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28. Mar 2023
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1. Question bank on function limit continuity & derivability There are 105 questions in this question bank. Select the correct alternative : (Only one is correct) Q.13 If both f(x) & g(x) are differentiable functions at x = x0, then the function defined as, h(x)=Maximum{f(x),g(x)}: (A) is always differentiable at x = x0 (B) is never differentiable at x = x0 (C*) is differentiable at x = x0 provided f(x0)  g(x0) (D) cannot be differentiable at x = x0 if f(x0) = g(x0) . [Hint: Consider the graph of h(x) = max(x, x2) at x = 0 and x = 1] Q.25 If 0 x Lim  (x3 sin 3x + ax2 + b) exists and is equal to zero then : (A*) a =  3 & b = 9/2 (B) a = 3 & b = 9/2 (C) a =  3 & b = 9/2 (D) a = 3 & b = 9/2 [Sol. b x a x x 3 sin Lim 2 3 0 x    = 0 x Lim  3 3 x bx ax x 3 sin   = 0 x Lim  2 2 x bx a x 3 x 3 sin 3   for existence of limit 3 + a = 0  a = – 3  l = 0 x Lim  3 3 x bx x 3 x 3 sin   = b t t t sin . 27 3   = 0 (3x = t) = b 6 27   = 0  b = 2 9 ] [ OR use L' Hospital's rule ] Q.36 A function f(x) is defined as f(x) = x x m N if x m x sin , 1 0 0 0       . Theleast value of m forwhich f(x)is continuous at x =0 is (A) 1 (B) 2 (C*) 3 (D) none [Sol. f ' (0+) = h x 1 sin h Lim m 0 h must exist  m > 1 for m > 1 h ' (x) =    0 x if 0 0 x x 1 cos x x 1 sin x m 2 m 1 m      now h 1 cos h h 1 sin h m Lim ) x ( h Lim 2 m 1 m 0 h 0 h       limit existif m>2  m  N  m = 3]
2. Q.410 For x > 0, let h(x) =        integers prime relatively are 0 q & p where irrational is x if 0 x if q p q 1 then which onedoes not hold good? (A*) h(x)isdiscontinuous for all xin (0,) (B)h(x)iscontinuous for eachirrationalin (0, ) (C)h(x)isdiscontinuous for each rationalin (0, ) (D) h(x) is not derivable for all x in (0, ). [Sol. Let x = 2 3 which is rational  3 1 3 2 h        Lim h t t  F H G I K J  0 2 3 0 discontinuous at x Q Let x = 2   Q h 2 d i = 0 consider 2 = 1.41401235839 h 2 d i = h 1414023583 1010 . F H G I K J= 1 1010  0 Hencehiscontinuous forallirrational  A A ] Q.512 The value of n e 1 x e 1 x x x 3 2 Limit x n x n                (where N n ) is (A) ln       3 2 (B*) 0 (C) n ln       3 2 (D)not defined [Hint: l = n e x e x x x 3 2 Limit x n x n    but 0 e x Limit x n x     l = 0 ] Q.613 For a certain value of c,   x Lim [(x5 + 7x4 + 2)C - x] is finite & non zero. The value of c and the value ofthelimitis (A*) 1/5, 7/5 (B) 0, 1 (C) 1, 7/5 (D) none [Sol.    x Lim                  x x 2 x 7 1 x c 5 c 5 =    x Lim                   1 x 2 x 7 1 x x c 5 1 c 5 This will beof theform  ×0onlyif 5c – 1 = 0  c = 1/ 5 substituting c = 1/5 , l becomes Hence l =    x Lim     1 X 1 x 5 / 1   where 5 x 2 x 7 X   =    x Lim          1 ..... 5 X 1 x =    x Lim 5 1 . x 2 x 7 x 5        = 5 7 Hence c = 5 1 and l = 5 7 ]
3. Q.714 Considerthepiecewisedefinedfunction f (x) =      4 x if 4 x 4 x 0 if 0 0 x if x       choose theanswerwhichbest describes thecontinuityofthis function (A)The functionis unbounded and thereforecannot be continuous. (B)The functionis right continuous atx =0 (C) Thefunction has a removablediscontinuityat 0 and4, but is continuous on the rest ofthe real line. (D*)Thefunctionis continuous ontheentirereal line [Hint: Refer tothe graph ] Q.817 If ,  arethe roots of thequadraticequation ax2 +bx+ c= 0 then   x Lim   1 2 2     cos ( ) ax bx c x  equals (A) 0 (B) 1 2 ()2 (C*) a2 2 ()2 (D)  a2 2 ()2 [Sol. Lim ax bx c ax bx c ax bx c x x     F H G I K J    0 2 2 2 2 2 2 2 2 2 2 4 sin ( ) c h c h c h  ; Lim a x b a x c a x b a x c a x x x   F H G I K J   F H G I K J      2 4 2 2 2 ( ) ( ) Lim a x x x x        2 2 2 2 2 ( ) ( ) ( ) ; Lim a C x      2 2 2 ( ) ] Q.918 Which one ofthe followingbest representsthe graph ofthefunctionf(x)=   nx tan 2 Lim 1 n     (A*) (B) (C) (D) Q.1019 1 x Lim    4 1 3 1 3 1 2 1 2 3 1 4 3 1 x x x x x x x x                          . = (A) 1 3 (B*) 3 (C) 1 2 (D) none
