JEE Mathematics Linear Inequalities MCQs Set 05

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

Explanation:
1. Case \( x \ge -3 \): the numerator is \( 2x + 3 \), so \( (2x+3)/(x+2) \gt 1 \), which gives \( (x+1)/(x+2) \gt 0 \), so \( x \lt -2 \) or \( x \gt -1 \). Combined with \( x \ge -3 \): [-3,-2) or (-1,inf).
2. Case \( x \lt -3 \): the numerator is \( -3 \), so \( -3/(x+2) \gt 1 \). Since \( x + 2 \lt 0 \), multiply and flip: \( -3 \lt x + 2 \)... giving \( x \gt -5 \) with \( x \lt -2 \), so (-5,-3).
3. Union of the two cases: \( (-5,-2) U \) (-1,inf).

Teacher's Note:
1. For absolute value inequalities, split into cases at the points where each modulus expression becomes zero.
2. Check the end points separately to decide between round and square brackets.

Answer: (c) \( (-\infty,2] \)

Explanation:
1. Left side: \( 37 - 3x - 5 = 32 - 3x \).
2. Right side: \( 9x - 8x + 24 = x + 24 \).
3. \( 32 - 3x \ge x + 24 \) gives \( 8 \ge 4x \), so \( x \le 2 \).

Teacher's Note:
1. Rewrite the inequality step by step, remembering that multiplying or dividing by a negative number reverses the sign.
2. Write the final answer as an interval and check one test value inside and outside it.

Answer: (b) \( \frac32 \)

Explanation:
1. Let \( [x] = n \) and \( {x} = f \), so \( x = n + f \).
2. First equation: \( 5f = 2n + f \), so \( f = n/2 \).
3. Second equation: \( n - f = 1/2 \), so \( n/2 = 1/2 \), giving \( n = 1 \) and \( f = 1/2 \).
4. \( x = 1 + 1/2 = 3/2 \).

Teacher's Note:
1. Always write \( x \) as integral part plus fractional part, with \( 0 \le f \lt 1 \).
2. Check: \( 5(1/2) = 5/2 = 3/2 + 1 \).

Answer: (b) 35

Explanation:
1. Let the third score be \( t \). Then \( (70 + 75 + t)/3 \ge 60 \).
2. \( 145 + t \ge 180 \), so \( t \ge 35 \).

Teacher's Note:
1. Average problems turn into a linear inequality in the unknown score.
2. The minimum is the boundary value 35.

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

Explanation:
1. Let \( (x) = m \). Then \( (x + 1) = m + 1 \), since adding 1 shifts the ceiling by 1.
2. \( m^2 + (m+1)^2 = 25 \) gives \( m^2 + m - 12 = 0 \), so \( m = 3 \) or \( m = -4 \).
3. \( (x) = 3 \) means \( 2 \lt x \le 3 \), and \( (x) = -4 \) means \( -5 \lt x \le -4 \).
4. Solution set: (-5,-4] U (2,3].

Teacher's Note:
1. The least integer greater than or equal to \( x \) is the ceiling function.
2. \( (x) = m \) exactly when \( m - 1 \lt x \le m \).

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

Explanation:
1. Let the shortest be \( x \). Lengths are \( x, x + 3 \) and \( 2x \).
2. Total cannot exceed 91: \( 4x + 3 \le 91 \), so \( x \le 22 \).
3. Third at least 5 more than second: \( 2x \ge x + 3 + 5 \), so \( x \ge 8 \).
4. So \( 8 \le x \le 22 \).

Teacher's Note:
1. Set up one inequality per condition and intersect the results.
2. Both end points are allowed because the conditions say at most and at least.

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

Explanation:
1. \( (p-a)(p-b)(p-c) = (b+c)(c+a)(a+b) \). By AM-GM, this is at most \( ((2p)/3)^3 = 8p^3/27 \), so \( (a) \) holds.
2. Also \( b + c \ge 2 \sqrt{bc} \) etc., so the product is at least 8abc, so \( (b) \) holds.
3. For \( (c) \): bc/a + ca/b + ab/c >= \( a + b + c = p \), so the claimed inequality with <= is false in general (\( a = b = c \) gives equality only).

Teacher's Note:
1. Check each option with \( a = b = c \), then with unequal values.
2. The sum bc/a + ca/b + ab/c is always at least \( a + b + c \).

Answer: (c) \( (4,\infty) \)

Explanation:
1. First: \( x + 7 \lt 2x + 3 \) gives \( x \gt 4 \).
2. Second: \( 2x + 4 \lt 5x + 3 \) gives \( 1 \lt 3x \), so \( x \gt 1/3 \).
3. Both hold only for \( x \gt 4 \).

Teacher's Note:
1. Rewrite the inequality step by step, remembering that multiplying or dividing by a negative number reverses the sign.
2. Write the final answer as an interval and check one test value inside and outside it.

