11Why are u assuming that limit exists?Can't it be none of these?
I m new here & dis is my 1st post So help me out.Some of these probs mite b vry easy fr u guys & nt of IIT level bt stl pls solve them.Give suitable xplanations.
1) If f is an odd function and limx→0f(x) exists then limx→0f(x) equals:
(A)0 (B)1 (C)-1 (D)None of these
2)Limx→0sin(6x2)ln[cos(2x2-x)] ;where[.] is a simple bracket & nt greatest int func. And ln= log base e
(A)12 (B)-12 (C)6 (D)-6
3)Limn→∞5n+1+3n-22n5n+2n+32n+3 is equal to
(A)5 (B)3 (C)1 (D)0
4)If Limx→0x3√a+x(bx-sinx)=1,then the constants 'a' and 'b' are(a>0)
(A)b=1,a=36 (B)a=1,b=6 (C)a=1,b=36 (D)b=1,a=6
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9 Answers
11) as f is odd function f(-x)=-f(x) or f(-h)=-f(h) ..........e.q.1
now as limit exists --- l.h.l = r.h.l.
ltx--0- f(x) = ltx--0+ f(x)
a) let x=o- h
then, l.h.l. = lth--0 f(-h) .....e.q.2
b) let x 0+h
then r.h.l. = lth--0 f(h).........e.q.3
now l.h.l. = r.h.l.
lth--0 f(-h) = lth--0 f(h).........from eq.2 and eq.3
so, f(-h)= f(h).........and,
- f(-h)=f(h) .....................from e.q.1
adding these
f(h)=0 hence,
lth--0 0 = 0
therefore , option A should be correct......
13) lt n---∞ = 5n+1+3n-22n5n+2n+32n+3
numerator = 5n.3n.22n{53n.22n + 15n.22n-1 5n.3n}
denominator = 5n.32n.22n{132n.22n + 2-n5n.32n+27 5n.22n}
lt n---∞ =numeratordenominator=5n.3n.22n{53n.22n + 15n.22n-1 5n.3n}
5n.32n.22n{132n.22n + 2-n5n.32n+27 5n.22n}
lt n---∞ ={53n.22n + 15n.22n-1 5n.3n}.3n{132n.22n + 2-n5n.32n+27 5n.22n}.3n.3n
=1∞=0
hence option D is correct......
12) LT x---0 sin(6x2){logecos(2x2-x)}
applying l hospital rule....
LT x---0 cos(6x2).12x- tan(2x2-x).(4x-1)
again applying l hospital rule....
LT x---0 144x3.sin(6x2)-12cos(6x2)sec2(2x2-x).(4x-1)2+4tan(2x2-x)
= 0.0 - 121+0=-12 .
so,option b) is correct...............
1114) ltx--0 x3√a+x.(bx-sinx)
applying l hospital rule...
ltx--0 3x2.2√a+xbx-sinx+2(a+x)[b-cosx]
=02a[b-1]=1 (given)
it should be converted in 0/0 form....
a cannot be equal to 0 as a>0 given.so, [b-1]=0
hence b=1
now,
again applying l hospital rule....
ltx--0 24x√a+x+6x2[2√a+x].2+2asinx-2cosx+2xsinx+1-cosx =0/0
again applying l hospital rule...
ltx--0 24√a+x++24x2√a+x+12x[2+2asinx-2cosx+2xsinx+1-cosx]√a+x+ 2√a+x{2acosx+2sinx+2xcosx+3sinx}
=24√a2√a{2a+0+0+0+0}
122a=1
a=6
hence b=1 and a=6.........so,d is correct
1Thanx for the solutions. Although u have given way too much xplanations, but its ok. Although the answer to ques no 4 is given as (A), but i will check with dat. Anyway thanx again. I've posted 2 more doubts in limits. i'll b happy if u can solve them too for me. But 2 save ur time and effort u can cut down on those big xplanations(especially da one dat u gave in ques no 1). But minilmal xplanation is reqd as i m nt dat a hi-fi student!!!...
Bye. Hey btw are u an IItian??