Eight persons are to be transported from city A to city B in three cars of different makes. If each car
can accommodate at most three persons, then the number of ways, in which they can be transported, is
(A) 3360 (B) 1120 (C) 1680 (D) 560
Let a1, a2, a3, .., an be n positive consecutive terms of an arithmetic progression. If d>0 is its common difference, then
limn→∞dn(1a1+a2+1a2+a3+...+1an-1+an) is
(c) : We have, a1, a2, ...., an are in A.P.
∴ a2- a1= a3- a2=......= an- an-1=d
Now, limn→∞dn(1a1+a2+1a2+a3+...+1an-1+an)
= limn→∞dn(a2-a1a2-a1+a3-a2a3-a2+...+an-an-1an-an-1)
= limn→∞dn(a2-a1d+a3-a2d+...+an-an-1d)
=limn→∞1nd(an-a1)
=limn→∞1nd(an-a1an+a1)=limn→∞1nd(n-1)dan+a1
=limn→∞1ndn(1-1n)dn(a1n+(1-1n)d+a1n)
=limn→∞(1-1n)dd(a1n+d-dn)+a1n=dd·d=1
Among
(S1): limn→∞1n2(2+4+6+...+2n)=1
(S2): limn→∞1n16(115+215+315+....+n15)=116
(d) : (S1) : limn→∞1n2(2+4+6+...+2n)
=limn→∞1n2×2(1+2+3+...+n)
=limn→∞1n2×2×n(n+1)2=limn→∞(1+1n)=1
∴ (S1) is true.
(S2) : limn→∞1n16(115+215+315+....+n15)=limn→∞1n16∑r=1nr15
=limn→∞1n∑r=1n(rn)15=∫01x15dx=|x1616|01=116
(S2) is also true.
limn→∞{(212-213)(212-215)....(212-212n+1)} is equal to
(c) : Let L = limn→∞{(212-213)(212-215)....(212-212n+1)}
By Sandwich Theorem,
(212-213)n<(212-213)(212-215)(212-217)....(212-212n+1)<(212-212n+1)n
⇒limn→∞ (212-213)n<L<limn→∞ (212-212n+1)n
As, limn→∞ (212-213)n=0 and limn→∞ (212-212n+1)n=0
∴ L = 0
limx→0((1-cos2(3x)cos3(4x))(sin3(4x)(loge(2x+1))5)) is equal to _____ .
(d) : limx→0[((1-cos23x)cos3(4x))(sin3(4x)(loge(2x+1))5)]
=limx→0[sin2(3x)cos34x×sin3(4x)[loge(2x+1)]5]
=limx→0[sin2(3x)(3x)2×sin3(4x)(4x)3×(3x)2×(4x)3cos3(4x)·[loge(2x+1)2x]5×(2x)5]
=9×6432=18
If α>β>0 are the roots of the equation ax2+bx+1=0, and limx→1α(1-cos(x2+bx+a)2(1-αx)2)12=1k(1β-1α), then k is equal to
(d) : Given, ax2+bx+1 = 0 has roots α, β, then x2+bx+a=0 has roots 1α,1β.
Now, limx→1α (1-cos(x2+bx+a)2(1-αx)2)1/2
=limx→1α (2sin2(x2+bx+a2)2(1-αx)2)1/2
=limx→1α (2sin2((x-1α)(x-1β)2)2(1-αx)2)1/2
=limx→1α (sin2((1-αx)(1-βx)2αβ)((1-αx)(1-βx)2αβ)2×((1-αx)(1-βx)2αβ)2(1-αx)2)1/2
=limx→1α (1-βx2αβ)=(α-β2α2β)=12α(1β-1α)
=1k(1β-1α) (⇒Given)
So, k = 2α