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 greater 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.

1. Start from AM–GM on the sum:

• Since abcd = 1, by AM–GM: (a+b+c+d)/4 ≥ (abcd)^1/4 = 1 ⇒ a+b+c+d ≥ 4.

2. Expand the product:
   (1+a)(1+b)(1+c)(1+d)
   = 1 + (a+b+c+d)

• (ab+ac+ad+bc+bd+cd)

• (abc+abd+acd+bcd)

• abcd.

3. Use abcd = 1 to rewrite terms in reciprocal pairs:

• abc = 1/d, abd = 1/c, acd = 1/b, bcd = 1/a.

• Similarly, cd = 1/ab, bd = 1/ac, bc = 1/ad.

Group as:
1 + abcd

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

• (ab + 1/ab) + (ac + 1/ac) + (ad + 1/ad).

4. Apply the inequality x + 1/x ≥ 2 for x>0:

• Each bracketed pair is ≥2.

• Also 1 + abcd = 1 + 1 = 2.
  So the whole expression ≥ 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 16.

5. Achievability:

• Equality in x + 1/x ≥ 2 holds when x=1.

• Set a=b=c=d=1 (which satisfies abcd=1), then (1+a)(1+b)(1+c)(1+d) = 2·2·2·2 = 16.

Conclusion
Minimum value = 16.

Quick verification / shortcut
Apply (1+x) ≥ 2√x to each factor and multiply:
(1+a)(1+b)(1+c)(1+d) ≥ 16√(abcd) = 16.
Equality at a=b=c=d=1.

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.

Stepwise reasoning

1. Write each option relative to xyz:

• A: (x−1)yz = xyz − yz

• B: x(y−1)z = xyz − xz

• C: xy(z−1) = xyz − xy

• D: x(y+1)z = xyz + xz

2. Compare distances from xyz:

• Distance for A: yz

• Distance for B: xz

• Distance for C: xy

• Distance for D: xz

3. Since x > y > z > 0,
   the products order as xy > xz > yz.
   Therefore, yz is the smallest distance.

Conclusion
The closest expression to xyz is (x−1)yz.

1. Bring all terms to one side:
   (x + 2)(x + 5) − 40 > 0
   x² + 7x + 10 − 40 > 0
   x² + 7x − 30 > 0.

2. Factor the quadratic:
   x² + 7x − 30 = (x + 10)(x − 3).

3. Sign analysis:

• Roots at x = −10 and x = 3.

• For a positive-leading quadratic, expression > 0 outside the roots:
  x < −10 or x > 3.

Answer: x < −10 or x > 3 ⇒ Option B.

  ↵

1. ma(x, y, z) = max(15, 10, 9) = 15

2. le(9, 8, ma(x, y, z)) = min(9, 8, 15) = 8

3. min(y, x - z) = min(10, 15 - 9) = min(10, 6) = 6

4. le(x, 6, 8) = min(15, 6, 8) = 6

Answer: b) 6

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