer by darkening
the appropriate bubbles in the 4  4 matrix given in the ORS.
Column I Column II
(A) If a = 1 & b = 0, then (x, y) (P) lies on the circle x2 + y2 = 1
(B) If a = 1 & b = 1, then (x, y) (Q) lies on (x2 - 1)(y2 - 1) = 0
(C) If a = 1 & b = 2, then (x, y) (R) lies on y = x
(D) If a = 2 & b = 2, then (x, y) (S) lies on (4x2 - 1)(y2 - 1) = 0 [JEE 2007,6]
Bansal Classes Functions & Trig.-- IV [26]
 BANSAL CLASSES
PRIVATE LIMITED
ANSWER KEY
FUNCTIONS
EXERCISE–I
 5  3      3 5    1
Q 1. (i)  ,   ,    ,  (ii)   4 ,   (2, )(iii) (–  , – 3]
 4 4   4 4  4 4  2

 1   1 1 
(iv) (– , – 1) [0, ) (v) (3  2 < x < 3  ) U (3 < x  4) (vi)  0,  , 
 100   100 10 
1  5  1  5 
(vii) (1 < x < 1/2) U (x > 1) (viii)  , 0   ,  (ix) (3, 1] U {0} U [ 1,3 )
 2   2 

 1    5 
(x) { 4 }  [ 5, ) (xi) (0 , 1/4) U (3/4 , 1) U {x : x  N, x  2} (xii)   ,    , 6 
 6 3  3 
(xiii) [– 3,– 2)  [ 3,4) (xiv) 
(xv) 2K < x < (2K + 1) but x  1 where K is nonnegative integer
 5
(xvi) {x 1000  x < 10000} (xvii) (–2, –1) U (–1, 0) U (1, 2) (xviii) (1, 2)   2, 
 2
(xix) ( , 3)  (3 , 1]  [4 , )

Q 2.
(i) D : x R R : [0 , 2] (ii) D = R ; range [ –1 , 1 ]
(iii) D : {xx R ; x  3 ; x  2} R : {f(x)f(x)R , f(x)  1/5 ; f(x)  1}
(iv) D : R ; R : (–1, 1) (v) D : 1  x  2 R :  3, 6 
(vi)  
D : x  (2n, (2n + 1))  2 n  6 , 2 n  2 , 2 n  56 , n  I and
R : loga 2 ; a  (0, )  {1}  Range is (–, ) – {0}
 1   1 1
(vii) D : [– 4, ) – {5}; R :  0,    , 
 6   6 3

Q.4 (a) neither surjective nor injective (b) surjective but not injective
(c) neither injective nor surjective
Q.5 f3n(x) = x ; Domain = R  {0 , 1}
Q.6 1 Q.7 (a) 2K  x  2K +  where K I (b) [3/2 , 1]
Q.8 (i) (a) odd, (b) even, (c) neither odd nor even, (d) odd, (e) neither odd nor even, (f) even,
1  5 1  5 3  5 3  5
(g) even, (h) even; (ii) , , ,
2 2 2 2
Q.9 (a) y = log (10  10x) ,   < x < 1
(b) y = x/3 when   < x < 0 & y = x when 0  x < + 
Q.10 f1(x) = (a  xn)1/n
Q.12 (a) f(x) = 1 for x < 1 & x for 1  x  0; (b) f(x) = 1 for x < 1 and x for 1  x  0

Bansal Classes Functions & Trig.-- IV [28]

 BANSAL CLASSES
PRIVATE LIMITED
INVERSE TRIGONOMETRY FUNCTIONS
EXERCISE–I
1 5  4 17 1  2 4
Q 1. (i) , (ii) 1, (iii) , (iv)  , (v) , (vi) Q 2. (i) , (ii) 1, (iii)  , (iv) , (v) , (vi) 
3 6 3 5 6 2 4 3 5
Q.3 (d) (–, sec 2)  [1, )
Q 4. (i) 1/3  x  1 (ii) {1, 1} (iii) 1 < x < 4
(iv) x (1/2 , 1/2), x  0 (v) (3/2 , 2]

(vi) {7/3, 25/9} (vii) (2, 2)  {1, 0, 1} (viii) {xx = 2n  + , n  I}
6
Q5. (i) D : x R R : [/4 , )
     2 
(ii) D: x   n, n   x x  n   n  I ; R:  ,    
 2  4 3 3  2

    
(iii) D: xR R : 0 ,  (iv) D: xR R :  , 
 2  2 4
 3  
Q 6.  , 1 Q 8. Q.11 
 2  3

1 3 1 1 3
Q.12 (a) x = (b) x = 3 (c) x = 0 , , (d) x =
2 7 2 2 10
1 ab
(e) x = 2  3 or 3 (f) x = , y = 1 (g) x =
2 1 ab
1 17
Q.13 57 Q.14 53 Q 19. x = 1 ; y = 2 & x = 2 ; y = 7 Q.20
2
EXERCISE–II
9 9
Q 4.  Q5. 6 cos2x – , so a = 6, b = –
2 2
  2n  5 
Q 6. (a) (b) (c) arc cot   (d) arc tan (x + n)  arc tan x (e)
2 4  n  4
4
Q 7. (a) x = n²  n + 1 or x = n (b) x = ab (c) x = Q 8. (2 + 2) (+ )
3

2 2
Q 9. K = 2 ; cos ,1 & cos , 1 Q 10. 720 Q.11 X = Y= 3  a 2
4 4

11 F 2 , 1OP  2 , 1   1 ,  2 
Q 12. k =
4
Q 14. (a) (cot 2 , )  (, cot 3) (b) GH 2 Q (c)  2    2 
 1 
Q15.  tan , cot 1 Q16. C1 is a bijective function, C2 is many to many correspondence, hence it is not a function
 2 
1 1
Q17. [e/6 , e] Q 18.(a) D : [0, 1] , R : [0, /2] (b)   x  (c) D : [ 1, 1] , R : [0, 2]
2 2
3
Q.19 Q.20 x  (–1, 1)
4
EXERCISE–III
1
Q.1 C Q.2  Q.3 x { 1, 0, 1} Q.4 x = Q.5 B Q.7 D Q.8 A
3
Q.9 (A) P; (B) Q; (C) P; (D) S Q.10 (1)D (2)B (3)A Q.11 (A) P (B) Q (C) Q (D) P
Bansal Classes Functions & Trig.-- IV [30]