The equation e4x+8e3x+13e2x-8ex+1=0,x∈R has
(a) four solutions two of which are negative
(b) two solutions and both are negative
(c) no solution
(d) two solutions and only one of them is negative
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