4. Q.1124 ABCis anisoscelestriangleinscribedinacircleofradiusr.IfAB=AC &histhealtitudefromAtoBC and Pbe the perimeter ofABC then 0 h Lim  3 P  equals (where  is the area of the triangle) (A) r 32 1 (B) r 64 1 (C*) r 128 1 (D) none [Sol. AB = 2 2 h rh 2 h     = rh 2 P = rh 2 2 + 2 h rh 2 2  P =         2 h rh 2 rh 2 2  = 2 h . h rh 2 2 2    = 2 h rh 2 h  3 0 h p Limit    = 3 2 2 0 h h rh 2 rh 2 8 h rh 2 h Limit           =  3 2 / 3 2 / 3 0 h h r 2 r 2 h 8 h r 2 h Limit     = r 128 1 ) r 2 ( ) r 2 ( 8 . 8 r 2 2 / 1   (C) ] Alternative: Note that as h  0 , b = 2 a or 2b = a Hence 3 p  = 3 ) b 2 a ( . R 4 ) b ( ab  (using R =  4 abc ) = 3 3 b 64 R 4 b 2 = R 128 1  (C) ] Q.1232 Let the function f, g and h be defined as follows : f (x) =    0 x for 0 0 x and 1 x 1 for x 1 sin x            g (x) =    0 x for 0 0 x and 1 x 1 for x 1 sin x2            h (x) = | x |3 for – 1  x  1 Which ofthesefunctions aredifferentiableatx =0? (A) f and g only (B) f and h only (C*) g and h only (D) none Q.1333 If [x] denotes the greatest integer  x, then Limit n   1 4 n         1 2 3 3 3 x x n x    ...... equals (A) x/2 (B) x/3 (C) x/6 (D*) x/4
5. Q.1436 Let f (x) = ) x ( ) x ( h g , where g and h are cotinuous functions on the open interval (a, b). Which of the followingstatements is true for a < x < b? (A) f is continuous at all x for which x is not zero. (B) f is continuous at all x for which g (x) = 0 (C) f is continuous at all x for which g (x) is not equal to zero. (D*) f is continuous at all x for which h (x) is not equal to zero. [Hint: Bytheorem,ifgandharecontinuous functions ontheopen interval (a,b),then g/h is alsocontinuous at all x in the open interval (a, b) where h(x) is not equal to zero. ] Q.1539 The period of thefunction f (x) = | x cos x sin | | x cos | | x sin |   is (A) /2 (B) /4 (C*)  (D) 2 Q.1645 If f(x) = 2 x x x 2 cos e x   , x  0 is continuous at x = 0, then (A) f (0) = 2 5 (B) [f(0)] = – 2 (C) {f(0)} = –0.5 (D*) [f(0)] . {f(0)} = –1.5 where [x] and{x} denotes greatest integerand fractional part function [Hint: 2 x 0 x x ) x 2 cos 1 ( 1 e x Lim      = – 2 1 – 2 = – 2 5 ; hence for continuity f (0) = – 2 5  [f (x)] = – 3 ; {f (0)} =        2 5 = 2 1 ; hence [f (0)] {f (0)} = – 2 3 = – 1.5 ] Q.1746 The valueof thelimit           2 n 2 n 1 1 is (A) 1 (B) 4 1 (C) 3 1 (D*) 2 1 [Hint:             2 n 2 2 n 1 n =                     2 n 2 n n 1 n n 1 n = n · ) 1 n ......( 4 · 3 · 2 ) 1 n ........( 3 · 2 · 1   = 2 1 n · n 1  = 2 n 1 1 Lim n    = 2 1 ] Q.1847 The function g (x) =    0 x , x cos 0 x , b x    can be made differentiable at x = 0. (A) if b is equal to zero (B) if b is not equal to zero (C) if b takes anyreal value (D*) for no value of b
6. [Hint: g' (0+) = h 1 cosh Lim 0 h   = 0 g ' (0–) = h 1 b h Lim 0 h      for existence of line b = 1 thus g ' (0–) = 1 Hence gcan not be madedifferentiable for anyvalueof b.] Q.1949 Let f be differentiable at x = 0 and f ' (0) = 1. Then h ) h 2 ( ) h ( Lim 0 h    f f = (A*) 3 (B) 2 (C) 1 (D) – 1 [Hint: reduces to 3 f ' (0) ] Q.2050 If f (x) = sin–1(sinx) ; R x then f is (A)continuousanddifferentiableforallx (B*) continuous for all x but not differentiable for all x = (2k + 1) 2  , I k (C) neither continuous nor differentiable for x = (2k – 1) 2  ; I k (D)neithercontinuousnordifferentiablefor ] 1 , 1 [ R x    [Hint:     2 3 k 2 x 2 k 2 for x ) 1 k 2 ( 2 k 2 x 2 k 2 for k 2 x                      ] Q.2153           ) x 3 sin x sin 3 ( 4 1 cos x sin Limit 1 2 x where [ ] denotes greatest integer function , is (A*)  2 (B) 1 (C)  4 (D) doesnot exist [Hint: ] x [sin cos x sin Limit 3 1 2 x    ax x  /2 , [sin3x]  0 and sinx  1  l =  2  (A) ] Q.2260 If 0 x Lim  x ) x 3 ( n ) x 3 ( n    l l = k , the value of k is (A*) 3 2 (B) – 3 1 (C) – 3 2 (D) 0
7. [Sol. 0 x Limit  x ) x 3 ( n ) x 3 ( n    l l = 0 x Limit  x 3 x 1 n 3 n 3 x 1 n 3 n                  l l l l = 0 x Limit  x / 1 x / 1 3 x 1 n 3 x 1 n                l l =         3 x x 1 3 x . x 1 = 3 2 3 1 3 1    (A) ] Q.2364 The function f (x) = 1 x 1 x Lim n 2 n 2 n     isidentical withthefunction (A) g (x) = sgn(x – 1) (B) h (x) = sgn (tan–1x) (C*) u (x) = sgn( | x | – 1) (D) v (x) = sgn (cot–1x) [Hint: f (x) =      1 | x | if 1 1 or 1 x if 0 1 x 1 if 1        and g (x) = sgn(|x| – 1) is same ] Q.2469 The functions defined by f(x) = max {x2, (x  1)2, 2x (1  x)}, 0  x  1 (A) isdifferentiableforallx (B) is differentiableforall x excetp at one point (C*) is differentiable for all x except at two points (D) is not differentiable at morethan two points. Q.2573 f (x) = nx x l and g (x) = x nx l . Then identifythe CORRECT statement (A*) ) x ( g 1 andf (x) areidentical functions (B) ) x ( f 1 and g(x)are identical functions (C) f (x) . g (x) = 1 0 x   (D) 1 ) x ( g . ) x ( f 1  0 x   [Hint: (A) nx x ) x ( f ; x / nx 1 ) x ( g 1 l l   x > 0 , x  1 for both (B) x nx ) x ( g ; nx / x 1 ) x ( f 1 l l    ) x ( f 1 is not defined at x =1 but g (1) = 0 (C) f (x) . g (x) = x nx . nx x l l = 1 if x > 0 , x  1  N. I. (D) 1 x nx . nx x 1 ) x ( g . ) x ( f 1   l l only for x > 0 and x  1 ] Q.2674 If f(3) = 6 & f(3) = 2, then Limit x  3 x f f x x ( ) ( ) 3 3 3   is given by : (A) 6 (B) 4 (C*) 0 (D) none of these
