The correct option is C (IV)→(R)
323∫01+x+√x2+x3x+√1+xdx
=323∫0√1+x(√1+x+x)x+√1+xdx
=32[23(1+x)32]30
=7
The denominator of f(x) will always be positive and will lie in [5,∞)
At x=0, we get f(x)=285
As x→∞, f(x)→0.
So, the range of f(x) is (0,285]
Number of integers will be 5 i.e., {1,2,3,4,5}
limx→π2tan2x(√4sin2x+sinx+3−√3sin2x+4sinx+1)
=limx→π2sin2xcos2x(4sin2x+sinx+3−3sin2x−4sinx−1√4sin2x+sinx+3+√3sin2x+4sinx+1)
=limx→π2sin2x(1−sinx)(1+sinx)((sinx−1)(sinx−2)√4sin2x+sinx+3+√3sin2x+4sinx+1)
=limx→π2sin2x(1+sinx)((2−sinx)√4sin2x+sinx+3+√3sin2x+4sinx+1)
=1×(2−1)(1+1)(√8+√8)
=18√2=1√128=1√32k
∴k=4
|→a+→b+→c|=10
⇒|→a+→b+→c|2=100
⇒|→a+→b+→c|2=|→a|2+|→b|2+|→c|2 +2(→a.→b+→b.→c+→c.→a)
⇒|→a+→b+→c|2=|→a|2+4|→a|2+9|→a|2 +2(2|→a|2cos(π3)+6|→a|2cos(π3)+3|→a|2cos(π3))
⇒25|→a|2=100
⇒|→a|=2