Mathematics-12.Unit-8 Function

FUNCTION SYLLABUS Function, into, onto and one to one. Sum, difference, product and quotient of two functions, composite function, absolute value, greatest integer. Polynomial, rational, trigonometric, exponential and logarithmic functions. Even and odd functions. Inverse of a function. 1. FUNCTION Function can be easily defined with the help of the concept of mapping. Let X and Y be any two non-empty sets. "A function from X to Y is a rule or correspondence that assigns to each element of set X, one and only one element of set Y". Let the correspondence be ‘f’ then mathematically we write f: X Y where y = f(x), x X and y Y. We say that 'y' is the image of 'x' under ‘f ’ (or x is the pre image of y). • A mapping f: X Y is said to be a function if each element in the set X has it's image in set Y. It is possible that a few elements in the set Y are present which are not the images of any element in set X. • Every element in set X should have one and only one image. That means it is impossible to have more than one image for a specific element in set X. Functions can't be multi-valued (A mapping that is multi-valued is called a relation from X to Y) Example 1. Let X = {1, –1, 2, 3, –3}, Y = {1, 4, 9, 10,11}. The Rule ¦ given by y = x2 is a function from X to Y. Domain ¦ = X = {1, –1, 2, 3, –3} Range ¦ = {1, 4, 9} Example 2. Let X = {1, –1, 2, 3}, Y = {2, –2, 4, 6}. The rule ¦ given by y = 2x is a function from X to Y. Domain = ¦ = X = {1, –1, 2, 3}, Range = {2, –2, 4, 6} Example 3. Let A = {1, 4, 9}, B = {-3, - 2, -1,1, 2, 3, 4}. The rule ¦ given by y2 = x is not a function. Graphical method: to check whether a relation between x and y is a function or not, draw a line parallel to y-axis and, if it interest the graph at one and only one point then the given relation represent a function. DOMAIN AND RANGE OF A FUNCTION The set X is called the domain of the function. ‘¦’ and set Y is called the co-domain. The set of the image of all elements of X under the function ‘¦’ is called the range of ‘¦’ and is denoted by ¦(x). It is obvious that range could be a subset of co-domain as we may have few elements in co-domain which are not the images of any element of the set X. Thus range of ‘¦’ i.e. ¦(x) = {¦(x): x Î X}. Clearly ¦(x) Í Y 1.1 PROBLEM BASED ON DOMAIN OF A REAL VALUED FUNCTION Illustration 1: Find the domain of Key concept: Denominator should be non-zero for any function Solution: x2 – 4 ¹ 0 Þ x ¹ ± 2 Hence domain is R – {–2, 2} Illustration 2: Find the domain of Key concept: Expression under even root (i.e. square root, fourth root, sixth root etc) should not be negative. Solution: f(x) is defined when ….(1) Case 1: x 0 For domain …….(2) Case 2: x < 0 For domain Rejecting the values of x because they don’t satisfy the inequality x < 0. We get ……(3) Taking union of (2) and (3) Domain = Illustration 3: Find the domain of . Key concept: is defined for values of x at which f(x)>0 and g(x)>0 1. Solution: and and Hence the domain is Illustration 4: Find the domain of . Key concept: are defined for are defined for all real x are defined for i.e. for all real x except those lying between –1 and 1. Solution: is defined when …………(2) Taking intersection of (1) and (2) 1.2 PROBLEM BASED ON RANGE OF A REAL VALUED FUNCTION Illustration 5: If then find the range of f(x). Key concept: If f(x) is in the form of , where p(x) and Q(x) are polynomial function of second degree we use the concept of quadratic equation. Solution: Let Illustration 6: Find the range of . Key concept: If in a function there is only one variable whose range if known, and can be easily found in terms of y, then we use the range of that variable to find the range of the given function. Solution: We know for any real