XAT Inequalities Questions

The function $$f(x)=\frac{\log{(3x-7)}}{\sqrt{2x^{2}-7x+6}}$$ is only defined when both the numerator and the denominator of the function are defined are the denominator is not equal to zero.

The logarithm of the function is only defined for positive values :

Hence 3x-7 is greater than zero. Hence 

x > 7/3.

The value inside square root are defined for positive values. The value of the quadratic equation in the square root must be positive.

Hence $$2x^2-7x+6\ =\ 0$$ has the roots :

$$\frac{\left(7+\sqrt{\ 49-48}\right)}{4},\ \frac{\left(7-\sqrt{\ 49-48}\right)}{4}$$    : 2, 3/2

The quadratic equation is positive for :

$$\left(-\infty\ \frac{3}{2}\right)\ U\ \left(2,\ \infty\ \right)$$

Since in order to be a part of the domain the values of x must be greater than 7/3  and 7/3 is greater than 2 all values of x which are greater than 7/3 must be a part of the domain for x.

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.