8. [Sol. f (3) = 6 ; f  (3)= 2 Limit x  3 x f f x x ( ) ( ) 3 3 3   , put x = 3 + h = h ) h 3 ( f 3 ) 3 ( f · ) h 3 ( Limit 0 h     = ) 3 ( f h ] ) 3 ( f ) h 3 ( f [ 3 Limit 0 h      = – 3f  (3) + f (3) = – 6 + 6 = 0 Ans ] Q.2776 Which oneof the followingfunctionsis continuous everywhere inits domain but has atleast onepoint whereitisnotdifferentiable? (A*) f (x) = x1/3 (B) f (x) = x | x | (C) f (x) = e–x (D) f (x) = tan x [Hint: x1/3 is not differentiable at x = 0] Q.2886 The limitingvalue of thefunction f(x) = x 2 sin 1 ) x sin x (cos 2 2 3    when x  4  is (A) 2 (B) 1 2 (C) 3 2 (D*) 3 2 [Hint: put x = 4  + h or use L'Hospital's rule ] Q.2989 Let f (x) =                  2 x if 2 x 3 x 4 x 2 x if 2 2 6 2 2 2 x 1 x x 3 x then (A) f (2) = 8  f is continuous at x = 2 (B) f (2) = 16 f is continuous at x = 2 (C*) f (2–)  f (2+)  f is discontinuous (D) f has a removable discontinuityat x = 2 [Hint: f (2+) = 8 ; f (2–) = 16 ] Q.3090 On the interval I = [2, 2], the function f(x) =   ( ) ( ) ( ) | | x e x x x x           1 0 0 0 1 1 then which oneof the followingdoesnot holdgood? (A*) is continuous for all values of x I (B) is continuous for x  I  (0) (C) assumes all intermediate values from f(2) & f(2) (D) has amaximum value equal to 3/e. [Sol. f (x) =           0 x if 0 0 x if 1 x 0 x if e ) 1 x ( x / 2 the graph off (x) is hence f can assume all values for f (– 2) to f (2) ]
9. Q.3193 Whichofthefollowingfunctionissurjectivebutnotinjective (A) f : R  R f (x) = x4 + 2x3 – x2 + 1 (B) f : R  R f (x) = x3 + x + 1 (C) f : R  R+ f (x) = 2 x 1 (D*) f : R  R f (x) = x3 + 2x2 – x + 1 [Sol. (A) f (x) = x4 + 2x3 – x2 + 1 Apolynomial of degree even will always be into say f (x) = a0x2n + a1 x2n–1 + a2 x2n – 2 + .... + a2n    x Limit f (x) =    x Limit [x2n             n 2 n 2 2 2 1 0 x a .... x a x a a =         0 a if 0 a if 0 0 Hence it will never approach  / –  (B) f (x) = x3 + x + 1  f (x) = 3x2 + 1 – injective as well as surjective (C) f (x) = 2 x 1 – neitherinjectivenor surjective f (x) = x3 + 2x2 – x + 1  f (x) =3x2 + 4x – 1  D > 0 Hencef(x) is surjectivebutnot injective.] Q.3294 Consider the function f (x) =       3 x 2 if x 6 2 x if 1 2 x 1 if ] x [ x       where [x] denotes step up functionthen at x =2 function (A)has missingpointremovablediscontinuity (B*)hasisolatedpointremovablediscontinuity (C)has nonremovablediscontinuityfinitetype (D)iscontinuous [Hint: fromthefigurethefunctionhasanobvious removableisolatedpointdiscontinuous. Q.3395 Suppose that f is continuous on [a, b] and that f (x) is an integer for each x in [a, b]. Then in [a, b] (A) f is injective (B) Range of f mayhavemanyelements (C) {x} is zero for all x  [a, b] where { } denotes fractional part function (D*) f(x) is constant [Hint: Explanation : Let f (x1) = n and f (x2) = m, x1, x2  (a, b) with n > m (say). According to the intermediate value theorem, between x1 and x2 there must be some value x for which f (x) = m + 2 1 which is impossiblesince m + 2 1 is not aninteger.] Q.34101 Thegraphof function f contains the point P (1, 2) andQ(s, r). Theequation ofthe secant line through P and Q is y =            1 s 3 s 2 s2 x – 1 – s. The value of f ' (1), is (A) 2 (B) 3 (C*) 4 (D)non existent
10. [Sol. I Bydefinition f '(1) is the limit of the slope of the secant line when s  1. Thus f '(1) = 1 s 3 s 2 s Lim 2 1 s     = 1 s ) 3 s )( 1 s ( Lim 1 s     = ) 3 s ( Lim 1 s   = 4  (D) II Bysubstitutingx = s intothe equation of thesecant line, and cancellingbys – 1 again, we get y = s2 + 2s – 1. This is f (s), and its derivative is f '(s) = 2s + 2, so f ' (1) = 4.] Q.35108 The range of the function f(x) = 12 x 11 x 2 ) 10 x 7 x ( 5 x n e 2 2 ) 2 x ( x 2      l is (A*) ) , (   (B) ) , 0 [  (C)        , 2 3 (D)       4 , 2 3 [Sol. f (x) = ) 4 x ( ) 3 x 2 ( ) 5 x ( ) 2 x ( . 5 nx e ) 2 x ( x 2      l Note that at x = 3/2 & x = 4function is not defined andin open interval (3/2,4)function is continuous.            ) ve ( ) ve ( ) ve ( ) ve ( ) ve ( Lim 2 3 x           ) ve ( ) ve ( ) ve ( ) ve ( ) ve ( Lim 4 x In the openinterval (3/2,4) the function iscontinuous & takes up all real values from (– ,) Hence range of the function is (– , ) or R] Q.36113 Consider f(x) =     2 2 3 3 3 3 sin sin sin sin sin sin sin sin x x x x x x x x               , x   2 for x  (0, ) f(/2) = 3 where [ ] denotes the greatest integer function then, (A*) f is continuous &differentiable at x = /2 (B) f is continuous but not differentiable at x = /2 (C) f is neither continuous nor differentiable at x = /2 (D) none of these [Sol. In the immediate neighborhood of x = /2 , sinx > sin3x  |sinx – sin3x|= sinx – sin3x Hence for x  /2 , f (x) =             x sin x sin ) x sin x (sin 2 x sin x sin ) x sin x (sin 2 3 3 3 3 = 3 x sin x sin x sin 3 x sin 3 3 3    Hence f is continuous and diff. at x = /2 ] Q.37114 The number of points at which the function, f(x) = x – 0.5 + x – 1 + tan x does not have a derivativeintheinterval (0,2) is : (A) 1 (B) 2 (C*) 3 (D) 4
11. Q.38117 Let [x] denote the integral part of x  R. g(x) = x  [x]. Let f(x) be any continuous function with f(0) =f(1) then the function h(x) =f(g(x)) : (A) hasfinitelymanydiscontinuities (B) is discontinuous at some x = c (C*) is continuous on R (D) is a constant function . [Sol. g(x) = x – [x] = {x} fis continuouswithf(0)=f(1) h(x) = f (g(x)) = f ({x}) Let the graphof f is as shownin the figure satisfying f(0) = f(1) now h(0) = f ({0}) = f(0) = f(1) h(0.2) = f ({0.2}) = f(0.2) h(1.5) = f ({1.5}) = f(0.5) etc. Hence the graphof h(x) will be periodicgraph as shown  h is continuous in R  C ] Q.39118 Given the function f(x) = 2x x3 1  + 5 x 1 4  x + 7x2 x  1 + 3x + 2 then : (A) thefunction is continuous but not differentiable at x = 1 (B*) thefunction is discontinuous at x = 1 (C) the function is both cont. & differentiable at x = 1 (D) the range of f(x) is R+. Q.40119 If f (x + y) = f (x) + f (y) + | x | y + xy2,  x, y  R and f ' (0) = 0, then (A) f need not be differentiable at everynon zero x (B*) f is differentiablefor all x R (C) f is twice differentiable at x = 0 (D) none [Hint: f '(x) = h xh h | x | ) h ( Lim h ) x ( ) h x ( Lim 2 0 h 0 h        f f f  f (0) = 0  f ' (x) =           xh | x | h ) 0 ( ) h ( Lim 0 h f f f ' (x) = f ' (0) + | x | = | x | ] Q.41131 For } x 10 { } 10 x sin{ Lim 8 x    (where { } denotes fractional part function) (A)LHLexist but RHLdoes not exist (B*) RHLexist but LHLdoes not exist. (C) neitherLHLnor RHLdoes not exist (D) both RHLand LHLexist and equals to 1
12. [Hint: } x { } x sin{ Lim 8 x    = 0 as {I + x } = {x} ; as 0 } x { Lim I x    and 1 } x { Lim I x     } x { } x sin{ Lim 8 x      as sin{x}  sin(1) and {–x}  0 ] Q.42134   n Lim 3 3 3 3 2 2 2 2 n ...... 3 2 1 1 . n ..... ) 2 n ( 3 ) 1 n ( 2 n 1           is equal to : (A*) 1 3 (B) 2 3 (C) 1 2 (D) 1 6 [Sol. Lim n n n n n n n n          1 2 1 3 2 1 2 2 2 2 3 ( ) ( ) ............ ( ( )) Nr. = n n n n ( ....... ) ( . . . .......... ( ) ) 1 2 12 2 3 34 1 2 2 2 2 2 2 2          = n n r r r n 2 2 2 1      ( ). = n n r r r n 2 3 2 1      ( ) = n n n n 2 3 2      = ( ) n n n     1 2 3          Lim n n n n n ( ) 1 2 3 3 Lim n n n n n n n n n       ( ) ( )( ) ( ) ( ) 1 1 2 1 6 1 1 1 4 3 1 1 3    A Alternatively:l =              1 n ...... 3 2 1 1 · n ...... ) 1 n ( 2 n 1 3 2 3 3 2 2 2         – 1 =        3 2 2 2 n ) 1 n ( n ...... ) 1 n ( 2 ) 1 n ( 1 – 1 l =    3 2 n n ) 1 n ( – 1 ] Q.43135 The domain of definition of the function f (x) = | 6 x x | log 2 x 1 x          + 16–xC2x–1 + 20–3xP2x–5 is (A*) {2} (B) } 3 , 2 { , 4 3         (C) {2, 3} (D)         , 4 1 Where [x] denotesgreatest integerfunction. Q.44136 If f (x) = 10 x 7 x 25 bx x 2 2     for x  5 and f is continuous at x = 5, then f (5) has the value equal to (A*) 0 (B) 5 (C) 10 (D) 25
13. [Hint: f (x) = ) 5 x )( 2 x ( 25 bx x2     . for existence of limit 25 – 5b + 25 = 0  b = 10 now f (x) = ) 5 x )( 2 x ( ) 5 x )( 5 x (     and ) x ( f Lim 5 x = 0  (A) ] Q.45139 Let f beadifferentiablefunctionontheopen interval(a,b). Whichof thefollowing statementsmustbe true? I. f is continuous on the closedinterval [a, b] II. f is bounded on the open interval (a, b) III. If a<a1<b1<b, and f (a1)<0< f (b1), then there is a number c such that a1<c< b1 and f (c)=0 (A) I and II only (B) I and III only (C) II and III only (D*) onlyIII [Sol. I and IIare false. The function f (x) = 1/x, 0 < x <1, is a counter example. Statement IIIis true.Applythe intermediate valuetheorem to f on the closed interval [a1, b1] ] Q.46145 The value of     a log a sec x log x cot x x 1 a a 1 x Limit      (a > 1) is equal to (A*) 1 (B) 0 (C) /2 (D) does not exist [Hint:                   x log a sec x x log cot Limit a x 1 a a 1 x ; 0 x x log as a a x          and           x log a a x (using L’opital rule)  l = 2 / 2 /   = 1 ] Q.47148 Let f : (1, 2 )  R satisfies the inequality 2 x | 8 x 4 | x ) x ( f 2 33 ) 4 x 2 cos( 2       , ) 2 , 1 ( x   . Then ) x ( f Lim 2 x   is equal to (A) 16 (B*) –16 (C)cannotbedeterminedfromthegiven information (D) doesnot exists [Hint: ) x ( f Lim 2 x   16 2 33 ) 4 x 2 cos(     ; ) x ( f Lim 2 x   16 ) 4 ( 4 2 x | 2 x | 4 . x2       Bysandwich theorem ) x ( f Lim 2 x   = –16 ]
