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Mere Teacher · Previous Year Questions
JEE Main Mathematics · Sets, Relations & Functions · Set 13 (Q361-390)
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For k>0,∑x=0∞x!kxlimn→∞(n−x)!n!(1−nk)n−x(n1)x=
(1)0
(2)k
(3)x
(4)1
363.
Let R1 and R2 be two relations defined on R by aR1b⇔ab≥0 and aR2b⇔a≥b, then
(1)R1 is an equivalence relation but not R2
(2)R2 is an equivalence relation but not R1
(3)both R1 and R2 are equivalence relations
(4)neither R1 nor R2 is an equivalence relation
364.
For any two real numbers, an operation * defined by a∗b=1+ab is
(1)neither commutative nor associative
(2)commutative but not associative
(3)both commutative and associative
(4)associative but not commutative.
365.
If x+y−x−y=c, then dx2d2y=
(1)y1(dxdy)2
(2)4y3−c4
(3)y(dxdy)2
(4)4y3−c2
366.
Find the function g(t) if f(t)=3t−2 and (gof)−1(t)=t−2.
(1)g(t)=3(t−8)
(2)g(t)=3(t+8)
(3)g(t)=3(8−t)
(4)g(t)=3t−8
367.
Let D be the domain of a twice differentiable function f. For all x∈D,f’’(x)+f(x)=0 and f(x)=∫g(x)dx+ constant. If h(x)=f(x)2+g(x)2 and h(0)=5, then h(2015)−h(2014) is equal to
(1)5
(2)3
(3)0
(4)1
368.
If X={4n−3n−1:n∈N} and Y={9(n−1):n∈N}, where N is the set of natural numbers, then X∪Y is equal to
(1)X
(2)Y
(3)N
(4)Y – X
369.
If y=f(x2+2) and f’(3)=5, then dxdy at x=1 is
(1)25
(2)10
(3)5
(4)15
370.
In P(X), the power set of a non-empty set X, a binary operation ∗ is defined by A∗B=A∪B∀A,B∈P(X). Under *, a TRUE statement is
(1)inverse law is not satisfied
(2)identity law is not satisfied
(3)associative law is not satisfied
(4)commutative law is not satisfied.
371.
Let f,g:R→R be two functions defined as f(x)=|x|+x and g(x)=|x|−x∀x∈R. Then (fog)(x) for x<0 is
(1)0
(2)4x
(3)−4x
(4)2x
372.
If x=1+y1−y, then (x+1)dx2d2y+(y3y+1)dxdy equals
(1)−2y
(2)0
(3)−y
(4)y
373.
If R=((x,y);x,y∈Z,x2+3y2≤8) is a relation on the set of integers Z, then the domain of R−1 is:
(1){0,1}
(2)(−2,−1,1,2)
(3)(−1,0,1)
(4)(−2,−1,0,1,2)
374.
Let the relation R be defined in N by aRb if 3a+2b=27 then R is
(1){(1,12),(3,9),(5,6),(7,3),(9,0)}
(2){(1,12),(3,9),(5,6),(7,3)}
(3){(2,1),(9,3),(6,5),(3,7)}
(4)((0,227),(1,12),(3,9),(5,6),(7,3))
375.
f:R→R and g:[0,∞)→R are defined by f(x)=x2 and g(x)=x. Which one of the following is not true?
(1)(fog)(2)=2
(2)(gof)(4)=4
(3)(gof)(−2)=2
(4)(fog)(−4)=4
376.
If f(x)=x3−x and g(x)=sin2x, then f(g(12π))=
(1)0
(2)1
(3)−83
(4)2
377.
The derivative of tan−1(1+1−x2x) with respect to sec−1(2x2−11) is
(1)21
(2)41
(3)4−1
(4)2−1
378.
If 2x+2y=2x+y, then dxdy=
(1)1−2y
(2)2y−1
(3)1+2y
(4)1+2−y
379.
Let A = {1, 3,4,6,9} and B = {2,4, 5, 8,10}. Let # be a relation defined on A x B such that R ={(a1, b1),(a2, b2):a1≤b2 and b1≤a2} - Then the number of elements in the set R is
(1)160
(2)52
(3)26
(4)180
380.
If f:R→R is defined by f(x)=|x|, then
(1)f−1(x)=−x
(2)f−1(x)=1|x|
(3)the function f−1(x) does not exist
(4)f−1(x)=1x
381.
If f:R→R is defined by f(x)=(|[x−5]|,forx<5[|x−5|],forx≥5) Then, ( fof )(−27)= (here, [x] is the greatest integer not exceeding x )
(1)(fof)(−211)
(2)(fof)(−29)
(3)(fof)(3)
(4)(fof)(29)
382.
Let [x] denote the greatest integer less than or equal to x and f(x)=2x−[2x]. If limx→2−f(x)=l1 and limx→2+f(x)=l2 then l1+l2=
(1)1
(2)2
(3)0
(4)4
383.
If xycos4α+yxsin4α=2sin2α⋅cos2α, then dxdy=
(1)sin3αcosα
(2)sin2αcos2α
(3)cos2αsin2α
(4)sinαcos3α
384.
If y=log(logx), then dx2d2y is equal to
(1)x2logx(1+logx)
(2)(xlogx)2−(1+logx)
(3)(xlogx)2(1+logx)
(4)x2logx−(1+logx)
385.
Let f:R→R be defined as f(x)=(2sin(−2πx),ifx<−1(ax2+x+b),if−1≤x≤1sin(πx)ifx>1)If f(x) is continuous on R, then a+bequals :
(1)3
(2)−1
(3)−3
(4)1
386.
If f(1)=1,f’(1)=3 then the derivative of f(f(f(x)))+(f(x))2 at x=1 is
(1)10
(2)35
(3)33
(4)12
387.
Let f :N → N be a function such that f(m + n) = f(m) + f(n) for every m, n ∈N . If f(6) = 18, then f(2) ⋅f(3) is equal to :
(1)6
(2)54
(3)18
(4)36
388.
Let S={1, 2, 3,4}. Then the number of elements in the set {f: S x S → S : f is onto and f(a, b) = f(b, a)≥a∀(a,b)∈S×S } is
389.
Consider the function f:R→R defined by f(x)=1+9x22x. If the composition of f,(f∘f∘f∘⋯∘f)10times(x)=1+9αx2210x, then the value of 3α+1 is equal to______
390.
The number of solutions of the equation, log(x+1)(2x2 + 7x + 5) + log(2x+5)(x+1)2 − 4 = 0 x > 0 is