IIT JEE , AIEEE Forum - Mathematics , Algebra

ashok_p_ap

03.02.2008 (to be sent)

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

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

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

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

3) 6 dice are thrown. Probability that different nos. will turn up is

a)5/36 b)5/324 c)3/324 d)1/324 ans.a)

4) 12 balls are distributed among 3 boxes. The probability that first box contains 3 balls is

a)110/9 X (2/3)¹⁰ b)9/110 X (2/3)¹⁰ c)¹²C₃/12 X2⁹ d)¹²C₃/3¹²ans.a)

jatinroxx

Dude, u shudnt have mentioned the answers.............

cos^2x +.........forms a GP....till infinity......

U can sum it up..........

e^(log(x)) = x

Hence x^(series summation) = 1,9 (roots of the given equation)

akhil_o

3)

No of combinations of diff nos.= 6!

Total 6⁶

hence prob= 6!/6⁶

=5/324

jatinroxx

Selecting 3 balls to put into the first box........

12C3

Next 9 balls remaining can be put in the 2 boxes in 2^9 ways.......

Hence 12C3 * 2^9 are the favourable outcomes...

Total no. of outcomes are 3^12

raulrag009

Q2 answer should be B

iitkgp_bipin

cos²x + cos⁴x + ..... is an infinite GP with 1st term cos²x and common ratio cos²x.

Sum = cos²x / (1-cos²x) = cot²x

roots of t²-8t+9 are 1,9.

so exp(cot²x) = 1,9

taking log :

cot²x = ln1 , ln9

since tanx = 1/cotx, tanx is not defined for 1st case.

for cot²x = ln9

tanx =

1/ln9

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