IIT JEE , AIEEE Forum - Mathematics , Trignometry

shivamk

Solve this for me.

1. If tan

= ntan

then find tan²(

-

) ans in terms of n only.

2.mod (mod(tanx+cotx) - mod(tanx-cotx)) is ( multiple ans )

a . differentiable eveywhere

b continuous everywhere

c not continuous at some pts

d not differentiable at 2 pts.

3.If 1 is a twice repeated root of ax³+bx²+cx+d=0 then which are correct( multiple ans)

a. a=d

b.a=b=d

c.a+b=0

d. a+d=0

pLZ HELP ME

shivamk

someone plzzzzzzzzzzzz help.

ankur.kkhurana

u want first ans in form of ..........

shivamk

I want the first answer in terms of n.

akku

In the 4th option atd what is "t"

shivamk

sorry it is not t . it is +

akku

Hi shivamk

I think the answers for the 3 one are a) b) c) d)

troy

ans 1
(n-1)^2/4n

akku

3)SOLUTION

Since 1 is a twice repeated root f(1)=0 and f'(1)=0

ie a+b+c+d=0 and 3a+2b+c=0

Hence 2a+b=d

f(x)=a(x-1)^2(x-p)

product of roots 1*1*p= -d/a ---->third root =p= -d/a

troy

tan^2(theta -phi)=(n-1)^2tan^2phi/1+tan^4phi n^2+2ntan^2phi

=n-1^2/cot^2phi+n^2tan^2phi+2n

=deno is min at tan^2phi=1/n

so max value is n-1^2/4n

shivamk

how can u get a+b=d after that step. u get 2a+b=d . Just explain to me.and if u take a =d in that step uget b = -a. so according to me only only a,c , d should be correct.

akku

q3)

apart from 2a+b=d ---1)

we get another eqn by sum of roots taken 2 at a time

p+p+1=c/a

--> c+2d=a ---2)

if we substitute p in f(x)

f(p)=0

d^2-bd+ca-a^2=0

d(d-b)+a(c-a)=0

putting values from 1) and 2)

we get

2ad+ac-a^2 => a=0 or (2d+c-a)=0

and

d^2-db-2ad=0 => d=0 or (2a+b-d)=0

IF WE TAKE a=d=0

we obtain b=0(from 1) and c=0 (from 2)

HENCE a) b)c) d) MAY be the correct ans

Sorry for the delay!

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