Test: Inequalities- 1 - CAT MCQ

Taking lowest possible positive value of m i.e. 1 .

Such that a + b + c + d = 5, so atleast one of them must be grater than 1, take a = b = c = 1 and d = 2

we get a² + b² + c² + d² = 7 which is equal to 4m² + 2m + 1 for other values it is greater
than 4m² + 2m +1. 

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

-4x = -24, 5y = -30

So none of the options out of a,b or c satisfies .

Since the product is constant, (a + b + c + d)/4 > = (abcd)^1/4

We know that abcd = 1.

Therefore, a + b + c + d > = 4

(a + 1)(b + 1)(c + 1)(d + 1)

= 1 + a + b + c + d + ab + ac + ad + bc + bd + cd + abc + bed + cda + dab + abcd

We know that abcd = 1

Therefore, a = 1/bcd, b = 1/acd, c = 1/bda and d = 1/abc

Also, cd = 1/ab, bd = 1/ac, bc = 1/ad

The expression can be clubbed together as

1 + abcd + (a+1/a)+(b+1/b)+(c+1/c)+(d+1/d) + (ab+1/ab) + (ac+1/ac) + (ad +1/ad)

For any positive real number x, x + 1/x ≥ 2

Therefore, the least value that (a+1/a), (b+1/b).... (ad + 1/ad) can take is 2.

(a+1)(b+1)(c+1)(d+1) > 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 2

=> (a + 1)(b + 1)(c + 1)(d + 1) ≥ 16

The least value that the given expression can take is 16.

The given equations are  x + y + z = 5 — (1) , xy + yz + zx = 3 — (2)

xy + yz + zx = 3

x(y + z) + yz = 3

⇒ x ( 5 - x ) +y ( 5 – x – y) = 3

⇒ -y² - y(5 - x) - x² + 5x = 3
⇒ y² + y(x - 5) + (x² - 5x + 3) = 0
The above equation should have real roots for y, => Determinant >= 0

⇒ b² - 4ac0

⇒ (x - 5)² - 4(x² - 5x + 3) ≥ 0

⇒ 3x² - 10x - 13 ≤ 0

⇒ -1 ≤ x ≤ 13/3

Hence maximum value x can take is 13/3, and the corresponding values for y,z are 1/3, 1/3

First inequality:

-n + 2 ≥ 0

-n ≥ -2

n ≤ 2

Second inequality:

2n ≥ 4

n ≥ 2

Only n = 2 satisfies both inequalities. So, there is only 1 integer that satisfies both the inequalities.

The correct option is A.

The expressions in the four options can be expanded as

xyz-yz; xyz-xz; xyz-xy and xyz+xz

The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.

Option a) is the correct answer.

x² + 7x + 10 > 40 or x² + 7x - 30 > 0
(x + 10) (x - 3) > 0
The roots are -10 and +3.
=> x should lie outside the roots.

Now, what is this based on?
There is a simple thumb rule for solving quadratic inequality

For any quadratic inequality ax² + bx + c < 0
Factorize it as a(x - p) ( x - q) < 0

Whenever a is greater than 0, the above inequality will hold good if x lies between p and q.
a(x - p) (x - q) will be greater than 0, whenever x does not lie between p and q. In other words x should lie in the range ( -∞, p) or (q, ∞).

Now, coming back to the question (x + 10) (x - 3) > 0 Or, x < -10 or x > 3

Given expression can be reduced to
le(15, min(10,15-9) , le(9,8,12))
Or le(15,6,1) = 9

The integral values of x for which y is an integer are 13, 30, 47,……

The values are in the form 17n + 13, where n ≥ 0

17n + 13 < 1000

⇒ 17n < 987

⇒ n < 58.05

⇒ n can take values from 0 to 58

⇒ Number of values = 59

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