JEE Mathematics Linear Inequalities MCQs Set 02

Answer: (c) 1

Explanation:
1. Let t = log₁₆ x, which takes every real value as x ranges over positive reals, so t² − t + log₁₆ k = 0 must have exactly one root.
2. Discriminant 1 − 4 log₁₆ k = 0 gives log₁₆ k = 1/4, so k = 16^1/4 = 2: a single value of k.

Teacher's Note:
1. One solution in x corresponds to one (repeated) root in t.

Answer: (a) \( x\in(-5,-2)\cup(-1,\infty) \)

Explanation:
1. For x ≥ −3: (2x + 3)/(x + 2) > 1 gives (x + 1)/(x + 2) > 0, so x > −1 or x < −2; with x ≥ −3 this is −3 ≤ x < −2 or x > −1.
2. For x < −3: −3/(x + 2) > 1 with x + 2 < 0 gives −3 < x + 2, i.e. −5 < x < −3.
3. Combining: (−5, −2) ∪ (−1, ∞).

Teacher's Note:
1. Split into the cases x ≥ −3 and x < −3.

Answer: (c) (−∞, 2]

Explanation:
1. The left side is 32 − 3x and the right side is x + 24.
2. 32 − 3x ≥ x + 24 gives 8 ≥ 4x, so x ≤ 2: the set (−∞, 2].

Teacher's Note:
1. Expand both sides and collect x on one side.

Answer: (b) 8

Explanation:
1. For x ≥ −7: 2x − 1 = x + 7 gives x = 8, which satisfies x ≥ −7 (check 15 = 15).
2. For x < −7: 2x − 1 = −x − 7 gives x = −2, which is not less than −7, so it is rejected (it gives 2x − 1 = −5 ≠ 5).

Teacher's Note:
1. Always check each case against its own domain.

Answer: (b) 3/2

Explanation:
1. Write x = n + f. Then 5f = 2n + f gives f = n/2.
2. n − f = 1/2 gives n/2 = 1/2, so n = 1 and f = 1/2, hence x = 3/2.

Teacher's Note:
1. Substitute x = [x] + {x} in the first equation.

Answer: (b) 35

Explanation:
1. (70 + 75 + m)/3 ≥ 60 gives 145 + m ≥ 180.
2. So m ≥ 35; the minimum is 35.

Teacher's Note:
1. Average of three tests ≥ 60 means total ≥ 180.

Answer: (a) is satisfied if |d − R| ≤ r

Explanation:
1. Since R ≥ r and d > 0, d² + R² − r² > 0, so the left inequality always holds.
2. The right inequality d² + R² − r² ≤ 2dR is (d − R)² ≤ r², i.e. |d − R| ≤ r.

Teacher's Note:
1. Rearrange the right inequality into a perfect square.

Answer: (c) breadth ≥ 20

Explanation:
1. With breadth b, the perimeter is 2(3b + b) = 8b ≥ 160.
2. So b ≥ 20.

Teacher's Note:
1. 'Minimum perimeter' gives a lower bound on 8b.

Answer: (b) \( (-5,-4]\cup(2,3] \)

Explanation:
1. Let (x) = c. Then c² + (c + 1)² = 25 gives c² + c − 12 = 0, so c = 3 or c = −4.
2. (x) = 3 means 2 < x ≤ 3 and (x) = −4 means −5 < x ≤ −4.

Teacher's Note:
1. Solve for the integer c first, then convert back to intervals.

Answer: (b) greater than or equal to 8 cm but less than or equal to 22 cm

Explanation:
1. Let the shortest be x; then the others are x + 3 and 2x.
2. x + (x + 3) + 2x ≤ 91 gives x ≤ 22. 2x ≥ (x + 3) + 5 gives x ≥ 8.
3. So 8 ≤ x ≤ 22.

Teacher's Note:
1. Form one inequality for the total length and another for the 5 cm condition.

Answer: (a) 2

Explanation:
1. Multiply by 60: 20(2x − 1) ≥ 15(3x − 2) − 12(2 − x), i.e. 40x − 20 ≥ 45x − 30 − 24 + 12x.
2. 40x − 20 ≥ 57x − 54 gives 34 ≥ 17x, so x ≤ 2 and a = 2.

Teacher's Note:
1. Clear the denominators with the LCM 60.

Answer: (c) \( \frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\le p \)

Explanation:
1. (a): the three numbers p − a, p − b, p − c add to 2p, so by AM–GM their product is at most (2p/3)³ = 8p³/27, which holds.
2. (b): (b + c)(c + a)(a + b) ≥ 2√(bc) · 2√(ca) · 2√(ab) = 8abc, which holds.
3. (c): bc/a + ca/b + ab/c ≥ a + b + c = p, so the reverse inequality fails in general (equality only if a = b = c).

Teacher's Note:
1. Compare each statement with AM–GM or the rearrangement-type inequality.

Answer: (c) (4, ∞)

Explanation:
1. x + 7 < 2x + 3 gives x > 4.
2. 2x + 4 < 5x + 3 gives x > 1/3.
3. Both hold together for x > 4.

