If five dices are tossed, then what is the probability that the five numbers shown will be different?
554
518
527
581
A.
Total cases, n(S) = (6)
Total number of favourable cases = 6!
⇒ nE = 720∴ Required probability = nEnS = 72062 = 554
If the events A and B are independent and if PA = 23, PB = 27, then PA ∩ B is equal to
421
321
521
221
∫1xlogxloglogxdx is equal to
loglogx + C
loglogxlogx + C
loglogloglogx + C
logloglogx + C
∫3x1 - 9xdx is equal to
log3sin-13x + C
13sin-13x + C
19sin-13x + C
The value of the integral ∫cosxsinx + cosxdx is equal to
x + logsinx + cosx + C
12x + logsinx + cosx + C
logsinx + cosx + C
x2 + logsinx + cosx + C
∫27e9x + e12x13dx is equal to
1427 + e3x13 + C
1427 + e3x23 + C
1327 + e3x43 + C
1427 + e3x43 + C
∫4dxx24 - 9x2 is equal to
4 - 9x2 + C
- 234 - 9x2 + C
- 4 - 9x2x + C
234 - 9x2x + C
∫e- xcscx1 + cotxdx is equal to
e- xcscx + C
- e- xcscx + C
- e- xcscx + cotx + C
- e- xcscx - tanx + C
∫sinx + cosxe- x + sinxdx is equal to
log1 - exsinx + C
log1 + exsinx + C
log1 - e- xsinx + C
The solution of ∫2xdtt2 - 1 = π12 is
1
2
3
4