XAT Inequalities Questions

Since $$x^2-y^2=20$$ and x,y,z are positive integers, 

(x+y)*(x-y) = 20, Hence x-y, x+y are factors of 20.

Since x, y are positive integers, x+y is always positive, and for the product of (x+y)*(x-y)  to be positive x-y must be positive.

x, y are positive integers and x-y is positive x must be greater than y.

The possible cases are : (x+y = 10, x-y = 2), (x+y = 5, x-y = 4).

The second case fails because we get x =9/2, y = 1/2 but x, y are integral values

For case one x = 6, y = 4.

$$y^3 - 2x^2 - 4z \geq -12$$

Substituting  the values of x and y, we have :

64 - 72 - 4*z $$\ge\ -12$$

 -8 - 4*z $$\ge\ -12$$

z$$\le\ 1$$

Since x, y, z are positive integers, the only possible value for z is 1.

To maximize Z, B has to be minimized and A,C,D are to be maximised.

The value of B can be 0 or 1.

Case 1:

B=0 => A=1=> C=8 and D=12.

Z= 15+8+6=29.

Case 2:

B=1 => A=0=> C=12 and D=15.

Z=-3+12+7.5=16.5.

.'. D=12 is correct answer. 

The critical points of the function are -4, -6, -8, ... , -98 ( 48 points).

For all integers less than -98  and greater than -4 f(x) > 0 always .

for x= -5, f(x) < 0 

Similarly, for x= -9, -13, ...., -97 (This is an AP with common difference -4)

Hence, in total there are 24 such integers satisfying f(x)< 0.

2 ≤ |x - 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x - 1| will be 2 and the maximum value of |x - 1| will be 5 as per the question.

When, |x - 1| = 2, |y + 3| can be either 1 or 2
So, for x =  -1, y can be - 4 or - 2 or - 5 or -1.
Thus, we get 4 pairs of (x, y) 

When |x - 1| = 3, |y + 3| can be 1 only
So, for x = - 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x - 1| = 4, |y + 3| can be 1 only
So, for x = - 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x - 1| = 5, |y + 3| can be 1 only
So, for x = - 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option E is the correct answer.

$$4(x - 2)^{2} + 4(y - 3)^{2} - 2(x - 3)^{2}$$
$$y$$ is an independent variable. The value of $$y$$ is unaffected by the value of $$x$$. Therefore, the least value that the expression $$4(y-3)^2$$ can take is $$0$$ (at $$y=3$$).

Let us expand the remaining terms. 
$$4(x-2)^2-2(x-3)^2$$=4*$$(x^2-4x+4)-2*(x^2-6x+9)$$
$$=2x^2-4x-2$$
=$$2(x^2-2x-1)$$
=$$2(x^2-2x+1-2)$$
=$$2((x-1)^2-2)$$
The least value that the expression $$(x-1)^2$$ can take is $$0$$ (at $$x$$ = $$1$$)
Therefore, the least value that the expression $$2((x-1)^2-2$$ can take is $$2*(0-2)=2*(-2) = -4$$
Therefore, option B is the right answer.

|a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10

Expression : $$[abc - (a + b + c)]$$

For maximum value, two of a,b and c should be negative, as all three negative will make abc negative.

Thus, max value will occur if $$a = -10 , b = -9 , c = 8$$

=> Max value = $$[(-10 \times -9 \times 8) - (-10 -9 + 8)]$$

= $$720 + 11 = 731$$

The given expression can be written as $$\frac{a^2+a+1}{a}*\frac{b^2+b+1}{b}*\frac{c^2+c+1}{c}*\frac{d^2+d+1}{d}*\frac{e^2+e+1}{e}$$.
$$\frac{a^2+a+1}{a}=a+\frac{1}{a}+1$$
We know that for positive values, AM $$\geq$$ GM. 
$$\frac{a+\frac{1}{a}}{2} \geq \sqrt{a*\frac{1}{a}}$$
$$a+\frac{1}{a} \geq 2$$
The least value that $$a+\frac{1}{a}$$ can take is $$2$$.
Therefore, the least value that the term $$a+\frac{1}{a}+1$$ can take is $$3$$.
Similarly, the least value that the other terms can take is also $$3$$.
=> The least value of the given expression = $$3*3*3*3*3$$ = $$243$$.
Therefore, option E is the right answer.

The product of 2 numbers A and B is maximum when A = B.
If we cannot equate the numbers, then we have to try to minimize the difference between the numbers as much as possible. 
pq will be maximum when p=q.
qr will be maximum when q=r.
qr will be maximum when r=p.
Therefore, p, q, and r should be as close to each other as possible. 
We know that p,q,and r are integers and p+q+r=10.
=> p,q, and r should be 3,3, and 4 in any order. 
Substituting the values in the expression, we get, 
pq+qr+pr+pqr = 3*3 + 3*4 + 3*4 + 3*3*4
= 9 + 12 + 12 + 36
= 69
Therefore, option C is the right answer.

It is given that $$x=(9+4\sqrt{5})^{48}$$ ... (1)

Let us assume that $$y=(9-4\sqrt{5})^{48}$$ ... (2)

We can see that  0 < y < 1. 

Also, x + y = 2($$48C0*(9)^{48}$$+$$48C2*(9)^{46}*(4\sqrt{5})^2$$+$$48C4(9)^{44}*(4\sqrt{5})^4$$+...+$$48C48*(4\sqrt{5})^{48}$$)

We can see that x + y is an integer therefore we can say that y + f = 1. Hence, y = (1 - f)

We can see that = x(1 - f) = x*y

$$\Rightarrow$$ $$(9+4\sqrt{5})^{48}(9-4\sqrt{5})^{48}$$

$$\Rightarrow$$ $$(9^2-(4\sqrt{5})^2)^{48}$$

$$\Rightarrow$$ $$(81-80)^{48}$$ = 1

Hence, option A is the correct answer.