x, Range Key concept: Using the method of substituting Alternate: Given (put x = tanq where , don’t put x= . Otherwise this will restrict the range of x) y = sin2q If q Î (–p/2, p/2), sin2q Î [0, 1) Hence range is [0, 1) Illustration ¬7: Find the range of ¦(x) = 2x3 – 9x2 + 12x + 1, x Î [–1, 3] Key concept: To find the range of a polynomial function generally we use the concept of calculus. Solution: f(x) is increasing in and decreasing in (1, 2). Hence the graph of the given function is Hence the range is [-22, 10]. Illustration 8: Find the range of . Key concept: Range of and is and the range of is . Also . Common mistake: students used to solve in this manner Hence range is , But this answer is wrong. Because range of is when and if we find the domain of f(x) it is [-1,1]. And in this domain belongs to . Hence f(x) = Hence the range is VALUE OF A FUNCTION The value of a function y = ¦(x) at x = a is denoted by ¦(a). It is obtained by putting x = a in ¦(x) Illustration 9: If , xÎ(–1, 1), then find the value of function for x=0,1/2. Solution. , 2. ALGEBRA OF FUNCTIONS Let two functions be f: D1® R and g: D2 ® R. We describe functions f + g, f - g, f. g and f /g as follows:  f + g : D ®R is a function defined by (f + g)x = f(x) + g(x) where D = D1ÇD2  f - g : D® R is a function defined by (f - g)x = f(x) –g(x) where D = D1ÇD2  f . g : D ® R is a function defined by by (f . g)x = f(x). g(x) where D = D1ÇD2  f / g : D ® R is a function defined by (f /g)x = where D = D1Ç{xÎD2: g(x) ¹0} Illustration 10: Let f(x) = , g(x) = . Find f + g, f - g, f . g and f /g Solution: (f+g)x = , defined on -1£ x£1 (f-g)x = , defined on -1£ x£1 (f.g)x = , defined on -1£ x£1 (f/g)x = , defined on -1 < x £1 Illustration 11: Solve for x, Key concept: First make the base of log same both the sides. Then solve.  Common Mistake: Given.  After this step students used to solve in this manner  They forget to check the base of log. Hence the answer is wrong. Solution: Given. We know The base of log lying between 0 and 1 hence inequality sign will change, hence . Also log x is defined only when x>0. Hence the answer is 3. GREATEST INTEGER AND FRACTIONAL PART GREATEST INTEGER: Any real number x can always think of lying between two consecutive integers say P and P+1. i.e. P £ x < (P + 1). That means, there always exist an integer, say ‘P’ which is just less than or equal to x. This unique ‘P’ is called the greatest integral value of x and is symbolically denoted as [x] i.e. [x] stands for the greatest integer that is less than or equal to x. e.g. x = 3.54 Þ 3 < x < 4 Þ [x] = 3, x = -2.95 Þ -3 < x < -2 Þ [x] = - 3 It is obvious that if x is integer, then [x] = x. Domain ® R; Range ® I; Period ® non periodic; Nature ® neither even nor odd IMPORTANT POINTS: • P £ x < P + 1 Û [x] = P • P1  [x]  P2  P1  x < P2 + 1  x  [P1, P2 + 1) e.g. -2 £ [x] £2 Þ -2 £ x < 3 Þ x Î [-2, 3) • x-1 < [x] £ x • [[x]] = [x] • [n + x] = n + [x] where n ÎI • FRACTIONAL PART: Fractional Part of any real number is defined as the difference between the number ‘x’ and it’s integral value ‘[x]’ and is symbolically denoted as {x}. Thus, {x} = x – [x], e.g. if x = 5.68, then [x] = 5 and {x} = 0.68. If x is an integer Þ x = [x] Þ {x} = 0 Þ {[x]} = 0 If , then [x]=0 {x}=x , then [x]=1 {x}=x-1 Domain ® R; Range ® [0,1); Period ® 1; Nature ® neither even nor odd IMPORTANT POINTS: • 0 £ {x} < 1 • [{x}] = 0, {[x]} = 0 • x – 1 < [x] £ x, 0 £ {x} < 1 • {x} + {–x} = Illustration 12: Solve for x,  Common Mistake: Student used to solve in this manner  First they cancel the common term both the sides. [x]=2 . Solution: First step is correct. Cancel common term. But The important point is that is not defined when x=2.7, hence exclude this from the solution set. 3.1 TYPE OF FUNCTIONS ONE – ONE OR INJECTIVE FUNCTION: A function f: X ®Y is said to be one – one or injective if each element in the domain of a function has a distinct image in the co – domain. Example: f:R R f(x) = 2x is one – one. MANY – ONE FUNCTION: A function f: X ®Y is said to be many one if there are at least two elements in the domain whose images are the same. EXAMPLE: f: R R given by f(x) = x2 is Many – one. METHODS TO DETERMINE ONE – ONE AND MANY – ONE:  If f(x1) = f (x2) x1 = x2 for every x1,x2 in the domain, then ‘f’ is one – one else many – one.  