Test: Inequalities (April 15) - 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. 

Let us take an example of  x=6 and y=-6 ,

• Substituting te values of x and y in Option A:
• 6-24 = -18 which is not greater than 1.
• And the given relation must be true for every value.

Therefore, option A is NOT TRUE

• Substituting te values of x and y in Option B:
• 6>-4(-6) = 6>24 which is not true
• And the given relation must be true for every value.

Therefore, option B is NOT TRUE

• Substituting the values of x and y in Option C:
• -4 x 6 = -24 is not lesser than -6 x 5 = -30 
• And the given relation must be true for every value.

Therefore, option C is NOT TRUE

So none of the options out of a,b or c satisfies for all values.
Hence, Option D is correct.

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.

Ma(10, 4, le((la10, 5, 3), 5, 3))

Or Ma(10, 4, le(8, 5, 3))

Or Ma(10, 4, 3)

Or 1/2(6 + 7) = 6.5

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