14. Q.48163 Let a = min [x2 + 2x + 3, x  R] and b = x x 0 x e e x cos x sin Lim    . Then the value of    n 0 r r n r b a is (A) n 1 n 2 · 3 1 2   (B) n 1 n 2 · 3 1 2   (C) n n 2 · 3 1 2  (D*) n 1 n 2 · 3 1 4   [Sol. a = (x + 1)2 + 2  a = 2 b = x 2 x 2 ). 1 e ( 2 x 2 sin Lim x 2 0 x   = 2 1 now    n 0 r r n r b a =         r n r 2 1 2 =   n 0 4 r 2 n 2 2 1 = n 2 1 [1 + 22 + 24 + ... + 22n ] = n 2 1          1 2 1 ) 2 ( 2 1 n 2 = n 2 1         3 1 4 1 n = n 1 n 2 . 3 1 4   ] Q.49165 Period of f(x) = nx + n  [nx + n], (n  N where [ ] denotes the greatest integer function is : (A) 1 (B*) 1/n (C) n (D) none of these Q.50171 Let f beareal valued functiondefined by f(x)=sin1 1 3        x +cos1 x        3 5 .Then domain of f(x) isgivenby: (A*) [ 4, 4] (B) [0, 4] (C) [ 3, 3] (D) [ 5, 5] Q.51172 For the functionf (x) = ) x ( sin n 1 1 Lim 2 n     , which of thefollowing holds? (A) The range of f is a singleton set (B)f iscontinuous on R (C*) f is discontinuous for all x I (D) f is discontinuous for some x R [Hint: f (x) =    I x if 0 I x if 1    fis discontinuous for all x I ] Q.52174 Domain of the function f(x) = x cot n 1 1  l is (A) (cot1 ,  ) (B) R – {cot1} (C) (– ,0) (0,cot1) (D*) (– , cot1) [Sol. ln (cot–1x) > 0 cot–1x > 1 x < cot 1  domain (– , cot1) Ans. ]
15. Q.53175 The function          Q x , 5 x 2 x Q x , 1 x 2 ) x ( f 2 is (A)continuousnowhere (B)differentiablenowhere (C)continuousbut not differentiableexactlyatonepoint (D*)differentiableandcontinuous onlyatonepointand discontinuouselsewhere [Hint: Diff. & cont. onlyat x = 2 which is the point of intersection of y= 2x + 1 and y= x2–2x +5 ] [Sol. y = (x – 1)2 + 4 = x2 – 2x + 5 = 2x + 1 = x2 – 4x + 4 = 0 = (x – 2)2 = 0  x = 2  line is tangent at x = 2 ] Q.54180 For the function f (x) = ) 2 x ( 1 2 x 1   , x  2 which of the following holds? (A) f (2) = 1/2 and f is continuous at x =2 (B) f (2)  0, 1/2 and f is continuous at x = 2 (C*) f can not be continuous at x = 2 (D) f (2) = 0 and f is continuous at x = 2. Q.55184 ) x tan(sin 1 ) x cos(sin x Lim 1 1 2 1 x      is (A) 2 1 (B*) – 2 1 (C) 2 (D) – 2 [Hint: put sin–1x =          tan 1 cos sin Lim 4 =      cos Lim 4 = – 2 1 ] Q.56186 Which oneof the followingis not bounded on theintervals as indicated (A) f(x) = 1 x 1 2  on (0, 1) (B*) g(x) = x cos 1 x on (–) (C) h(x) = xe–x on (0, ) (D) l (x) = arc tan2x on (–, ) [Sol. (A)  0 x Limit f (x) = 2 1 2 Limit 1 h 1 0 h    ;  1 x Limit f (x) = 0 2 Limit h 1 0 h            2 1 , 0 ) x ( f  bounded (C) 0 h Limit  x e–x = 0 h Limit  h e –h = 0 ;   x Limit x e–x =   x Limit 0 e x x   Also x e x y  y = x 2 x x e xe e  e x (1 – x)         e 1 , 0 ) x ( h ]
16. Q.57190 The domainof thefunction f(x)=   arc x x x cot 2 2  , where [x] denotesthe greatest integernotgreater than x,is: (A) R (B) R  {0} (C*) R        n n I : { } 0 (D) R  {n : n  I} Hint: x2 – [x2] = {x2} > 0; but 0  {y} < 1  {x2}  0  x  ± n ] Q.58192 If f(x) = cos x, x =n  , n = 0, 1, 2, 3, ..... = 3, otherwise and (x) = x when x x when x when x 2 1 3 0 3 0 5 3           , then Limit x  0 f((x)) = (A) 1 (B*) 3 (C) 5 (D) none Q.59195 Let 0 x Lim  sec–1 x x sin       = l and 0 x Lim  sec–1 x x tan       = m, then (A*) l exists but m does not (B) m exists but l does not (C) l and m both exist (D) neitherl norm exists Q.60200 Range of the function f (x) = 2 2 x 1 1 ) e x ( n 1          l is , where [*] denotes the greatest integer function and e =      / 1 0 ) 1 ( Limit (A)        e 1 e , 0 {2} (B) (0, 1) (C) (0, 1]  {2} (D*) (0, 1)  {2} [Sol.        ) e x ( n 1 2 l –      0 x 1 0 x 0         0 x x 1 1 0 x 2 ) x ( f 2 Hence range of f (x) is (0, 1)  {2}] Q.61201 ] x [tan sin Lim 1 0 x    = l then { l } is equal to (A) 0 (B) 2 1   (C) 1 2   (D*) 2 2   where [ ] and { } denotes greatest integer andfractional part function. [Hint: 2 ] x [tan sin Lim 1 0 x       ; hence         2 =          2 2 2 = 2 – 2  Ans.]