If the function is entirely increasing or decreasing in the domain, then ‘f’ is one – one else many – one. GRAPHICAL METHOD: If we draw a line parallel to the x – axis intersect the graph of y = f(x) at one and only one point, then f(x) is one – one and if the line parallel to the x – axis cuts the graph at more than one different points then f(x) is a many – one function.  Any continuous function f(x) which has at least one local maxima or local minima is many – one.  All even functions are many one.  All polynomials of even degree defined on R have at least one local maximum or minima and hence are many one on the domain R. Polynomials of odd degree can be one – one or many – one. ONTO FUNCTION OR SURJECTIVE FUNCTION: A function f: X®Y is said to be a onto function or Surjective function if and only if each element of Y is the image of some element of X i. e. if and only if for every y Î Y there exists some x Î X such that y = f(x). Thus ‘f’ is onto if f(x) = Y i. e. range = co – domain of function. Example: The map f: R ®[ –1,1] given by f(x) = sin x is an onto map. INTO FUNCTION: A function f: X®Y is said to be an into function if there exists at least one element in the co – domain Y which is not an image of any element in the domain X. Example: The map f: R ®R given by f(x) = x2 is an into map Onto Into ONE – ONE ONTO MAP OR BIJECTIVE FUNCTION: A function f: X ®Y is said to be one – one onto or bijective function if and only if (i) distinct elements of X have distinct images in Y (ii) each element of Y has at least one pre – image in X. Example: The map f: X ®Y given by f(x) = 2x is a one – one onto map. Illustration 13: If f: R R where , find whether f(x) is one – one or many one. Key concept : If f(x) is a rational function then f(x1) = f(x2) will always be satisfied when x1 = x2 in the domain. Hence we can write where is some function in x1 and x2. Now if = 0 gives some solution which is different from x1 = x2 and which lies in the domain, then f is many – one else one – one. Solution: One solution of this is obviously x1 = x2. Also we have got a relation in x1 and x2 and for each value of x1 in the domain we get a corresponding value of x2 which may or may not be same as x1. If x1 = 1 we get , and both lies in the domain of f. Hence we have two different values x1 and x2 for which f(x) has the same value. Hence is many one. Illustration 14: Let f: R [2, ) defined by f(x) = is an onto function, then find the value of b. Key concept: Given function is an onto function. Hence range should be to codomain. Solution: Given f(x) = = . Hence range of f(x) is [b–1, ). Since f(x) is an onto function b – 1 = 2 b = 3 3.2 INVERSE FUNCTION If f: X®Y be a function defined by y = f(x) such that f is both one – one and onto, then there exists a unique function g: Y®X such that for each y Î Y, g(y) = x if and only if y = f(x). The function g so defined is called the inverse of f and denoted by f – 1. f(a) = a1 f – 1(a1) = a SOME IMPORTANT POINTS: • The condition for existence of inverse of a function is that the function must be one – one and onto. • Whenever an inverse function is defined, the range of the original function becomes the domain of the inverse function and