Let A=[115101]. If B=[12-1-1] A[-1-211], then the sum of all the elements of the matrix ∑n=150Bn is equal to
(a) 50 (b) 100 (c) 75 (d) 125
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