17. Q.62206 Number of points where the function f (x) = (x2 – 1) | x2 – x – 2 | + sin( | x | ) is not differentiable, is (A) 0 (B) 1 (C*) 2 (D) 3 [Hint: not derivable at x = 0 and 2 ] Q.63219                           x 1 1 x 1 x 1 x 2 sec x 1 x cot Limit is equal to (A*) 1 (B) 0 (C) /2 (D)nonexistent [Hint: x 1 x Limit x     = 0  cot–1(0) = /2             x x 1 x 1 x 2 Limit  sec–1 () = /2  l = 1 Ans ] Q.64223 If f (x) =       0 0 2 x x if b ax x x if x derivable R x  then the values of a and b are respectively (A*) 2x0 , – 2 0 x (B) – x0 , 2 2 0 x (C) – 2x0 , – 2 0 x (D) 2 2 0 x , – x0 Q.65231 Let f(x) = 1 2 1 1 2 1 2 2 1 4 2 1 2 1 2                  cos sin , , ,   x x x p x x x x . If f(x) is discontinuous at x = 1 2 , then (A*) p  R  {4} (B) p  R  1 4       (C) p  R0 (D) p  R [Hint: f 1 2          = 4 = f 1 2          ] Q.66235 Let f(x) beadifferentiable functionwhichsatisfies theequation f(xy) = f(x) + f(y) for all x > 0, y> 0 then f (x) is equal to (A*) f x '( ) 1 (B) 1 x (C) f (1) (D)f (1).(lnx) [Sol. f (x) = Limit h f x h f x h    0 ( ) ( ) = Limit h f x h x f x h                0 1 . ( ) =   Limit h f x f h x f x h           0 1 ( ) = Limit h f h x x h x         0 1 =   1 0 1 1 x Limit t f t f t    ( ) (note that f(1) = 0) = f x '( ) 1 Ans. ]
18. Q.67240 Given f(x) = b ([x]2 + [x]) + 1 for x  1 = Sin ((x+a) ) for x < 1 where [x] denotes the integral part of x, then for what values of a, b the function is continuous at x = 1? (A*) a = 2n + (3/2) ; b  R ; n  I (B) a = 4n + 2 ; b  R ; n  I (C) a = 4n + (3/2) ; b  R+ ; n  I (D) a = 4n + 1 ; b  R+ ; n  I [Hint: f (–1) = b (1 – 1) + 1 = 1 ) h 1 ( Lim 0 h    = 1   a ) h 1 ( Lim ) h 1 ( Lim 0 h 0 h           forcontinuous sin a = – 1 = sin          2 3 n 2  a = 2 3 n 2     a = 2 3 n 2  hence a = 2 3 n 2  , n  I and b  R] Q.68247 Let f(x) = ) e x ( n ) e x ( n x 2 4 x 2   l l . If Limit x   f(x) = l and Limit x    f(x) = m then : (A*) l = m (B) l = 2m (C) 2 l = m (D) l + m = 0 [Sol.   x Lim       n x e n x e x x 2 4 2   =   x Lim                   x 2 4 x 2 x 2 x e x 1 e n e x 1 e n l l =   x Lim                     x 2 4 x 2 e x 1 n x 2 e x 1 n x l l = x / 1 x 2 4 x / 1 x 2 e x 1 n 2 e x 1 n 1                     l l note that   x as 0 e x x 2  and   x as 0 e x x 2 2  =   x Lim                       1 e x 1 x 1 2 1 e x 1 x 1 1 x 2 4 x 2 =   x Lim x 2 3 x e x 2 e x 1   |||ly 2 1 Limit x     ] Q.69248   n Lim cos         n n2 when n is an integer : (A) is equal to 1 (B) is equal to  1 (C*) is equal to zero (D) does not exist [Hint:  n 1 1 1 2        n / = n  1 1 2 1 2 1 2 1 1 2 1 2                 n n ! ...... =  n n                 1 2 1 2 1 2 1 1 2 1 ! ...... as n  ;  2 . 2 1 1 2 1 1 2 1 n n                 ! ....... = (2n + 1)  2
19. Alternatively (1): Take A = (2n +1) 2  and B = 2                .... n 1 ! 2 1 1 2 1 now cos (A + B) = cosA cosB – sinA sinB = 0 – (1)   n Limit sin                  ... n 1 ! 2 1 1 2 1 2 = 0 ] Alternatively (2) Best l =              n n n cos Limit 2 n                    n n n cos Limit 2 n                 n n n ) n ( cos Limit 2 n =                  n 1 1 n n n cos Limit n =                  n 1 1 1 cos Limit n = cos 2   0 ] Q.70250 0 x Limit  x sin 3 ) x (sin ) x .(tan 7 x ) x 2 cos 1 ( ) x tan x (sin 5 6 1 7 1 5 4 2         is equal to (A) 0 (B) 7 1 (C*) 3 1 (D) 1 [Hint: DivideNr. andDr. byx5 andevaluate limit] Q.71264 Range of the function , f (x) = cot1   log / ( ) 4 5 2 5 8 4 x x   is : (A) (0 , ) (B*)   4 ,       (C) 0 4 ,        (D) 0 2 ,        Q.72266 Let Limit x  0 [ ] x x 2 2 = l & Limit x  0 [ ] x x 2 2 = m , where [ ] denotes greatest integer , then: (A) l exists but m does not (B*) m exists but l does not (C) l & m both exist (D) neither l nor m exists . [Hint: [ ] x x 2 2 = 0 0 1 1 1 0 2 if x x if x           l does not exist . [ ] x x 2 2 = 0 0 1 0 1 0 if x if x          m exists and is equal to zero. ] Q.73268 The value of Limit x  0       tan { } sin { } { } { } x x x x   1 1 where {x} denotes the fractional part function: (A) is 1 (B) is tan 1 (C) is sin 1 (D*) is non existent [Sol. Limit x  0       tan { } sin { } { } { } x x x x   1 1 =  0 x Limit f (x) = ) 1 h ( h sinh . ) 1 h tan( Limit 0 h     = 1 tan 1 ) 1 tan(    1 1 sin ) 1 h 1 ( ) h 1 ( ) h 1 sin( ) 1 ) h 1 tan(( Limit 0 x          = sin 1 Hence limit x  0 f (x) does not exist ]
20. Q.74270 If f(x) = n x e x x2 2        tan is continuous at x = 0 , then f (0) must be equal to : (A) 0 (B) 1 (C) e2 (D*) 2 [Hint: f (0) = Limit x  0 n x e x x2 2        = Limit x  0 l n e x x x 2 1 2        = Limit x  0 1 x e x x2 2 1         = Limit x  0 ex x 2 1 2           = 2 ] Q.75272   x Lim x sin e ) x 2 sin x 2 ( x 2 sin x 2 2    is : (A) equal to zero (B) equal to 1 (C) equal to 1 (D*)non existent [Sol. Limit x   x sin e x x 2 sin 2 x x 2 sin 2 x 2          as x  l = Limit x   x sin e . 2 2 = oscillatory between e 1 to 1 e 1   non existent ] Q.76275 The value of   x ec bx ax 0 2 lim cos cos is (A) e b a          8 2 2 (B) e a b          8 2 2 (C*) e a b          2 2 2 (D) e b a          2 2 2 [Sol. l = ) 1 ax (cos bx 2 ec cos 0 x Limit e   now – bx sin ax cos 1 Limit 2 0 x   = – ax cos 1 1 . bx sin ax sin Limit 2 2 0 x   = – 2 2 b 2 a l = 2 b 2 2 a e  ] Select the correct alternative : (More than one are correct) Q.77503 c x Lim  f(x) does not exist when : (A) f(x) = [[x]]  [2x 1], c = 3 (B*) f(x) = [x]  x, c = 1 (C*) f(x) = {x}2  {x}2, c = 0 (D) f(x) = tan (sgn ) sgn x x , c = 0 . where[x] denotes step up function& {x} fractional part function.