domain of the original function becomes the range of the inverse function. • Note that fof –1(x) = f -1of(x) = x always and roots of the equation f(x) = f –1(x) would always lie on the line y = x. • f and f –1 are symmetric about the line y = x. Illustration 15: The function ¦: [1, ¥) ® [1, ¥) is defined by ¦(x)=2x(x – 1), find ¦ – 1(x). Key concept : First check the function for one – one and onto. And if function is one – one and onto then find inverse using the identity Solution: Given, ¦(x) = 2x(x – 1)Þlog ¦(x) = x(x – 1) loge2 Þ ¦’(x) = 2x(x – 1) loge2 (2x – 1) Thus ¦(x) is an increasing function in [1, ¥), therefore, ¦(x) is a one – on function. Also range of f(x) is [1, ) which is equal to co – domain. Hence the function is also onto. TO FIND ¦ – 1(x): Let f – 1 be the inverse function of f, then by rule of identity = x \ 3.3 EVEN AND ODD FUNCTIONS: If f: X®Y is a real valued function such that for all x Î D Þ – xÎD (where D = domain of f) and if f(– x) = f(x) for every x Î D then f is said to be an even function and if f(– x) = – f(x) then f is said to be odd function. IMPORTANT POINTS: • Even functions are symmetric about the y – axis (i.e. if (x, y) lies on the curve, then (– x, y) also lies on the curve). • Odd functions are symmetric about the origin and it is placed either in the first and third quadrant or in the second and fourth quadrant. (i.e. if (x, y) lies on the curve, then (– x, – y) also lies on the curve). • f(x) = 0 is the only function which is both even and odd. • If f(x) is an odd function , then f(x) is an odd function provided f(x) is differentiable on R. • To express a given function f(x) as the sum of an even and odd function, we write , where is an even function and is an odd function. • If x = 0 domain of f, then for odd function f(x), f(0) = 0 i.e. if for a function, f(0) ¹ 0, then that function can not be odd. Illustration 16: Which of the following functions is (are) even, odd or neither : (i). f(x) = (ii). (iii). (iv). f (x) = sinx – cosx Solution: (i) f(x) = f(– x) = = = f(– x) Þ f(x) is even. (ii) = – f(x).Hence f(x) is odd. (iii) f(– x) = = = = – f(x). Hence f(x) is odd. (iv) f(– x) = sin(– x) – cos(– x) = – sinx – cos x Hence f(x) is neither even nor odd. Illustration 17: Identify the given function as odd, even or neither f(x + y) = f (x) + f (y) for all x, yÎR. Key concept : In such type of questions we use manipulation such as replace x = 0 or x = y or x = –x and try to find out relation in f(x) and f(–x) or function f(x). Solution: f(x + y) = f(x) + f (y) for all x, yÎR Replacing x, y by zero, we get f(0) = 2 f (0) Þ f (0) = 0 Replacing y by –x, we get f (x) + f (–x) = f (0) = 0 Þ f (x) = – f (–x) Hence f (x) is an odd function. 3.4 PERIODIC FUNCTION A function f: X®Y is said to be a periodic function provided there exists a positive real number T such that f(x + T) = f(x), for all x Î X. The least of all such positive numbers T is called the principal period or fundamental period or simply period of f. • To check the periodicity of a function put f(T+x)=f(x) and solve this equation to find the positive values of t independent of x. If positive values of T independent of x are obtained, then f(x) is a periodic function and the least positive value of T is the period of the function f(x). If no positive value of T independent of x is obtained then f(x) is non-periodic function. • A constant function is periodic but does not have a well-defined period. • If f(x) is periodic with period p, then f(ax + b) where a, b Î R (a ¹ 0) is also period with period p/|a|. • If f(x) is periodic with period p, then a f(x) + b where a, b Î R (a ¹ 0) is also periodic with period p. • If f(x) is periodic with period p, then f (ax + b) where a, b R (a ¹ 0) is also period with