1) If f(x+y2) = f(x) + f(y)2 for x,y belongs to R anf f'(0) exists and equal to -1 and f(0) = 1 then the value of f(2) is
1) 0
2) 1
3)-1
4) 2
answer 3
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2)If f(2m + n8 , 2m - n8) = mn then f(x,y)+f(y,x) = 0 when
1)only when x=y
2)only when x≠y
3)only when x=-y
4)for all x and y
answer 4
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3)The value of
[20082]+[20094]+[20118]+[201516]+[203232]....∞ equals...([] represents the GIF)
1)2006
2)2007
3)2008
4)2009
answer 2
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4) Find the domain.
f(x) = √cos-1(cos x) - [x] ([] = GIF)
is_____________
1) (-∞,2(pi)+3]
2)(-∞,(pi)-3]
3)(-∞,(pi)+3]
4)(-∞,2(pi)-3]
answer 4
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5)If q2 - 4pr = 0 p>0 then the domain of the function f(x) = log (px3 + (p+q)x2 + (q+r)x + r) is
1)R - {-(q/2p)}
2)R - { {-(q/2p)} U {x : x ≥ 1} }
3)R - { {-(q/2p)} ∩ (-∞ , -1] }
4)N.O.T.
answer 3
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6)The domain of the function
f(x) = sin-1(2-|x|4)+cos-1(2-|x|4)+tan-1(2-|x|4) is
1) [-6,6]
2) [-2,6]
3) [-6,2]
4) [6,-2]
answer 1
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67 Let f(x+y) = f(x).f(y) for x,y belongs to R and f(0) ≠0 . Then the function g defined by g(x) = f(x)1+(f(x))2 is
1)An odd function
2)An even function
3)Constant function
4) Neither odd nor even function
answer 2
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8) The function f defined as
f(x) = {x if x is rational
0 if x is irrational
is
1)Continuous at all rational numbers
2)Continuous at all irrational numbers
3)Continuous at only one point x = 0
4)N.O.T.
answer 3
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9) If f(x) = sin[(pi)3]x + sin [-(pi)3]x then f((pi)/2) is
1)0
2)-1
3)1
4)N.O.T.
answer 3
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10)Ltn --> ∞ xnn!
1)is =0
2) tends to infinity
3)is =1
4) does not exist.
answer 1
(Please give the detailed solution of this.)
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11) Ltx ---> 0 [a sin xx]+[b tan xx] where a,b are integers and [] = GIF is
1) a+b
2)a+b+1
3)a+b-1
4)a+b-2
answer 3
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12) Ltx --->0 sin √x√x ln (1+5x)
1)Exists and is equal to 1/5
2)exists and is equal to 5
3) Exists and is equal to 0
4)Does not exists.
answer 4
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Please explain your answers....
Source - AAKASH SUCCESS MAGNET
1questions are very good.......
for question no 8 i think its gonna be a disontinuous function........ because for rational number its goona be x and for irrational number its goona be 0..... after a rational there is always aa irrational number.... so the function is discontinuous through out...........
if i have any misconception please post what mistake i have done......
trhanks..............
1for question number 12 the limit does not exist....... because if you put 0-h for lhl you will find the its going to be an imaginary number............. but if you put 0+h for right hand limit you going to find a real number........ imaginary number not equalto real number... so lhl not equal to rhl............ so limit does not exists............
if i have any misconception please post what mistake i have done......
trhanks..............
1for question number 10 the question is fully written or not. because if f lies between 0and 1 then answer is different than for x>1
if i have any misconception please post what mistake i have done......
trhanks..............
11for question 10 its ok i think coz other than finding this question in Aakash SM it was done as part of a question in colg.When i asked mam, she told that its valid not only for -1<x<1 but for all x. But she did'nt solve the problem...!!! :( so i think that the question is ok....
and thanks for the other two solutions....
1lt n--∞ xnn!
apply l hospital rule...
lt n--∞ xn-1(n-1)!
..
..
..
.
applying l hospital rule
up to n terms
lt n--∞ xn-n(n-n)!=1/1=1=ans...
1does not exists because l.h.l. is indefinite and thus not equals to r.h.l.