∫1+x41-x432dx=
x1-x4+C
-x1-x4+C
2x1-x4+C
-2x1-x4+C
Explanation for Correct answer:
Finding the value of the given integral:
∫1+x41-x432dx=∫1+x4x2321x2-x232dx=∫x-31+x41x2-x232dx=∫1x3+x1x2-x232dx
Solving this by using the substitution method
t=1x2-x2dt=-2x3-2xdx∵ddxxn=nxn-1dt=-21x3+xdx
∫1x3+x1x2-x232dx=-12∫dtt32=-12×-2t-12+C=t-12+C=1x2-x2-12+C=1-x4x2-12+C=x1-x4+C
Hence, option (A) is the correct answer.
Write = or ≠ in the place holder.18□34
Consider two events A and B such that P(A)=14, P(BA)=12, P(AB)=14. For each of the following statements, which is true.
I.P(A'B')=34
II. The events A and B are mutually exclusive
III.P(AB)+P(AB')=1