period p/|a| • Let f(x) has period p = m/n (m, n N and co-prime) and g(x) has period q = r/s (r, sÎN and co-prime) and let t be the LCM of p and q i.e. , then t shall be the period of f + g provided there does not exist a positive number) k (< t) for which f(k + x) + g(k + x) = f(x)+ g(x), else k will be the period. The same rule is c applicable for any other algebraic combination of f(x) and g(x). SOME IMPORTANT POINTS: • LCM of p and q always exist if p/q is a rational quantity. If p/q is irrational then algebraic) combination of f and g is non-periodic. • sinnx, cosnx, cosecnx and Secnx have period 2p if n is odd and p if n is even. • tannx and cotnx have period p whether n is odd or even. • If g is periodic then fog will always be a periodic function. Period of fog may or may not be the period of g. • If f is periodic and g is strictly monotonic (other than linear) then fog is non-periodic. There are two types of questions asked in the examination. You may be asked to test for periodicity of the function or to find the period of the function. In the former case you just need to show that f(x + T) = f(x) for same T (>0) independent of x whereas in the latter, you are required to find a least positive number T independent of x for which f(x +T)= f(x) is satisfied Illustration 18: Find the period of function sin4x + tan2x. Solution: Period of sin4x is , also period of tan 2x is . Hence period of f(x) is 3.5 COMPOSITE FUNCTION If ¦: x ® y and g: y ® z then we define the composite function (go¦): x ® z by (go¦) (x) º g {f(x)}. To obtain (go¦)(x), we first take the f-image of an element x x so that f(x) y, which is the domain of g(x). Then take g-image of f(x), i.e.g (f(x)) which would be an element of z. • (g(¦( ))= g( )= . • Range (go¦) = { } But Range (g) = { } • Clearly domain (gof) = {x : x Î Domain(¦),¦(x) Î domain(g)} • Similarly we can define, (fog)x = f(g(x)) and domain (fog) = {x : x Î Domain(g), g(x) Î domain(f)}. In general fog ¹ gof. 4. OBJECTIVE ASSIGNMENTS 1. The domain of is (A) (B) (C) (D) None of these. Solution: Given domain of sin x is R. But domain of log x is x > 0. Hence domain of given function is values of x such that Hence (A) is the correct answer. 2. The range of is (A) (B) (C) (D) Solution: Given clearly domain of f(x) is R. Now f(x) will be minimum when x = 2/3 Hence range is Hence (B) is the correct answer. 3. For x, the solution of [x + 2] + [x – 8] > 0 is : (A) [3, ) (B) [4, ] (C) [4, ) (D) (3, ) Solution: given [x + 2] + [x ¬– 8] > 0  [x] + 2 + [x] ¬– 8 > 0  [x] > 3  x  [4, ) Hence (C) is the correct answer. 4. For x, the solution of x – 3 = {x} is : (A) [2, 4] (B) [3, 5] (C) (3, 4) (D) [3, 4] Solution: given x – 3 = [x]  x – {x} = 3  [x] = 3  x  [3, 4] Hence (D) is the correct answer. 5. If the function defined by f(x) = is onto, then B is equal to : (A) (–, 1] (B) (–, 1) (C) (–, 2] (D) None of these. Solution: We have to find B such that f(x) because onto  B should be equal to range of f(x). Now f(x) = 1 – (x – 3)2  Range = (–, 1] = B Hence (A) is the correct answer. 6. Function is : (A) even (B) odd (C) neither even nor odd (D) None of these. Solution: Hence f(x) is neither even nor odd Hence (C) is the correct answer. 7. Function f(x) = cos is : (A) even (B) odd (C) neither even nor odd (D) None of these. Solution: Hence f(x) is even Hence (A) is the correct answer. 8. The period of is : (A) 0 (B) 1 (C)  (D) None of these. Solution: tan  {x} = 0 because  {x} will always be integral multiple of   f(x) = sin 3 {x} Hence period of f(x) is 1 Hence (B) is the correct answer. 9. The period of is : (A) (B) (C) (D) None of these. Solution: Period of sin 5x is 2 / 5. Period of for the period of f(x) LCM of 2 / 5 and , which doesn’t exist. Hence f(x) is non-periodic Hence (D) is the correct answer. 