Inequalities and Modulus: IPMAT practice questions

Solution

By AM ≥ GM: 4x + 16/x ≥ 2√(4 × 16) = 16.

Equality holds when 4x = 16/x, that is x = 2.

Minimum value = 16.

Solution

The expression factors as (x − 4)(x − 11).

It is zero or negative for 4 ≤ x ≤ 11.

That gives 8 integers.

Solution

The expression factors as (x + 4)(x − 4).

It is negative for −4 < x < 4.

That gives 7 integers.

Solution

For x = 0 there are 2 values of y (±6); for x = ±6, one (y = 0).

For each x with 0 < |x| < 6 there are 2 values of y, giving 2 × 2 × 5 = 20.

Total = 2 + 2 + 20 = 24.

Solution

The critical points are 3, 5 and 7; x = 5 is excluded (the denominator would be 0).

For x > 7 all three factors are positive, so the expression is positive. The sign alternates across each critical point.

Negative on (5, 7) and on (−∞, 3); zero at x = 3 and x = 7. So x ≤ 3 or 5 < x ≤ 7.

Solution

The expression factors as (x + 1)(x − 4).

It is negative for −1 < x < 4.

That gives 4 integers.

Solution

|2x − 6| ≤ 13 means −13 ≤ 2x − 6 ≤ 13.

So −72 ≤ x ≤ 192.

That range contains 13 integers.

Solution

The critical points are −5, −1 and 7; x = −1 is excluded (the denominator would be 0).

For x > 7 all three factors are positive, so the expression is positive. The sign alternates across each critical point.

Negative on (−1, 7) and on (−∞, −5); zero at x = −5 and x = 7. So x ≤ −5 or −1 < x ≤ 7.

Solution

A sum of distances |x − p| is least when x is at the median of the points.

The points are −7, −6, 9, 11; the minimum is reached for any x between −6 and 9.

Value = 1 + 0 + 15 + 17 = 33.

Solution

A sum of distances |x − p| is least when x is at the median of the points.

The points are 3, 4, 8; the minimum is reached at x = 4 (the median).

Value = 1 + 0 + 4 = 5.

Solution

|x − 9| < 10 means −10 < x − 9 < 10.

So −1 < x < 19.

That range contains 19 integers.

Solution

The critical points are −4, 1 and 4; x = 1 is excluded (the denominator would be 0).

For x > 4 all three factors are positive, so the expression is positive. The sign alternates across each critical point.

Negative on (1, 4) and on (−∞, −4); zero at x = −4 and x = 4. So x ≤ −4 or 1 < x ≤ 4.

Solution

By AM ≥ GM: 9x + 36/x ≥ 2√(9 × 36) = 36.

Equality holds when 9x = 36/x, that is x = 2.

Minimum value = 36.

Solution

The points with |x| + |y| = k number 4k for k ≥ 1, plus the origin for k = 0.

Total = 1 + 4(1 + 2 + … + 8) = 1 + 4 × 36 = 145.

Solution

The points with |x| + |y| = k number 4k for k ≥ 1, plus the origin for k = 0.

Total = 1 + 4(1 + 2 + … + 2) = 1 + 4 × 3 = 13.

Solution

By AM ≥ GM: 4x + 4/x ≥ 2√(4 × 4) = 8.

Equality holds when 4x = 4/x, that is x = 1.

Minimum value = 8.

Solution

A sum of distances |x − p| is least when x is at the median of the points.

The points are −8, −6, 2, 11; the minimum is reached for any x between −6 and 2.

Value = 2 + 0 + 8 + 17 = 27.

Solution

|x − 4| ≤ 6 means −6 ≤ x − 4 ≤ 6.

So −2 ≤ x ≤ 10.

That range contains 13 integers.

Solution

The critical points are −3, 0 and 7; x = 0 is excluded (the denominator would be 0).

For x > 7 all three factors are positive, so the expression is positive. The sign alternates across each critical point.

Negative on (0, 7) and on (−∞, −3); zero at x = −3 and x = 7. So x ≤ −3 or 0 < x ≤ 7.

Solution

By AM ≥ GM: 9x + 25/x ≥ 2√(9 × 25) = 30.

Equality holds when 9x = 25/x, that is x = 53.

Minimum value = 30.

Solution

A sum of distances |x − p| is least when x is at the median of the points.

The points are −3, 3, 6, 7; the minimum is reached for any x between 3 and 6.

Value = 6 + 0 + 3 + 4 = 13.

Solution

The points with |x| + |y| = k number 4k for k ≥ 1, plus the origin for k = 0.

Total = 1 + 4(1 + 2 + … + 4) = 1 + 4 × 10 = 41.

Solution

|x − 9| ≤ 6 means −6 ≤ x − 9 ≤ 6.

So 3 ≤ x ≤ 15.

That range contains 13 integers.

Solution

The expression factors as (x − 2)(x − 5).

It is zero or negative for 2 ≤ x ≤ 5.

That gives 4 integers.