21. Q.78505 Let f(x) = tan { } [ ] { } cot { } 2 2 2 1 0 0 0 x x x x x for x for x for x            where [x] is the step up function and {x} is the fractional part function of x , then : (A*) Limit x   0 f(x) = 1 (B) Limit x   0 f (x) = 1 (C*) cot -1 Limit f x x         0 2 ( ) = 1 (D) f is continuous at x = 1 . Q.79506 If f(x) =   x n x n x x x . (cos )   1 2 0 0 0         then : (A*) f is continuous at x = 0 (B)fiscontinuous atx=0 but notdifferentiableatx=0 (C*) f is differentiable at x = 0 (D) f is not continuous at x = 0. Q.80508 Whichofthefollowingfunction(s)is/areTranscidental? (A*) f (x) = 5 sin x (B*) f (x) = 2 3 2 1 2 sin x x x   (C) f (x) = x x 2 2 1   (D*) f (x) = (x2 + 3).2x [Hint: functions whicharenotalgebraicareknownastranscidental function] Q.81513 Which ofthefollowingfunction(s) is/areperiodic? (A*) f(x) = x  [x] (B) g(x) = sin (1/x) , x  0 & g(0) = 0 (C) h(x) = x cosx (D*) w(x) = sin1 (sinx) Q.82515 Which offollowingpairs offunctionsareidentical : (A) f(x) = e n x  sec1 & g(x) = sec1 x (B*) f(x) = tan(tan1 x) & g(x) = cot(cot1 x) (C*) f(x) = sgn(x)& g(x) = sgn(sgn(x)) (D*) f(x) = cot2 x.cos2 x & g(x)= cot2 x  cos2 x Q.83516 Whichofthefollowingfunctionsarehomogeneous ? (A) x siny+ ysinx (B*) x ey/x + y ex/y (C*) x2  xy (D) arc sin xy Q.84521 If  is small & positivenumber then which of the following is/are correct ? (A) sin  = 1 (B)  < sin < tan (C*) sin <  < tan (D*) tan  > sin  Q.85522 Let f(x) = x x x x . cos 2 1   & g(x) = 2x sin n x 2 2       then : (A) Limit x  0 f(x) = ln 2 (B) Limit x   g(x) = ln 4 (C*) Limit x  0 f(x) = ln 4 (D*) Limit x   g(x) = ln 2
22. Q.86524 Let f(x) = x x x    1 2 7 5 2 . Then : (A*) Limit x  1 f(x) =  1 3 (B*) Limit x  0 f(x) =  1 5 (C*) Limit x   f(x) = 0 (D*) Limit x  5 2 / does not exist Q.87527 Whichofthefollowinglimitsvanish? (A*) Limit x   x 1 4 sin 1 x (B*) Limit x  /2 (1  sinx) . tanx (C) Limit x   2 3 5 2 2 x x x    . sgn(x) (D*) Limit x   3 [ ] x x 2 2 9 9   where [ ] denotes greatest integer function Q.88528 If x is a real number in [0, 1] then the value of Limit m   Limit n   [1 + cos2m (n!  x)] is given by (A) 1 or 2 according as x is rational or irrational (B*) 2 or 1 according as x is rational or irrational (C) 1 for all x (D*) 2 for all x . [Sol.     n m Limit Limit 1 + cos2m(n ! n) Let ] 1 , 0 [ x is a rational then n!. x if n  becomes integral  cos2 (n ! x)  = 1    m Limit ( cos2 n ! x  )m = 1  l = 1 + 1 = 2 ifxisrational if x is irrational , then n ! x is not an integer  cos2( n! x)  is less than 1 cos2 ( ( n! x)  )m  0  l = 1 + 0 = 1  the result ] Q.89529 If f(x) is a polynomialfunction satisfyingthe condition f(x) . f(1/x) = f(x) + f(1/x)and f(2) = 9 then : (A) 2 f(4) = 3 f(6) (B*) 14 f(1) = f(3) (C*) 9 f(3) = 2 f(5) (D) f(10) = f(11) [Hint: f (x) = 1 + xn or 1  xn  n = 3  f(x) = x3 + 1 ] Q.90531 Which ofthe followingfunction(s) not defined at x = 0 has/have removablediscontinuityat x =0 ? (A) f(x) = 1 1 2  cotx (B*) f(x) = cos       x | x sin | (C*) f(x) = x sin  x (D*) f(x) = 1 n x Q.91535 The function f(x) =       x x x x x          3 1 1 2 4 3 2 13 4 , , is : (A*) continuous at x = 1 (B*) diff. at x = 1 (C*) continuous at x = 3 (D) differentiableat x = 3
23. [Sol. f (x) =                1 x if 4 13 2 x 3 4 x 3 x 1 if x 3 3 x if 3 x 2 f (1+) = h ) 1 ( f ) h 1 ( f Limit 0 h    = 1 h 2 ) h 1 ( 3 Limit 0 h       f (1–) = h 2 4 13 ) h 1 ( 2 3 4 ) h 1 ( Limit 2 0 h        = h 4 5 ) h 1 ( 6 ) h 1 ( Limit 2 0 h       = h 4 h 6 h 2 h Limit 2 0 h     = – 1  f is continuous at x =1 ] Q.92539 If f(x) = cos  x       cos    2 1 x        ; where [x] is the greatest integerr function of x, then f(x) is continuous at : (A) x = 0 (B*) x = 1 (C*) x = 2 (D) none of these [Hint: Not defined at x = 0 ; f(x) = cos 3 ; f(2) = 0 and both the limits exist  B and C ] Q.93540 Identifythepair(s) of functions which are identical . (A*) y = tan (cos 1 x) ; y = 1 2  x x (B*) y = tan (cot 1 x) ; y = 1 x (C*) y = sin (arc tan x) ; y = x x 1 2  (D*) y = cos (arc tan x) ; y = sin (arc cot x) [ Ans. : (A) T domain [ 1, 0)  (0, 1] (B) T (C) T (D) T , 1 1 2  x ] (A) y= tan(cos–1x) note that 0 < sin–1x < p but tan(/2) is not defined hence cos–1x  2   x  0 hence domain is [–1,1] – {0} and range is R. The graphis as shown.