10. If f(x)= , then f(f(x)) = (A) x (B) x2 (C) x7 (D) x – x7 Solution: f(x) = (3 – x7)1/7 Hence (A) is the correct answer. 11. The number of solution of |x2-4|=2 is : (A) 0 (B) 1 (C) 2 (D) 3 Solution: Hence the range |x2 - 4|=2 Hence (C) is the correct answer. 12. For x, the solution of |log x|<1 is : (A) (1/e, e) (B) (e, 1/e) (C) (1/e, e] (D) [1/e, e) Solution: Hence |log x|<1  x  (1/e, e) Hence (A) is the correct answer. 13. If y = , then the domain of y is (A) [–2, 3] (B) [1, 2] (C) [2, 3] (D) [–2, 2] Solution: x2 – 4 ³ 0 Þ x ³ 2 or x £ –2 –1 £ 2 – x £ 1 Þ x £ 3 and x > 1 \ x Î [2, 3]. Hence (C) is the correct answer. 14. The range of the function f(x) = is (A) (– 1, 2) (B) (– 1, 3) (C) (– 1, 4) (D) none of these Solution: f(x) = Þ x2(1 – y) + x(1 – 4y) + 1 – 3y = 0 Since x is real Þ discriminant ³ 0 Þ 4y2 + 8y – 3 ³ 0 Þ y Î Hence (D) is the correct answer. 15. The inverse of the function f(x) = is (A) (B) (C) (D) none of these Solution: Since f(x) is defined for all x Î R and range of f(x) is (-1, 1), Now fof -1 = x, where f -1(x) : (-1, 1) ® R . Þ Þ Þ Þ f -1(x) = , where -1< x < 1 Hence (A) is the correct answer. 16. Period of cos(x2) is (A) (B) 2p (C) p (D) none of these Solution: Let cosx2 has periodic T cos(x+T)2 = cosx2 Þ (x+T)2 = 2np ± x2 Þ (x+T)2 ± x2 = 2np Since it is a quadratic in T and if solved, T will never be independent of x. Hence it is not periodic function. Hence (D) is the correct answer. 17. The set of all x values for which |f(x) + g(x)| < |f(x)| + |g(x)| is true if f(x) = x – 3 and g(x) = 4 – x is given by (A) R (B) R -]3, 4[ (C) R –[3, 4] (D) None of these. Solution: Given |f(x) + g(x)| < |f(x)| + |g(x)| This holds true if f(x) g(x) < 0 Þ (x – 3) (4 – x) < 0 (x –3) (x – 4) > 0 Þ x > 4 or x < 3 Þ x Î R – [3, 4]. Hence (C) is the correct answer. 18. Which of the following functions from f: A ® A are invertible, where A = [–1, 1] (A) f(x) = (B) g(x) = sin (C) h(x) = |x| (D) k(x) = x2 Solution: g(x) = sin is both one–one and onto. Hence (B) is the correct answer. 19. Period of |sin2x| + |cos8x| is (A) (B) (C) (D) None of these. Solution: Clearly period of |sin 2x| = and period of |cos8x| = Hence period = Hence (A) is the correct answer. 20. If f is a function such that f(0) = 2, f(1) = 3 and f(x + 2) = 2 f(x) – f(x + 1) for every real x, then f(5) is (A) 7 (B) 13 (C) 1 (D) 5 Solution: Put x = 0 f (2) = 2 f (0) – f (1) = 1 f (3) = 5, f (4) = –3, f (5) = 13. Hence (B) is the correct answer. 21. If f(x) + 2 f(1 –x) = x2 + 2 " x Î R, then f(x) is given as (A) (B) x2 –2 (C) 1 (D) None of these. Solution: f(1 – x) + 2 f(x) = (1 – x)2 + 2 f(x) + 2f (1 – x) = x2 + 2 Solving, we get f(x) = . Hence (A) is the correct answer. 22. If f(x) is a function which is odd and even simultaneously, then f(3) – f(2) is equal to (A) 1 (B) –1 (C) 0 (D) None of these. Solution: f(x) = 0 " x Î R Þ f(3) – f(2) = 0 Hence (C) is the correct answer. 23. If f(x) = and g(x) = f(f(x)) then for x > 50, (x) is equal to (A) 0 (B) 1 (C) 25 (D) None of these. Solution: g(x) = f(f(x)) = f( ) = for x > 50, g(x) = = x – 50, (x) = 1 Hence (A) is the correct answer. 24. Let ¦ be a function with domain [–3, 5] and let g(x) = |3x + 4|. Then the domain of (¦og) (x) is (A) (B) (C) (D) None of these. Solution: (fog) (x) = ¦[g(x)] = ¦(|3x + 4|) Þ –5 £ 3x + 4 £ 5 Þ –9 £ 3x £ 1 Þ –3 £ x £ 1/3 \ Domain of ¦og is Hence (B) is the correct answer. 25. If then the mapping ¦: A®B is (A) one-one but not onto (B) onto but not one-one (C) both one-one and onto (D) neither one-one nor onto Solution: Let t = 5x + 2, then A = {t: 0 £ t £ p} \ ¦(t) = cost which is bijective in [0, p] Hence, ¦(x) is bijective. Hence (C) is the correct answer. 