Question 32

-3x + 1 cannot equal 5 when |x| < 1.

Given the condition |x| < 1, the values of x must lie within the interval (-1, 1). Therefore, when substituting values of x into the expression -3x + 1, it is impossible for the result to reach 5.

A) 5

This option cannot be achieved since if we set -3x + 1 = 5, we would solve for x and find that x = -4/3, which lies outside the range of |x| < 1. Thus, 5 is not a possible value.

B) 3

Setting -3x + 1 = 3 leads to -3x = 2, or x = -2/3. This value is within the acceptable range of (-1, 1), meaning 3 can be a potential value.

C) 2

If we set -3x + 1 = 2, we find -3x = 1, which gives x = -1/3. This value is also within the interval (-1, 1), confirming that 2 can be a valid result.

E) -1

When we set -3x + 1 = -1, we solve for x and get -3x = -2, or x = 2/3. This value also lies within the range of |x| < 1, indicating that -1 is a feasible outcome.

Conclusion

The only option that cannot be the value of -3x + 1 when |x| < 1 is 5, as it requires an x value outside the allowed range. In contrast, options 3, 2, and -1 all yield valid x values within the specified limits. Thus, 5 is definitively the only impossible result.