Answer: (a) \( [1,\infty) \)

Explanation:
1. Multiply by 4: \( 2(3x - 4) \ge x + 1 - 4 \).
2. \( 6x - 8 \ge x - 3 \), so \( 5x \ge 5 \), so \( x \ge 1 \).

Teacher's Note:
1. Rewrite the inequality step by step, remembering that multiplying or dividing by a negative number reverses the sign.
2. Write the final answer as an interval and check one test value inside and outside it.

Answer: (a) \( \frac32 \)

Explanation:
1. Since \( 4x^2 + 9 \gt 0 \), the inequality becomes \( 12|x| \ge 4x^2 + 9 \).
2. Rearrange: \( 4x^2 - 12|x| + 9 \le 0 \), i.e. \( (2|x| - 3)^2 \le 0 \).
3. A square is at most 0 only when it equals 0, so \( |x| = 3/2 \).

Teacher's Note:
1. Look for a perfect square when a quadratic inequality has discriminant zero.
2. Here the discriminant 144 - 144 is 0.

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

Explanation:
1. Let \( y \) litres of 30% solution be added. Acid = \( 0.12(600) + 0.30y = 72 + 0.3y \); volume = \( 600 + y \).
2. Greater than 15%: \( 72 + 0.3y \gt 90 + 0.15y \) gives \( y \gt 120 \).
3. Less than 18%: \( 72 + 0.3y \lt 108 + 0.18y \) gives \( y \lt 300 \).
4. So \( 120 \lt y \lt 300 \).

Teacher's Note:
1. Mixture problems: acid amount divided by total volume gives the concentration.
2. Write one inequality for each bound and combine.

Answer: (b) \( x\ge2 \)

Explanation:
1. Both sides are non-negative, so square: \( (x-1)^2 \ge (x-3)^2 \).
2. \( x^2 - 2x + 1 \ge x^2 - 6x + 9 \) gives \( 4x \ge 8 \), so \( x \ge 2 \).

Teacher's Note:
1. The points equidistant from 1 and 3 is \( x = 2 \); \( |x-1| \ge |x-3| \) means \( x \) is closer to 3.
2. Squaring is safe when both sides are non-negative.

Answer: (c) non-positive

Explanation:
1. Known inequality (\( a \) form of Schur's inequality): \( (b+c-a)(c+a-b)(a+b-c) \) <= abc for positive a, \( b, c \).
2. Equality holds only when \( a = b = c \).
3. So the expression is at most 0, i.e. non-positive (it is in fact negative for distinct numbers, but non-positive is the general statement).

Teacher's Note:
1. Test with \( a = 1, b = 2, c = 2 \): \( (3)(1)(1) - 4 = -1 \).
2. Option \( (c) \) is the safest correct choice.

Answer: (c) 7

Explanation:
1. \( 27/64 = (3/4)^3 \).
2. Base 3/4 is less than 1, so the inequality reverses: \( 6x + 10 - x^2 \gt 3 \).
3. \( x^2 - 6x - 7 \lt 0 \) gives \( (x-7)(x+1) \lt 0 \), so \( -1 \lt x \lt 7 \).
4. Integers: 0, 1, 2, 3, 4, 5, 6, which is 7 values.

Teacher's Note:
1. When the base is between 0 and 1, comparing exponents reverses the inequality.
2. Exclude both end points because the inequality is strict.

Answer: (a) Only I and II are true.

Explanation:
1. I: \( -75 \lt 3(x-2) \le 0 \) gives \( -25 \lt x - 2 \le 0 \), so \( -23 \lt x \le 2 \). True.
2. II: \( 14 \le 3x + 11 \le 22 \) gives \( 3 \le 3x \le 11 \), so \( 1 \le x \le 11/3 \). True.
3. III: \( -20 \le 2 - 3x \le 36 \) gives \( -34 \le -3x \)... i.e. \( -34/3 \le x \le 22/3, a \) single interval, not \( [-1,1] U [3,5] \). False.

Teacher's Note:
1. Solve each compound inequality fully before comparing.
2. A compound inequality of this form always gives a single interval.

Answer: (d) \( (0,\log_2(4/3))\cup(1,\infty) \)

Explanation:
1. Let \( u = 2^x \gt 0 \). Then \( 1/(u-1) \gt 1/(1-u/2) = 2/(2-u) \).
2. Denominators vanish at \( u = 1 \) and \( u = 2 \), so \( x = 0 \) and \( x = 1 \) are excluded.
3. Bring to one side and simplify: \( 1/(u-1) - 2/(2-u) \gt 0 \) gives \( (2 - u - 2u + 2)/((u-1)(2-u)) = (4 - 3u)/((u-1)(2-u)) \gt 0 \).
4. Sign chart in \( u \): positive for \( 1 \lt u \lt 4/3 \) and for \( u \gt 2 \).
5. So \( 0 \lt x \lt \log2(4/3) \) or \( x \gt 1 \).

Teacher's Note:
1. Substituting \( u = 2^x \) turns the problem into a rational inequality in \( u \).
2. Use a sign chart with critical points \( 1, 4/3 \) and 2.