26. If the real valued function is even then n equals (A) 2 (B) 2/3 (C) 1/4 (D) 1/3 Solution: f is an even function, hence f(– x) = f(x)  – (– x)n = xn Hence (D) is the correct answer. 27. If xf(x) – yf(y) = x – y then " x, y Î R then f(1) + f(2) is (A) 3 (B) -1 (C) 0 (D) 2 Solution: put y = 0 xf(x) = x  f (x) =1 Hence (D) is the correct answer. 28. If f : I ® I be defined by f(x) = [x + 1], where [.] denotes the greatest integer function, then f -1(x) is equal to (A) x – 1 (B) [x + 1] (C) (D) Solution: since the domain of f is I, hence f(x) is one– one and onto and [x] = x  f(x) = x+1  f– 1 (x) = x– 1 Hence (A) is the correct answer. 29. Which pair of functions is identical ? (A) sin-1(sinx), sin(sin-1x) (B) lnex, elnx (C) lnx2, 2 lnx (D) None of these. Solution : Two function are identical. If there domain and corresponding range are qual. Hence (D) is the correct answer. 30. The function f: R®R defined by f(x) = 2|x| - 2-x is (A) one one, onto (B) one-one, into (C) many one, onto (D) many-one, into Solution : Its clear from the graph, that the given function is many one and into. Hence (D) is the correct answer. 31. Let f(x) = |x – 2| + |x – 3| + |x – 4| ,then the minimum value of f(x+1) occur at x equal to (A) 1 (B) 2 (C) 3 (D) 4 Solution : f (x+1) =|x– 1| + |x – 2| + |x – 3| Hence (B) is the correct answer. 32. The domain of the real valued function f(x) = loge|logex| is (A) (0, 1) È (1, ¥) (B) (0, ¥) (C) (e, ¥) (D) (1, ¥) Solution : For the domain |log x| > 0  log x  0  x > 0  1 Hence (A) is the correct answer. 33. The range of the function f(x) = x2 + is (A) [1, ¥) (B) [2, ¥) (C) (D) None of these. Solution : f(x) = 1 + x2 + = +1  f (x) ≥ 1 Hence (A) is the correct answer. 34. If f(x) = sin2x + sin2 + cos x cos and g = 1, then (gof)(x) is equal to (A) 2 (B) 1 (C) 3 (D) 4 Solution : on simplifying f(x) we get f(x) = Hence (B) is the correct answer. 35. Let f(x) = Then f(x) + f(1 – x) is equal to (A) 0 (B) 1 (C) –1 (D) None of these. Solution : (B) is the correct answer. 36. The value of b and c for which the identity f(x + 1) – f(x) = 8x + 3 is satisfied, where f(x) = bx2 + cx + d, are (A) b = 2, c = 1 (B) b = 4, c = -1 (C) b = -1, c = 4 (D) None of these. Solution : From the given identity b(x + 1)2 + c(x + 1) +d- (bx2 + cx + d) = 8x + 3  2bx + b+c = 8x+3  b = 4, c = – 1 Hence (B) is the correct answer. 37. The period of the function f(x) = sin4x + tan2x is (A) 2p (B) p (C) (D) None of these. Solution : Period of sin4x is Period of tan 2x is Hence the period of f(x) is Hence (C) is the correct answer. 38. The domain of definition of the function f(x) = is : (A) [1, ¥) (B) [2, ¥) (C) [3, ¥) (D) None of these. Solution : For the domain 4x + 52 – 22(x ¬¬– 1) x ≥ 3 Hence (C) is the correct answer. 39. The period of f(x) = is : (A) p (B) p/2 (C) p/4 (D) 2p Solution : (B) is the correct answer. 40. If f(x) = , then f of(x) is : (A) x2; x ; x ³ 0 (B) x4, x ³ 0 ; – x2 ; x < 0 (C) x4, x ³ 0 ; x2 < 0 (D) x4 , x ³ 0 ; x, x < 0 Solution : (D) is the correct answer. 41. If the equation e ||x| – 2| + b = 2 has four solution then b lies in (A) (In2 – 2, In2) (B) (– 2, In2) (C) (0, In2) (D) None of these. Solution : Given equation can be written as ||x|– 2| = ln2 – b. Plot the graph and find the range of b such that there exist four points of intersection. Hence (A) is the correct answer. 42. The range of the function ¦(x) = sin(sin–1{x}) where {.} is a fractional part of x is (A) [0, 1) (B) [0, 1] (C) (–1, 1) (D) None of these. Solution : We know 0  {x} < 1 0  sin– 1 {x} < Hence (A) is the correct answer. 43. Let ¦(x) and g(x) be bijective functions where ¦: {a, b, c, d} ® {1, 2, 3 ,4} and g: {3 ,4, 5 ,6 } ® {w, x ,y ,z} respectively. The number of elements in the range set of g (¦(x)) are (A) 1 (B) 2 (C) 3 (D) 4 Solution : (D) is the correct answer. 