ISI - Most urgent Maths questions!

[Ridhika]

ISI 2013
The minimum number of real roots of f(x) = |x|^3 + a|x|^2 + b|x| + c, where a, b and c are real, is
(A) 0,
(B) 2,
(C) 3,
(D) 6.

16. Let N = {1, 2, . . .} be a set of natural numbers. For each x ∈ N , define
An = {(n + 1)k, k ∈ N }. Then A1 ∩ A2 equals
(A) A2,
(B) A4,
(C) A5,
(D) A6.

 The number of permutations of the letters a, b, c, and d such that b
does not follow a, c does not follow b, and d does not follow c is
(A) 14,
(B) 13,
(C) 12,
(D) 11.

ISI 2012

Q9) The function f(.) is increasing over [a,b] . then [f(.)]^n where n is an odd integer greater that 1 is neccesarily:
a) increasing over [a,b]
b) decreasing over [a,b]
c) increasing over [a,b] iff f(.) is positive over [a,b]
d0 none of the above

ISI 2011

16. Number of continuous functions characterized by the equation
xf(x)+ 2 f (−x) = −1 , where x is any real number, is
 
(a) 1, (b) 2, (c) 3, (d) None of these.

25)In a sample of households actually invaded by small pox, 70% of
the inhabitants are attacked and 85% had been vaccinated. The
minimum percentage of households (among those vaccinated) that  
must have been attacked [Numbers expressed as nearest integer
value] is
 
(a) 55, (b) 65, (c) 30, (d) 15.

Please respond guys..!

[Amit Goyal]

The minimum number of real roots of f(x) = |x|^3 + a|x|^2 + b|x| + c, where a, b and c are real, is
(A) 0,
(B) 2,
(C) 3,
(D) 6.

For a, b, c > 0, clearly the above will have no real root. Hence, the minimum possible roots are zero.

[Amit Goyal]

Let N = {1, 2, . . .} be a set of natural numbers. For each x ∈ N , define
An = {(n + 1)k, k ∈ N }. Then A1 ∩ A2 equals
(A) A2,
(B) A4,
(C) A5,
(D) A6.

A1 has all positive multiples of 2. A2 has all positive multiples of 3. Hence, A1 ∩ A2 has those numbers that are positive multiples of 2 as well as 3 which is equivalent to saying that it has all positive multiples of 6. But thats A5. So, A1 ∩ A2 = A5

[Ridhika]

Thanks a lot sir ! I got these :) 

----- Reply message -----
From: "Amit Goyal [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI - Most urgent Maths questions!
Date: Sat, May 10, 2014 12:31 PM

Let N = {1, 2, . . .} be a set of natural numbers. For each x ∈ N , define
An = {(n + 1)k, k ∈ N }. Then A1 ∩ A2 equals
(A) A2,
(B) A4,
(C) A5,
(D) A6.

A1 has all positive multiples of 2. A2 has all positive multiples of 3. Hence, A1 ∩ A2 has those numbers that are positive multiples of 2 as well as 3 which is equivalent to saying that it has all positive multiples of 6. But thats A5. So, A1 ∩ A2 = A5

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[Amit Goyal]

The number of permutations of the letters a, b, c, and d such that b
does not follow a, c does not follow b, and d does not follow c is
(A) 14,
(B) 13,
(C) 12,
(D) 11.

For any set A, we use A' to denote its complement and n[A] to denote the number of elements in set A.

Let P denote the set of all permutations of the letters a, b, c, d.
Let P(ab) denote the set of all permutations in P in which b follows a.
Let P(bc) denote the set of all permutations in P in which c follows b.
Let P(cd) denote the set of all permutations in P in which d follows c.

