Let `A={x in R : [x+3]+[x+4]le3} and B={x in R :3^x(sum_(r=1)^oo3/(1 0^r))^(x-3)<3^(-3x)}` then

int(sum_(r)=0^(oo)(x^(r)3^(r))/(r!))dx

Let A = {(x,y) in R xx R : y = 5^(x) + 3^(x)} B = {(x,y) in R xx R : y = 4^(x)} , Then

(4(x+3))/(4-x)le3,x ne4 and x in R

If f: A->B given by 3^(f(x))+2^(-x)=4 is a bijection, then A={x in R :-1

lim_(x rarr oo) (1)/(n^(4)) sum_(r = 1)^(n) r^(3) is :

Let A = R -{3} " and " B =R -{1}. " Then " f : A to B : f (x) = ((x-2))/((x-3)) is

Let f(x)=(r-2)(r+1)x^(2)+r(r-2)x-r(r-2)=0, where r in I,x in R. If roots of the equation f(x_(i)) are real and sum_(i=1)^(3)(f(x_(i)))^(2)=0, then the value of r is

Express each of the following sts as an interval : (i) A={x:x in R, -4 , lt x lt 0} (ii) B={x:x in R, 0 le x lt 3} (iii) C= {x:x in R, 2 lt x le 6} (iv) D= {x:x in R, -5 le x le 2 } (v) E= {x:x in R, -3 le x lt2) (vi) F={x:x in R, -2, le x lt 0}