Hi pav,

 

u know.._[x][0]  (1+x)^1/x = e ;_[x][a]  (f(x))^g(x) = e^{lim(as x->a) [f(x) - 1]*g(x)} where g(x) tends to infinity and f(x)->1 for x->a ;

 

given..f(x) = cosx ; g(x) = cot^2 x ;

 

so,limit  = e^{[ - 2sin^2(x/2) ] * [ cos^2(x) / 4sin^2(x/2)cos^2(x/2) ]}  ;

 

as x->0 ; lim = 1/e.....ans

Hie shubhu,

3) [xf(a) - af(x)] = [xf(a) -a{f(x) - f(a)} - af(a)] ;

 

u know ,  _[x][a]   f(x) - f(a) / (x-a) = f ' (a)

 

so, limit  = { _[x][a]   [(x - a)f(a)] / (x-a) } - af ' (a) + af ' (a)  = f(a) .....ans.

 

 

1] given a(1+r) = 1 ;

sum (S) = a/(1-r) ;

now , u've a = 2[S-a] ;

so, r = 1/3 ;

hence,frm the first eq,......a = 3/4..ans.

 

2]

for the 5th Q,

u can simply use L'Hospital's rule....jus one step solu...

 

4.) Use L'Hospital's rule.....

 

limit as x->0 = 0......ans.

6.)

u know ,

_[x][a] [f(x)]^g(x) = e^{lim [f(x) - 1] *g(x) as x->a} ;

if g(x)->infinity , f(x)->1 as x->a  ;

 

so,given = e^{lim {(a^x+b^x+c^x - 3) / 3} * (2/x)} ;

 

             = e^{{(2/3) * lim {(a^x-1)+(b^x-1)+(c^x -1) / x}}  ;

 

now , lim as x->0 (a^x-1)/x = log a ;

 

So, given lim as x->0 = e^{(2/3) * log(abc)}  = (abc)^2/3............ans.

5.)

u know..._[x][0] [log(x+1)] / x =1 ;

 

given = (1/e)* _[x][e] [log(x/e)] / [(x/e)-1] ;

 

        =(1/e)*_{[(x/e) - 1][0]} [log(x/e)] / [(x/e)-1] ;

       

        = 1/e.................ans.

Hi shubhu,

2)

num = sinx[5-14cosx+3(3-4sin²x)];

       = sinx[14(2sin²x/2)-12sin²x];

 

now limit = 7{ (sin²x/2) / (x/2)² } - 12{sin²x/x²} ;

 

as x->0 , lim->-5.........ans.

Hi shubhu,

 

4)its very easy....

jus differentiate 'y' two times and divide both sides by 'y'...u get the result...

 

 

2)this one too...yr = a^r * cos[ax+(rpi/2)] ;

u'll observe tht all the rows in the det r same takin a³,a⁶ common frm 2nd,3rd rows resp.....

hence,det  = 0 .

 

 

 

4) Put z = lzlcis  ;

so LHS = lcis -1l = 2*l sin /2 l * l i*cis /2 l..........{also, l i*cis /2 l = 1 } ;

            = 2*lsin /2l ;

            <= l  l = l arg(z) l.............ans.

 

3) given :  x+i(y+2)+*[(x-1)²+y²] = 0........

also,given tht  is a real no...

so, y = -2 ; x+*[(x-1)²+y²] = 0 ;

i.e; x²(²-1) - 2x² + 5 = 0 ;

x is real...so, discriminant >=0.........

⁴-5² + 5 >= 0 ...u can find range of  frm this.......ans.

Hi shubham,

 

2)if p were tht real root...then p²+p+ = 0 ;

so, Im() / Im() = -p ; p²+p*Re() = -Re() ;

substitute value of p frm 1st relation into the 2nd relation....ans.

 

1) if u meant lz+1/zl = p(say)....

then check out the ans....

u know lz+1/zl <= lzl+1/lzl ....{let lzl be x}

so,x²-px+1 >= 0

therefore min value of x is ....u find it out