THESE ARE VERY SIMILAR TO ORIGINAL BITSAT STANDARD [ I'm serious ]
I HAVE ADDED 2 MORE QUESTIONS...
Here are Some questions taken DIRECTLY FROM MTG-2010 BITSAT EXPLORER , please try to give the steps for these questions :
1.If r>p>q , the number of different selections of p+q things taking r at a time , where p things are identical and q other things are identical , is :
[A] p+q-r
[B] p+q-r+1
[C] r-p-q+1
[D] None of these
Answer given : [B]
2. If
Δ = 
> 0
[A] abc > 1
[B] abc > -8
[C] abc < -8
[D] abc > -2
Answer given : [B]
3. Id a and d are two complex numbers , then the sum to (n+1) terms of the series :
aC0 - (a+d)C1 + (a+2d)C2 - (a+3d)C3 +.....is :
[A]a2n
[B] na
[C] 0
[D] None of these
Answer given : [B]
4. Let P(x) and Q(x) be 2 polynmials .Suppose that f(x)=P(x3) + xQ(x3) is divisible by x2+x+1 , then :
[A] P(x) is divisible by (x-1) but Q(x) is not divisible by (x-1)
[B] P(x) is not divisible by (x-1) but Q(x) is divisible by (x-1)
[C] Both P(x) and Q(x) are divisible by (x-1)
[D] f(x) is divisible by (x-1)
Answer given : [C]
5. The value of the expression
32 10
S=Σ (3p+2)( Σ (sin2qΠ11-i cos2qΠ11))4p
p=1 q=1
is :
[A] 1648
[B] 1646
[C] 1645
[D] 0
Answer given : [A]
6.Let S be the solution set of all real x such that
(tanÎ 8)x -(tanÎ 8)-x < 3
[A] (0,1)
[B] (1,2)
[C] (-1,1)
[D] (-1,0)
Answer given : [A]
7.Three six-faced fair dice are thrown together . The probability that sum of the numbers appearing on the dice is k ( 3 ≤ k ≤ 8) is :
[A] (k-1)(k-2)432
[B] k(k-1)432
[C] k2432
[D] None of these
Answer given : [A]
8.If A3 = O , then I+A+A2 is :
[A] (I-A)
[B] (I-A)-1
[C] (I+A)-1
[D] (I+A)
Answer given : [B]
My Doubt :
Let y=I+A+A2
yA=A+A2+A3
yA=A+A2+O
yA=A+A2
As we know from options that y≠O , we cancel A to get
y=I+A..!!!! ??? [7]
9.Solution set of the inequality
√x - 3 ≤ 2√x - 2
[A] 2 ≤ x ≤ 4
[B] [0,1] U [2,16]
[C] [0,1] U [4,16]
[D] None of these
Answer given : [C]
ADDED QUESTIONS :
10.This is a probability question to which I got an answer [now , dont ask me how ! [7] ] but I am not able to get a proper explanation ?
Three mothers each with her one child is called for an interview for the admission of the child in a school.The principal proposes to meet only one [either child or mother at a time ] .The probability that NO child is called before his mother if the calling process is random is :
[A] 2!2!2!6!
[B] 16!
[C] 18
[D] None of these
Answer given : [C]
11.The sum 3C0 +7C1 +11C2+....+(4n+3)Cn equals :
[A] (2n+3).2n-1
[B] (4n+6).2n
[C] (2n+3).2n
[D] None of these
Answer given : [C]
11)The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of xr in the expansion (1 + x1 + x2 + x3+.....+ xp)(1 + x1 + x2 + x3+.....+ xq)
3)m not sure about coz if u put n=0 or maybe 2 answer is coming out to be 0.......but again for a n=1 itz different ...u may check
1whoa..you are a whiz kid , archana..[ OR I'M A fool .?!!] , Thanx..!
1for 11) we use reversing technique
if a0 a1.......an are in A.P
then
S=a0C0+a1C1+........+anCn
now Using Cr=Cn-r we obtain
S=anC0+an-1C0...........+a0Cn
ading both summation
2S=(a0+an)C0+(a1+an-1)C1+...........+(a0+an)Cn
As a0,a1,a2,......an are in A.P
a0+an=a1+an-1=....
2S=a0+an(c0+C1+C2....+Cn)
S=(a0+an)2n-1.
so answer comes out to be C
1@ Archana , I think your formulation of Q.1 is not right [7]
Thanks for that new general formula !
@ Ray , thanks for giving me the initial steps for Q.7..yes , they seem correct and I have cracked it !
And for that Inequality question..I'M A FOOL..I forgot to putt he denominator [[4]]..thanx Ray..
1Dunno probability much...but I worked out this solution
Total no. of case =216
then find...coeff of x^{k} in
coeff. of x^{k-3} in (1-x^{6})^{3}(1-x)^{-3}
is this correct?
1Q8)
I don't understand your doubt...maybe A=O...
But here goes the solution...
(I-A)(I+A+A^{2})
Expand..
\Rightarrow I-O=I
\Rightarrow (I-A)^{-1}=I+A+A^{2}
1Okk..now starting from the last one
The inequality can be written as \frac{(\sqrt{x}-3)(\sqrt{x}-2)-2}{\sqrt{x}-2}\leq 0
Expand, break and simplify. U'll get
\frac{\sqrt{x}(\sqrt{x}-4)-1(\sqrt{x}-4)}{\sqrt{x}-2}\leq 0
Now this will give u the desired answer.
If the answer dont match, the answer is wrong
17.Three six-faced fair dice are thrown together . The probability that sum of the numbers appearing on the dice is k ( 3 ≤ k ≤ 8) is :
this is general result
suppose for calculating the probability of getting sum S on n number of dice with x sides.
(s-n)/x
SUM (-1)k * nCk * s-(n*k)-1Cn-1
k=0
now take sum to be anything beween 3 and 8
1Δ=abc-(a+b+c)+2
Δ>0(given)
abc+2>a+b+c
abc+2>3(abc)1/3
x3+2>3x, where x=(abc)1/3
..
...
....
further solving we get:
(abc)1/3 > -2
abc>-8
1there is a mistake in my 2nd solution. it should be greater than and not greater than or equal to
1since P(1) = Q(1)
SUBSTITUTE it any of the equation
P(1) + ωP(1) = 0
P(1)(1 + ω) = 0
Therefore P(1) =0
12)
on expanding the det, we get
(abc)≥ a + b + c - 2
since a,b and c ≥0
a + b + c ≥ 3(abc)1/3
substituting
(abc) ≥ (abc)1/3 - 2
let abc=t
t≥t3 - 2
t3 - t - 2 ≥0
(t-1)2 (t + 2) ≥ 0
t ≥ -2
abc ≥ -8
1@ Sridhar..didnt follow how P(1)=0..can u explain pls ?
14)
f(ω) = P(1) + ωQ(1) = 0
f(ω2)= P(1) + ω2Q(1) = 0
adding we get
P(1) = Q(1)
Subsituting in any of the equations, we get P(1)=0. HENCE P(x) is divisible by x-1 also Q(x) is divisible by x-1.