44. Let f:[1/2,1] [-1,1] is given by then f -1(x) is given by (A) (B) (C) (D) Solution : (A) is the correct answer. 45. Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is (A) 14 (B) 16 (C) 12 (D) 8 Solution : (A) is the correct answer. 46. The domain of definition of f(x) = is : (A) R – {-1, -2} (B) (-2, ¥) (C) R – {-1, -2, -3} (D) (-3, ¥) – {-1, -2} Solution : x +3 > 0 and x2 + 3x +2  0 Hence (D) is the correct answer. 47. If f:[1, ¥) ® [2, ¥) is given by f(x) = x + then f -1 (x) equals : (A) (B) (C) (D) 1 + Solution : use the identity f (f– 1(x)) = x replace x by f– 1 (x), in the given function we get f(f– 1(x)) = f– 1(x) +  x = f– 1 (x) + , solve to find f-1x. Hence (A) is the correct answer. 48. Let g(x) = 1+ x – [x] and f(x) = . Then for all x, f {g(x)} is equal to : (A) x (B) 1 (C) f(x) (D) g(x) Solution : f(g(x)) = Since g(x) ≥ 1 >0 Hence f (g(x)) =1 Hence (B) is the correct answer. 49. The number of solutions of log4 (x - 1) = log2(x – 3) is : (A) 3 (B) 1 (C) 2 (D) 0 Solution : Given  (x – 1) = (x – 3)2  x= 2, 5 But at x = 2, given log is not defined. Hence (B) is the correct answer. 50. 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. Solution : 2y = 2 ¬– 2x  for domain 2 – 2x  x < 1 Hence (D) is the correct answer. 51. If f(x) = 3x – 5, then f -1 (x) is : (A) if given by (B) is given by (C) does not exist because f is not one – one (D) does not exist because f is not onto. Solution : (B) is the correct answer. 52. If f(x) is an odd periodic with period 2, then f(4) is : (A) 0 (B) 2 (C) 4 (D) – 4 Solution : Given f(x + 2) = f(x) and f(-x) and f(-x) = -f(x) and clearly f(0) = 0, f(2) = 0, f(4) = 0 Hence (A) is the correct answer. 53. If f(x) = 1 + ax, (a ¹ 0) is the inverse of itself then the value of a is : (A) -2 (B) -1 (C) 0 (D) 2 Solution : 1 + X = Hence (B) is the correct answer. 54. Let f: R R be defined by f(x)=2x+sinx for x R. Then f is : (A) one to one and onto (B) one to one but not onto (C) onto but not one to one (D) neither one to one nor onto Solution : f’(x) = 2 + cos x > x Hence f(x) is always increasing, hence it will be one-one. Range is R. Hence f will be onto. Hence (A) is the correct answer. 55. Let f(x) be a function whose domain is [-5, 7]. Then the domain of is : (A) [-6, 1] (B) (-6, 1] (C) [-6, 1) (D) None of these Key concept: If domain of f(x) is [a, b] then the domain of f(g(x)) is values of x for which g(x) [a, b] Solution : |2x+5| -5 and |2x+5| 7 always true and -7 2x+5 7 -6 x 1 Hence domain is [–6, 1]. Hence (A) is the correct answer. 56. The range of ‘a’ for which (x) = ax + cos x is one-one, is : (A) (– , –1) (1, ) (B) (–, –1] [ 1, ) (C) (– , –1) (D) (1, ) Common mistake : If (x) in continuously increasing or decreasing the (x) will be a one-one function `(x) = a – sinx > 0 or a – sin x < 0  either a > sin x or a < sinx  either a > 1 or a < –1  (– , –1) (1, ) Solution : If a function is continuously increasing or decreasing the ’(x)  0 or `(x)  0 and equal to zero only at one point. `(x) = a – sinx  0 or a – sinx  0  a  sinx or a  sinx  a  1 or a  –1 (–, –1] [ 1, ) Hence (B) is the correct answer. 57. is : (A) even (B) odd (C) neither even nor odd (D) None of these. Solution : f(– x) = = = – f(x). Hence f(x) is odd. Hence (B) is the correct answer. 58. f (x) = x is : (A) even (B) odd (C) neither even nor odd (D) None of these. Solution : f(x) = x f(– x) = (– x) = f(x) So f(x) is even. Hence (A) is the correct answer. 59. Domain of is equal to : (If [x] denotes the greatest integer function] (A) [0, 2] (B) [0, 1] (C) [1, 2] (D) (1, 2] Solution : For the do main, [x]  0 and (2– x)  0 x (0,1) and x(x – 2)  0 Hence (C) is the correct answer. 60. If y = , then the domain of y is : (A) [ –2, 3] (B) [1, 2] (C) [2, 3] (D) [–2, 2] Solution : x2 -4≥ 0 and – 1  2– x <1 Hence (C) is the correct answer.