5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?

A. 2/7
B. 1/3
C. 7/20
D. 2/5
E. 3/7

\(x^2 + 2x -15 = -m\)
\(=> x^2 + 2x + 1 -16 = -m\)
\(=> (x+1)^2 = 16 -m\)
\(=> 16 -m >=0\) and a square number

If m varies between [-10,10]

\(=> 16 - 10 <= 16 - m <= 16 - (-10)\)
\(=> 6 <= 16 - m <= 26\)

Between 6 and 16, there are 3 square numbers i.e. 9, 16, 25

For \(16 - m = 9 => m = 7\)
For\(16 - m = 16 => m = 0\)
For\(16 - m = 25 => m = -9\)

So number of possible values of m = 3 i.e. [7, 0, -9]
And for m > 0, number of possible values = 1 i.e only for m=7

So probability = 1/3
Ans = B
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1. \(x^4=x^3 + 6x^2\)

So \(x^2(x^2-x-6)=0\)
Or \(x=0, -2, 3\)
However -2 cannot be the solution as x is a positive root. So sum of the roots is 3 i.e D
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3. If a and b are positive numbers, such that a^2 + b^2 = m and a^2 - b^2 = n, then ab in terms of m and n equals to:


2a^2 = m+n
2b^2 = m-n

4a^2b^2= m^2 - n^2
or 2ab= √(m^2-n^2)
C
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5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?

A. 2/7
B. 1/3
C. 7/20
D. 2/5
E. 3/7

x^2 + 2x - 15 =-m or x^2 + 2x + (m-15) = 0
or (x+5)(x-3) = -m
Now since m is an integer so x must also be an integer.
x=1 m= 12 NO
x=2 m= 7 YES
x=3 m= 0 YES
x=4 m= -9 YES
x=5 m= -20 NO
No further values will satisfy!
So m has 3 values so far!
x= 0 m= 15 NO
x= -1 m= 16 NO
x= -2 m= 15 NO
x= -3 m= 12 NO
x= -4 m= 7 YES
x= -5 m= 0 YES
x= -6 m= -9 YES
x= -7 m= -20 NO

SO m can take 3 values 7, 0 and -9 and only ONE of them is greater than 0
SO the probability is 1/3
B
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8. If m is a negative integer and m^3 + 380 = 381m , then what is the value of m?

A. -21
B. -20
C. -19
D. -1
E. None of the above

Lets just start with the bottom and solve
-1 clearly does not work
lets see for m = -19 and I have a good feeling because 380 is divisible by 19

\(-19*-19*-19 + 380\)

\(= -19(19^2-20)\)

\(=-19(361-20)\)
Not working

Lets try -20

\(-20^3 + 380\)

\(= -20(400-19)\)

\(=-20(381)\)

\(=381m\)

BINGO
B
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10. If f(x) = 2x - 1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?

A. -145
B. -24
C. 24
D. 145
E. None of the above

\(f(n^2) = 2n^2 - 1\)

\(g(n+12) = (n+12)^2\)

\(= n^2 + 24n +144\)

So if they are equal then

\(2n^2 - 1 = n^2 +24n+144\)

OR \(n^2-24n-145=0\)

Product of the roots of this is -145

A it is
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Hi Souvik
I liked ur approach cn u pls expand on it...i mean is there any hard and fast rule that u r following.....completely missed this question....

Archit

I used calculus maxima minima. However you are free to use the "completing the square" method where you are left with 2 perfect squares and a -9 as the guy above me did.
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9. x = \((\sqrt{5}-\sqrt{7})^2\)

As from the given options we know that x>0. Thus, I can safely write that

\(\sqrt{x} = \sqrt{7}-\sqrt{5}\)

or \(\sqrt{7} = \sqrt{x} +\sqrt{5}\)

x can not be 1 or more than 1 as then \(\sqrt{7}\)>3 which is not true. Thus, on approximation, x=0.

A.
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Hello Bunuel,

Nice set of Questions

Here goes my solutions

10. If f(x) = 2x - 1 and g(x) = x^2, then what is the product of all values of n for which f(n^2)=g(n+12) ?[/b]

Sol: From the given function we have

f(n^2)=g(n+12) ----> 2n^2-1= (n+12)^2
Solving this eqn we get
n^2 - 24n - 145= 0-----> n^2 -29n+ 5n - 145 =0
Possible values of n are 29 and -5.
Product will be -145 and hence ans is option A
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