Answer: (a) \( [-1,3] \)

Explanation:
1. \( |A| \le 1 \) gives \( -1 \le |x-1| - 1 \le 1 \), so \( 0 \le |x-1| \le 2 \).
2. \( |x-1| \le 2 \) gives \( -1 \le x \le 3 \).

Teacher's Note:
1. For absolute value inequalities, split into cases at the points where each modulus expression becomes zero.
2. Check the end points separately to decide between round and square brackets.

Answer: (a) \( \frac{a}{1+a}+\frac{b}{1+b}\ge\frac{c}{1+c} \)

Explanation:
1. \( a/(1+a) + b/(1+b) \ge a/(1+a+b) + b/(1+a+b) = (a+b)/(1+a+b) \), since each denominator decreases.
2. The function \( t/(1+t) \) is increasing, and \( a + b \ge c \), so \( (a+b)/(1+a+b) \ge c/(1+c) \).
3. Therefore the sum is at least \( c/(1+c) \). Equality needs both \( a + b = c \) and \( a = b = 0 \) type cases, so only the non-strict form is guaranteed.

Teacher's Note:
1. The function \( t/(1+t) \) is increasing and sub-additive.
2. Equality in the stated direction cannot always be strict, so choose the non-strict option.

Answer: (b) 7

Explanation:
1. The inequality becomes \( \log5(5(x^2+1)) \le \log5(ax^2 + 4x + a) \), so \( 5x^2 + 5 \le ax^2 + 4x + a \) for all \( x \).
2. \( (a - 5)x^2 + 4x + (a - 5) \ge 0 \) for all \( x \) needs \( a - 5 \gt 0 \) and discriminant \( 16 - 4(a-5)^2 \le 0 \).
3. \( (a - 5)^2 \ge 4 \) gives \( a - 5 \ge 2 \), so \( a \ge 7 \).
4. The least integer is 7.

Teacher's Note:
1. For a quadratic to stay non-negative for all \( x \): leading coefficient positive and discriminant at most 0.
2. The \( \log \) is increasing, so the inequality passes to the arguments.

Answer: (c) \( (5,\infty) \)

Explanation:
1. First: \( 3x - 7 \gt 2x - 12 \) gives \( x \gt -5 \).
2. Second: \( 6 - x \gt 11 - 2x \) gives \( x \gt 5 \).
3. Both together: \( x \gt 5 \).

Teacher's Note:
1. Rewrite the inequality step by step, remembering that multiplying or dividing by a negative number reverses the sign.
2. Write the final answer as an interval and check one test value inside and outside it.

Answer: (c) 1

Explanation:
1. Write the equation as \( 4^{x/4} + 9^{x/4} = 13^{x/4} \); let \( t = x/4 \).
2. Divide by \( 13^t \): \( (4/13)^t + (9/13)^t = 1 \).
3. The left side is a strictly decreasing function of \( t \) (both bases are less than 1), so it equals 1 for exactly one value of \( t \).
4. So there is exactly one value of \( x \).

Teacher's Note:
1. Strictly monotonic functions take each value at most once.
2. Check: \( t \) about 0.64... \( x \) about 2.57 is the single solution; there is no need to find it exactly.

Answer: (b) 1

Explanation:
1. Domain: \( 0 \le x \le 1 \) and \( x \ge \sqrt{1-x} \), which requires \( x^2 + x - 1 \ge 0 \), so \( x \ge 0.618 \).
2. On this domain the left side is increasing in \( x \) (both square roots grow), going from about 0.79 at \( x = 0.618 \) to \( 1 + 1 = 2 \) at \( x = 1 \).
3. The value 1 is attained once, near \( x = 0.64 \), so there is exactly one real root.

Teacher's Note:
1. Find the domain first; many root equations have a very restricted domain.
2. Monotonic left side means at most one root.

Answer: (c) 5

Explanation:
1. Let \( u = |x| \). Then \( |2 - |1 - u|| = 1 \) gives \( |1 - u| = 1 \) or \( |1 - u| = 3 \).
2. \( |1 - u| = 1 \) gives \( u = 0 \) or \( u = 2 \); \( |1 - u| = 3 \) gives u = 4 (u = -2 is rejected).
3. So \( |x| = 0, 2, 4 \) and \( x = 0, 2, -2, 4, -4 \), which is 5 roots.

Teacher's Note:
1. For absolute value inequalities, split into cases at the points where each modulus expression becomes zero.
2. Check the end points separately to decide between round and square brackets.

Answer: (b) x = 2

Explanation:
1. \( 3^x - 8 = 3^{2-x} = 9/3^x \).
2. Let \( u = 3^x \): \( u^2 - 8u - 9 = 0 \), so \( (u - 9)(u + 1) = 0 \).
3. u = 9 (u = -1 is impossible), so \( x = 2 \). Check: \( 9 - 8 = 1 \) and \( 3^0 = 1 \).

Teacher's Note:
1. Convert the logarithmic equation to exponential form first.
2. A substitution \( u = 3^x \) gives a quadratic.