We want to find n[P(ab)' ∩ P(bc)'∩ P(cd)'].
By De Morgan's law,
P(ab)' ∩ P(bc)'∩ P(cd)' = (P(ab) U P(bc) U P(cd))'
Thus,
n[P(ab)' ∩ P(bc)'∩ P(cd)'] = n[P] - n[P(ab) U P(bc) U P(cd)] = 4! - n[P(ab) U P(bc) U P(cd)]

So, we just need to find n[P(ab) U P(bc) U P(cd)]. Here is how we can find it:
n[P(ab) U P(bc) U P(cd)]
= n[P(ab)] + n[P(bc)] + n[P(cd)] - n[P(ab) ∩ P(bc)] - n[P(ab) ∩ P(cd)] - n[P(bc)∩ P(cd)] + n[P(ab) ∩ P(bc)∩ P(cd)]
= 3! + 3! + 3! - 2! - 2! - 2! + 1
= 13

Thus, n[P(ab)' ∩ P(bc)'∩ P(cd)'] = 24 - 13 = 11

[Amit Goyal]

The function f(.) is increasing over [a,b] . then [f(.)]^n where n is an odd integer greater that 1 is necessarily:
a) increasing over [a,b]
b) decreasing over [a,b]
c) increasing over [a,b] iff f(.) is positive over [a,b]
d0 none of the above

Consider two points x and y such that a <= x < y <= b,
Since f is increasing f(x) < f(y). Since n is an odd integer greater than 1, f(x) < f(y) further implies
[f(x)]^n < [f(y)]^n
So, [f(.)]^n is necessarily increasing over [a,b].

[Amit Goyal]

Number of continuous functions characterized by the equation
xf(x)+ 2 f (−x) = −1 , where x is any real number, is
 
(a) 1, (b) 2, (c) 3, (d) None of these.

Given that  xf(x)+ 2 f (−x) = −1 holds for any real x. So, fix any x and consider
xf(x)+ 2 f (−x) = −1.
Now consider -x and we also get (by replacing x by -x in above):
-xf(-x)+ 2 f (x) = −1

Treating f(x) and f(-x) as unknowns in the above two linear (in f(x) and f(-x)) equations, we can now solve for f(x) by eliminating f(-x) and obtain a single continuous function.

[Amit Goyal]

In a sample of households actually invaded by small pox, 70% of
the inhabitants are attacked and 85% had been vaccinated. The
minimum percentage of households (among those vaccinated) that  
must have been attacked [Numbers expressed as nearest integer
value] is
 
(a) 55, (b) 65, (c) 30, (d) 15.

Let A be the event that the randomly selected person has been attacked.  We are given that Pr(A) = 0.7.
Let V be the event that the randomly selected person has been vaccinated.  We are given that Pr(V) = 0.85.
We want to find min Pr(A|V) = min Pr(A ∩ V)/Pr(V).  This can be found by minimizing Pr(A ∩ V) which is clearly
0.55. (55%) This is because we can have at most 15% who are not vaccinated among those who have been attacked (70%). So clearly, Pr(A ∩ V)/Pr(V) = 0.55/0.85 = 0.647 (65%). So, answer is (b).

[Ridhika]

Thanks a lot sir! This was hugely helpful !

----- Reply message -----
From: "Amit Goyal [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI - Most urgent Maths questions!
Date: Sat, May 10, 2014 1:12 PM

In a sample of households actually invaded by small pox, 70% of
the inhabitants are attacked and 85% had been vaccinated. The
minimum percentage of households (among those vaccinated) that  
must have been attacked [Numbers expressed as nearest integer
value] is
 
(a) 55, (b) 65, (c) 30, (d) 15.

Let A be the event that the randomly selected person has been attacked.  We are given that Pr(A) = 0.7.
Let V be the event that the randomly selected person has been vaccinated.  We are given that Pr(V) = 0.85.
We want to find min Pr(A|V) = min Pr(A ∩ V)/Pr(V).  This can be found by minimizing Pr(A ∩ V) which is clearly
0.55. (55%) This is because we can have at most 15% who are not vaccinated among those who have been attacked (70%). So clearly, Pr(A ∩ V)/Pr(V) = 0.55/0.85 = 0.647 (65%). So, answer is (b).

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[don123]

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