more than one correct
Q1 if a,b,c in Ap and a≠b then b-c/a-b= ??
a)√2
b)√3
c)1
d)3
Q2 if x1,x2,x3 are positive and unequal real numbers such that 2x1,2x2,2x3are in gp then find range of (x1.x3x22)1/2
Q3 if if p,q,r in gp
and lnp-ln2q,ln2q-ln3r,ln3r-lnp in ap..then p,q,r are lengths of ____triangle ???
Q4 if sin(πcosθ)=cos(πsinθ),then find values of sin2θ
Q5 if tan(πcosθ)=cot(πsinθ),then find eqn whose roots are cosθ and sinθ
106Q3. lnp-ln2q,ln2q-ln3r,ln3r-lnp in ap
=> 2(ln2q-ln3r) = lnp - ln2q + ln3r - lnp
=> 2ln2 + 2lnq - 2ln3 - 2lnr = -ln2 -lnq +ln3 +lnr
=> 3ln2 - 3ln3 + 3lnq -3lnr = 0
=> ln(2q/3r) = 0
=> 2q = 3r
p,q,r in gp
=> r/q = q/p
=> 2/3 = q/p
=> p=p q = 2p/3 ... r = 4p/9
cosP = (36+16-81)p2/2qr < 0
=> triangle is obtuse
106Q4. cos(pisin@) = cos(pi/2 - picos@)
=> pisin@ = 2npi ± (pi/2 - picos@)
=> sin@ = 2n ± (1/2 -cos@)
=> sin@ ± cos@ = 2n ± 1/2
Now 2n±1/2 is (-√2,√2)
=> n=0
=> sin@ ± cos@ = ±1/2
squaring,
1 ± sin2@ = 1/4
=> sin2@ = ± 3/4
106tan(πcosθ)=cot(πsinθ),then find eqn whose roots are cosθ and sinθ
=> tan(picos@) = tan(pi/2 - pisin@)
=> picos@ = npi + pi/2 - pisin@
=> cos@ + sin@ = n+1/2 = ± 0.5
squaring,
1 - 2sin@cos@ = 1/4
=> sin@cos@ = 3/8
So the equations are
x2 - (±0.5)x + 3/8 = 0