24. (B) y= tan(cot–1x) note that 0 < cot–1x <   tan(/2) is not defined hence cos–1x  2   x  0 hence domain = R – {0} since cot–1x  0 ,   y  0 hence range is R – {0} The graphis as shown. (C) y=sin(tan–1x) – 2  < tan–1x < 2   y  (–1, 1) since tan–1x is defined for all x  R and sin is also defined for all x  R  domain is R y = 2 x 1 x  The graphis as shown. (D) y= cos(tan–1x) tan–1           2 , 2  y  (0, 1] DomainisR y = 2 x 1 1  Note that sin–1(cot–1x) also has the same graph The graphis as shown.] Q.94544 The function, f(x) = [x][x] where [x] denotes greatest integer function (A*)is continuousforallpositiveintegers (B*)is discontinuous forallnonpositiveintegers (C*)hasfinitenumberofelementsinits range (D*) is such that its graph does not lie above the xaxis . [Sol. [ | x | ] – | [x] | =               2 x 1 0 1 x 0 0 0 x 1 1 1 x 0  range is {0, –1} Thegraphis ] Q.95545 Let f (x + y) = f(x) + f(y) for all x , y  R. Then : (A) f(x) must be continuous  x  R (B*) f(x) may be continuous  x  R (C) f(x) must be discontinuous  x  R (D*) f(x) maybe discontinuous  x  R
25. [Hint: Limit h0 f (x + h) = Limit h0 f(x) + f(h) = f (x) + Limit h0 f (h) Hence if h  0 f (h) = 0  'f' is continuous otherwise discontinuous ] Q.96548 The function f(x) = 1 1 2   x (A*) has its domain –1 < x < 1. (B*) has finite one sided derivates at the point x = 0. (C)iscontinuous and differentiable atx = 0. (D*)is continuous but not differentiable atx = 0. [Hint: f (0+) = 1 2 ; f (0–) = – 1 2 ] Q.97553 Let f(x) be defined in [–2, 2] by f(x) = max (4 – x2, 1 + x2), –2 < x < 0 = min (4 – x2, 1 + x2), 0 < x < 2 Thef(x) (A)iscontinuousatallpoints (B*)has apoint ofdiscontinuity (C) isnot differentiable onlyat onepoint. (D*) is not differentiableat morethan onepoint [Hint: f(x) max. (4–x2, 1+x2), –2 < x < 0 min. (4 - x2, 1+x2), 0 < x < 2] Q.98555 Thefunctionf(x)=sgnx.sinx is (A*)discontinuousno where. (B*) an evenfunction (C*) aperiodic (D)differentiableforallx [Hint: f (x) =    sin sin x x x x    0 0 ] Q.99556 The function f(x) = x n 1 xl (A*)is aconstant function (B) has a domain (0, 1) U (e, ) (C*) is such that limit x1 f(x) exist (D)is aperiodic [Hint: y = x logx e = e (constant)  Aand C as limit x1 f(x) = e] Q.100557 Which pair(s)of function(s) is/are equal? (A*) f(x) = cos(2tan–1x) ; g(x) = 1 1 2 2   x x (B*) f(x) = 2 1 2 x x  ; g(x) = sin(2cot–1x) (C*) f(x) = e n x  (sgn cot ) 1 ; g(x) =     e n x  1 (D) f(x) = a X , a > 0; g(x) = a x 1 , a > 0 where {x}and [x] denotes thefractional part &integral part functions.
26. Fill in the blanks: Q.101801 Afunction f is defined as follows, f(x) = sinx if x c ax b if x c       where c is a known quantity. If f is derivable at x = c, then the values of 'a' & 'b' are _____ &______ respectively . [ Ans. : cos c & sin c  c cos c ] Q.102802Aweight hangs byaspring& is caused to vibrate by a sinusoidal force. Its displacement s(t)at time t is given by an equation of the form, s(t) = A c k 2 2  (sin kt  sin ct) where A, c & k are positive constants with c  k, then the limiting value of the displacement as c  k is ______. [Ans. : At kt k cos 2 ] Q.103805 Limit x  4 (cos ) (sin ) cos    x x x    2 4 where 0 <  <  2 is ______ . [Ans. : cos4  ln cos  sin4  ln sin ] [Hint: put cos 2  = cos4   sin4  ] Q.104809 Limit x  0   cos / 2 3 2 x x has the value equal to ______ . [ Ans. : e6 ] Q.105812 If f(x) = sin x, x  n, n = 0, ±1, ±2, ±3,.... = 2, otherwise and g(x) = x² + 1, x  0, 2 = 4, x = 0 = 5, x = 2 then Limit x0 g [f(x)] is ______ [Ans. : 1 ] [ IIT’86,2 ]
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