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Zorluk: ZorLinear Inequalities and Absolute Value

If xx is a real number such that 43x13|4 - 3x| \leq 13, and yy is an integer such that 5<12y33-5 < \frac{1 - 2y}{3} \leq 3, what is the least possible integer value of x2yx^2 - y?

  1. A
    8-8
  2. 7-7Cevap
  3. C
    4-4
  4. D
    22
  5. E
    99

Cevap

The least possible integer value of x2yx^2 - y is 7-7.
To find the minimum value of x2yx^2 - y, we must minimize x2x^2 and maximize yy. The absolute value inequality 43x13|4 - 3x| \leq 13 simplifies to 3x173-3 \leq x \leq \frac{17}{3}, which contains 00, so the minimum of x2x^2 is 00. The inequality 5<12y33-5 < \frac{1 - 2y}{3} \leq 3 simplifies to 4y<8-4 \leq y < 8. Since yy is an integer, its maximum value is 77. Therefore, the minimum possible value of x2yx^2 - y is 07=70 - 7 = -7.

Adım Adım Çözüm

1
Solve the absolute value inequality 43x13|4 - 3x| \leq 13 for xx.
1343x13    173x9    173x3    3x173-13 \leq 4 - 3x \leq 13 \implies -17 \leq -3x \leq 9 \implies \frac{17}{3} \geq x \geq -3 \implies -3 \leq x \leq \frac{17}{3}.
An absolute value inequality uk|u| \leq k expands to kuk-k \leq u \leq k. Dividing by a negative number flips the inequality signs.
2
Determine the minimum possible value of x2x^2.
Minimum x2=0x^2 = 0.
Since xx can take any real value in the interval [3,173][-3, \frac{17}{3}], which contains 00, the minimum square of any real number in this interval is 00 at x=0x = 0.
3
Solve the double inequality 5<12y33-5 < \frac{1 - 2y}{3} \leq 3 for yy.
15<12y9    16<2y8    8>y4    4y<8-15 < 1 - 2y \leq 9 \implies -16 < -2y \leq 8 \implies 8 > y \geq -4 \implies -4 \leq y < 8.
Multiplying by 33 preserves inequalities, subtracting 11 preserves inequalities, and dividing by 2-2 reverses all inequality signs.
4
Find the maximum integer value of yy.
Maximum integer y=7y = 7.
The solution set for yy is the half-open interval [4,8)[-4, 8). Since yy is constrained to be an integer, the largest integer strictly less than 88 is 77.
5
Minimize the expression x2yx^2 - y.
Minimum (x2y)=Minimum (x2)Maximum (y)=07=7\text{Minimum } (x^2 - y) = \text{Minimum } (x^2) - \text{Maximum } (y) = 0 - 7 = -7.
To minimize a difference ABA - B, one must minimize the minuend AA and maximize the subtrahend BB.

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Linear Inequalities and Absolute Value
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