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प्रश्न
Solve the following inequality and write the solution set using interval notation
6x2 + 1 ≤ 5x
उत्तर
6x2 + 1 ≤ 5x
∴ 6x2 − 5x + 1 ≤ 0
∴ (3x − 1)(2x − 1) ≤ 0
∴ `3(x - 1/3).2(x - 1/2) ≤ 0`
∴ `(x - 1/3)(x - 1/2) ≤ 0` ...[Dividing both side by 6]
If `x > 1/2, x - 1/2 > 0, x - 1/3 > 0`
∴ `(x - 1/3)(x - 1/2) > 0`
∴ `x ≱ 1/2`
If `x < 1/3, x - 1/3 < 0, x - 1/2 < 0`
∴ `(x - 1/3)(x - 1/2) > 0`
∴ `x ≰ 1/3`
If `1/3 ≤ x ≤ 1/2, x - 1/2 ≥ 0, x - 1/3 ≤ 0`
∴ `(x - 1/3)(x - 1/2) ≤ 0`
∴ `1/3 ≤ x ≤ 1/2 "i.e.", x ∈ [1/3, 1/2]`
∴ the solution set is `[1/3, 1/2]`
Alternative Method:
6x2 + 1 ≤ 5x
∴ 6x2 – 5x ≤ – 1
∴ `x^2 - 5/6x ≤ - 1/6`
∴ `x^2 - 5/6x + 25/144 ≤ - 1/6 + 25/144`
∴ `(x - 5/12)^2 ≤ 1/144`
∴ `-1/12 ≤ x - 5/12 ≤ 1/12`
∴ `-1/12 + 5/12 ≤ x ≤ 1/12 + 5/12`
∴ `1/3 ≤ x ≤ 1/2`, i.e., `x ∈ [1/3, 1/2]`
∴ the solution set is `[1/3, 1/2]`
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