IIT JEE , AIEEE Forum - Mathematics , Algebra

ashok_p_ap

26.01.2008 (To be sent)

1.If (exp)(cos²x+cos⁴x+??????.)logx is a root of t2-8t-9=0, then tanx is

a)?3 b)?2 c)1 d)1/?2

Ans.a)

2. (cos p -1)x²+ (cos p) +sin p=0 has real roots, then p belongs to ????

a) (0,2?) b)(- ?,0) c)(- ?/2, ?/2) d)(0,?)

Ans.d)

3) x2-ax+b=0 and x2+bx-a=0 have a common root ,then a+b=o and a-b=1. Prove.

4)x²-2xcos0+1=0, then x²ⁿ-2xⁿcosn0+1=?

a) cos2n0 b)sin2n0 c)0 d)some real no.other than 0.

Ans.c)

5) If x=c is a root of order 2 of a polynomial f(x), then x=c is also a root of

a) f ? (x) b) f ??(x) c) f ???(x) d)none

Ans) a)

6)An n-digit no. is a positive no. with exactly n-digits. 900 distinct n-digit nos. are to be formed using only the three digits 2,5 and 7. The smallest value of n for which this is possible is

a)6 b)7 c)8 d)9

Ans. d)

mukulss

i m trying........

raulrag009

Q2

Answer is B not d where ?=pie

Q3

Let R be the common root

since it is a root of both eqn ,surely it will satisfy them

R²-aR+b=0 and

R²+bR-a=0 subtract both equations

We get

b+a-aR-bR=0

b+a=R(b+a)

(b+a)(1-R)=0

hence R=1 or a+b=0 .....(I part proved)

Put R=1 in any one of eqn

1-a+b=0

hence a-b=1...........(II part proved)

raulrag009

What are these question marks in Q1

hsbhatt

Q4: x²-2xcos0+1=0,

This can be rewritten as x+1/x = cos

So, if x = cos

+isin

x+1/x = 2cos

=>

= 2k

+

where k is an integer.

xⁿ = (cos

+isin

)ⁿ = cosn

+isinn

=cosn

+isinn

.

x^-n = 1/xⁿ = cosn

-isinn

then x^-n + 1/xⁿ = 2ncosn

.

x²ⁿ-2xⁿcosn

+1 = 0.

hsbhatt

Q5: (x-2) is a root of order 2 of f(x) => (x-2)² is a factor of f(x).

That is f(x) = (x-2)² Q(x)

f'(x) = 2(x-2)Q(x) + (x-2)²Q'(x)

From this you can see that f'(x) also has (x-2) as a factor. f''(x) is however not divisible by 2.

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