Teacher's Note:
1. Solve each inequality and take the intersection.

Answer: (d) (0, ∞)

Explanation:
1. The right side is 5^−2/x, so we need x + 2 > −2/x (x ≠ 0).
2. For x > 0: x² + 2x + 2 > 0, which is always true. For x < 0: x² + 2x + 2 < 0, which never holds (discriminant < 0).
3. The solution set is (0, ∞).

Teacher's Note:
1. Write 1/25 as 5⁻² and treat x > 0 and x < 0 separately.

Answer: (a) \( \left(\frac92,\infty\right) \)

Explanation:
1. x ≥ 1: (x + 2) − (x − 1) = 3 < x − 3/2 gives x > 9/2.
2. −2 ≤ x < 1: (x + 2) + (x − 1) = 2x + 1 < x − 3/2 gives x < −5/2, which has no solution in this range.
3. x < −2: −3 < x − 3/2 gives x > −3/2, which is impossible here.
4. So the solution set is (9/2, ∞).

Teacher's Note:
1. Break the real line at x = −2 and x = 1.

Answer: (a) [1, ∞)

Explanation:
1. Multiply by 4: 2(3x − 4) ≥ x + 1 − 4, i.e. 6x − 8 ≥ x − 3.
2. 5x ≥ 5, so x ≥ 1.

Teacher's Note:
1. Multiply through by 4 to clear fractions.

Answer: (b) (−2, 3)

Explanation:
1. The domain is x > −2. Since log_1/3(x + 2) = −log₃(x + 2), the left side is log₃(x + 4).
2. The right side is (1/2) · 2 log₃ 7 = log₃ 7.
3. So x + 4 < 7, i.e. x < 3, and the solution set is (−2, 3).

Teacher's Note:
1. Convert the base 1/3 and base √3 logarithms to base 3.

Answer: (d) P² ≥ 16A

Explanation:
1. For sides a and b, P = 2(a + b) and A = ab.
2. By AM–GM, (a + b)² ≥ 4ab, so P² = 4(a + b)² ≥ 16ab = 16A.

Teacher's Note:
1. Use (a + b)² ≥ 4ab.

Answer: (a) \( \frac32 \)

Explanation:
1. 4x² + 9 ≥ 2 · 2|x| · 3 = 12|x| by AM–GM, so |12x/(4x² + 9)| ≤ 1 always.
2. Equality (needed for ≥ 1) holds only when 2|x| = 3, i.e. |x| = 3/2.

Teacher's Note:
1. Compare 4x² + 9 with 12|x| using AM–GM.

Answer: (a) more than 120 litres but less than 300 litres

Explanation:
1. With x litres added, acid = 72 + 0.3x in 600 + x litres.
2. 72 + 0.3x > 0.15(600 + x) gives x > 120.
3. 72 + 0.3x < 0.18(600 + x) gives x < 300.

Teacher's Note:
1. Write the two percentage conditions as inequalities in x.

Answer: (c) 2

Explanation:
1. For x ≥ 3 the equation is x + 4 = 4, so x = 0 (rejected). For 1 ≤ x < 3 it is (x − 1) + (x + 2) + (x − 3) = 3x − 2 = 4, so x = 2 (valid).
2. For −2 ≤ x < 1: (1 − x) + (x + 2) + (x − 3) = x = 4 is rejected. For x < −2: (1 − x) − (x + 2) + (x − 3) = −x − 4 = 4 gives x = −8 (valid).
3. The integral roots are x = 2 and x = −8, so there are 2.

Teacher's Note:
1. Take the intervals determined by −2, 1 and 3.

Answer: (b) x ≥ 2

Explanation:
1. Squaring: (x − 1)² ≥ (x − 3)², i.e. −2x + 1 ≥ −6x + 9.
2. 4x ≥ 8, so x ≥ 2.

Teacher's Note:
1. Square both sides since both are non-negative.

Answer: (b) negative

Explanation:
1. The known inequality (b + c − a)(c + a − b)(a + b − c) ≤ abc holds for positive reals, with equality only when a = b = c.
2. Since a, b, c are distinct, the inequality is strict, so the expression is negative.

Teacher's Note:
1. Compare with the triangle-type inequality; equality needs a = b = c.

Answer: (c) [−2, 0] ∪ [2, 4]

Explanation:
1. |x − 1| ≤ 3 gives −2 ≤ x ≤ 4. |x − 1| ≥ 1 gives x ≤ 0 or x ≥ 2.
2. The intersection is [−2, 0] ∪ [2, 4].

Teacher's Note:
1. Intersect the two solution sets.

Answer: (b) \( (-7,-4)\cup(-4,1) \)

Explanation:
1. |x + 4| > 0 for x ≠ −4, so the sign is decided by x² + 6x − 7 = (x + 7)(x − 1) < 0, i.e. −7 < x < 1.
2. The point x = −4 makes the denominator zero and must be excluded: (−7, −4) ∪ (−4, 1).

Teacher's Note:
1. Do not forget to remove the point where the denominator vanishes.