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Question 1:

If f(x) = 100 – x², is f(a) > f(b) ?

a² > b²
a/b > 1

Answer:

f(x) = 100 - x²

Therefore, f(a) = 100 - a² and f(b) = 100 - b²

Is f(a) > f(b)?

Or, is 100 - a² > 100 - b² ?

Or, is - a² > - b² ?

Or, is a² < b² ?

1) a²> b² , clearly its sufficient to tell that a² is not less than b² ;SUFFICIENT.

2) a/b > 1

=> (a/b) -1 > 0

=> (a-b)/b > 0, which means either both a-b and b are positive, or both a-b and b are negative.

=> If b>0, a-b>0 means a>b. So a²> b².

And if b<0, a-b<0 means a2 > b, which means a is bigger negative number than b.(For eg. a=-4, b=-2). So, a²> b².

So, either way we can say that a²> b² ; SUFFICIENT.

Alternatively, the statement 2 could also have been analyzed in the following manner.

a/b > 1 → that the magnitude of numerator is more than denominator which would mean a²> b².

The correct answer is D; each statement alone is sufficient.

Question 2:

a’ =10/a for all positive prime numbers. And a’ = -a²for all positive non-primes, which of the following is the largest?

(A) 5’ + 6’ + 2’

(B) 7’ – 4’

(D) 4’ + 8’ – 5’

(E) 4’ + 8’ – 2’

Ans:

Since, for all positive prime numbers;

a' = 10/a

And, for all positive non-prime numbers;

a' = -a²

Thus, using these functions to find out the values given in the question, we get;

(A) 5' + 6' + 2' = (10/5 - 6²+ 10/2) = (5 - 36 +2) = -29

(B) 7' - 4' = (10/7 - (-4²)) = (10/7 + 16) = 122/7

(D) 4' + 8' - 5' = ( -4²- 8²- 10/5) = ( -16 -64 - 2) = -82

(E) 4' + 8' - 2' = (-4²- 8²- 10/2) = (-16 - 64 -5) = -85

Thus, among all the options we can see the largest value is obtained in option B

Alternatively, you could have also analyzed that all the options will generate a negative value except option B. hence that is the largest.

The correct answer is B.

Question 3:

If [a] represents least integer greater than or equal to a, what is the value of [–3.3] + [–2.9] + [0.1] + [1] + [1.9] ?

(A) –1

(B) –2

(D) –4

(E) –5

Since [a] represents the least integer greater than or equal to a, this implies that if a is an integer than [a] = a only. But if a is a fraction than, [a] = the least integer greater than a.

Thus, [-3.3]=-3

And,[-2.9]=-2

And,[0.1]=1

And,[1]=1

And,[1.9]=2

Hence, [-3.3]+[-2.9]+[0.1]+[1]+[1.9]

=-3-2+1+1+2

=-1

The correct answer is A.

Question 4:

In which of the following functions, is f(x) = f(1/x), for all values of x greater than 1?

(A) f(x) = 1/x + 1/ (x + 1)

(B) f(x) = |x|

(D) f(x) = (1 - x² )²/x²

(E) f(x) = (1 - x²)/x²

Answer:

(A) f(x) = 1/x + 1/(x + 1) and f(1/x) = 1/(1/x) + 1/(1/x + 1) = x + x/(1 + x)

Thus, f(x) ≠ f(1/x) for all values of x > 1

(B) f(x) = |x| and f(1/x) = |1/x| = 1/(|x|)

Thus, f(x) ≠ f(1/x) for all values of x >1

Thus, f(x)≠f(1/x) for all values of x>1

(D) f(x)=(1 - x² )²/x² and f(1/x)=[1 - (1/x)² ]²/(1/x)² =[1 - 1/x² ]²/(1/x² )=[(x²- 1)/x² ]²/(1/x² )=(x²- 1)²/(x⁴× 1/x² )=(x²- 1)²/x² =(1 - x² )²/x²

Thus, f(x) = f(1/x) for all values of x > 1

(E) f(x) = (1 - x²)/x² and f(1/x) = (1 - (1/x)²)/(1/x)² = (1 - 1/x² )/(1/x² ) = ((x²- 1)/x² )/(1/x² ) = (x²- 1)/(x²×1/x² ) = x²- 1

Thus, f(x) ≠ f(1/x) for all values of x>1

Thus we can see only for (D); f(x) = f(1/x)

The correct answer is D.

Question 5:

In a sequence, a_{n + 2} = a_{n + 1} – a_n + (n + 2) for all n ≥1, where a_{n + 2} represents the (n + 2)^th term. If the second and the fourth terms of the sequence are 2 and 5 respectively then what is the value of the 6th term of the sequence?

(A) 5

(B) 7

(D) 9

(E) None of these

For n≥1,

a_{(n + 2)} = a_{(n + 1)} - a_n + (n + 2) ……Equation(1)

Putting value of n = 2 in the above equation, we get:

a₄= a₃- a₂+ (2 + 2)

In the question, it’s also mentioned that a₂= 2 , a₄= 5. Putting in the above equation, we get the value of a₃= 3

Now by putting value of n = 3 in equation(1), we get:

a₅ = a₄ - a₃ + (3 + 2)

a₅= 5 - 3 + (3 + 2)