If f(x)=⎧⎪⎨⎪⎩x2sin(πx2),|x|<1x|x|,|x|≥1, then f(x) is
View Solution
Q2
If f(x)={x2sinπx2|x|<1x|x||x|≥1 then f(x) is
View Solution
Q3
If f(x) is an odd function then- (i)f(−x)+f(x)2 is an even function (ii)[∣f(x)∣+1] is even where [.] denotes greatest integer function. (iii)f(x)−f(−x)2 is neither even nor odd (iv)f(x)+f(−x) is neither even nor odd Which of these statements are correct
View Solution
Q4
f(x) = x2|x| is an even function
View Solution
Q5
Assertion :f is even, g is odd then fg(g≠0) is an odd function. Reason: If f(−x)=−f(x) for every x of its domain then f(x) is called odd function and if f(−x)=f(x) for every x of its domain, then f(x) is called even function.