IIT JEE , AIEEE Forum - Mathematics , Algebra

ashok_p_ap

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)

akhil_o

3)

for different numbers to show up

we have

6x5x4x3x2x1=6! ways

total ways is 6*6*6*6*6*6=6⁶ ways

so prob=6!/6⁶=

=5/324

aditi_g

for the equation to have real roots D>=0

so u have cos^2 p - 4(cos p-1) sin p>=0
=> cos^2p - 4 sinp cos p + 4 sinp>=0
now this is a quadratic in cos and for this to be gretaer than 0 its D should be <0 ...( as cos^2 p is always >0)
=> 16 sin^2p - 16 sinp<0
=>sinp(sinp -1)<0
u know that sinp <=1 so sinp - 1<0 so for the whole term to be less than zero, sinp>0...
and for this p belongs to (0,pi)....
hope u have got it...nudge me if u have any doubt

priyesh

for 4)

total ways of distributing

= 3^12

no .of ways in which only 3 balls go into first box = 12C3 * 2^9

prob = [12C3 * 2^9] / [(3)^12] = 12*11*10/6 * 2^9/3^12 = 110/9 * (2/3)^10 = A)

priyesh

for 3)

there is no condition on the first throw
on second throw any no. other than the first throw can appear

on third any no. other than first or second can appear

and so on so forth

hence prob = 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 5/324

aditi_g

hey wat r ur question marks yaar???and in first question is log x also in e to the power term....and are u sure its + 9...jst chk if its -9 or +9

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