JEE Main Maths: Linear Inequalities questions with solutions

Q1. For which values of x does the inequality log base 1/2 of ((x² + 6x + 9) / (2(x + 1))) become less than −log base 2 of (x + 1)?

1. (-1, -1 + 2√2)
2. (1 − 2√2, 2)
3. (-1, ∞)
4. None of these

Answer: None of these

Since (x^2+6x+9)=(x+3)^2 and log_(1/2)(A)=-log_2(A), the inequality becomes log_2((x+3)^2/(2(x+1)))>log_2(x+1) with x>-1. This gives (x+3)^2>2(x+1)^2 -> x^2-2x-7<0 -> x in (1-2*sqrt(2), 1+2*sqrt(2)). Intersecting x>-1 yields (-1, 1+2*sqrt(2)) approx (-1, 3.83), which matches none of the options.

Q2. Find the set of all real values of x that satisfy the inequality |9^x − 3^(x+1) − 15| < 2·9^x − 3^x.

1. (−∞, 1]
2. (1, ∞)
3. (−∞, 1]
4. None of these

Answer: (1, ∞)

Let t=3^x>0. RHS 2t^2-t must be positive (t>1/2). For 1/2<t<5.65 the inside is negative, so |t^2-3t-15|=15+3t-t^2; the inequality becomes 3t^2-4t-15>0 -> t>3 -> 3^x>3 -> x>1. Hence (1, infinity).

Q3. How many positive-integer triples (x,y,z) satisfy the system 3x+2y-z=4, 2x-y+2z=6, x+y+z<7?

1. 0
2. 1
3. 3
4. None of these

Answer: 1

From z=3x+2y-4 substituted into 2x-y+2z=6 gives 8x+3y=14. Positive integers: x=1,y=2,z=3 (x+y+z=6<7). x>=2 gives 8x>14, impossible. So exactly 1 triple.

Q4. Given that −7 < x < 18 and 9 < y < 20, what is the interval of possible values of x + y?

1. [2, 38]
2. (2, 38)
3. [2, 38)
4. (2, 38]

Answer: (2, 38)

The minimum value of x + y occurs when x is at its lowest (-7) and y at its lowest (9), resulting in -7 + 9 = 2. The maximum occurs when x is at its highest (18) and y at its highest (20), giving 18 + 20 = 38. Since x and y do not include their endpoints, the sum x + y will also not include 2 and 38, leading to the interval (2, 38).

Q5. A person's intelligence quotient is defined by IQ = (MA/CA) × 100, where MA denotes mental age and CA denotes chronological age. For a set of 12 children, if 80 ≤ IQ ≤ 140, then the possible interval for their mental age is

1. 9.8 ≤ MA ≤ 16.8
2. 10 ≤ MA ≤ 16
3. 9.6 ≤ MA ≤ 16.8
4. 9.6 ≤ MA ≤ 16.6

Answer: 9.6 ≤ MA ≤ 16.8

From IQ = (MA/CA)*100, MA = IQ*CA/100. With CA = 12: lower MA = 80*12/100 = 9.6 and upper MA = 140*12/100 = 16.8. Hence 9.6 <= MA <= 16.8.

Q6. Determine the set of x-values that satisfy all of the following inequalities at the same time: −17 < 3x + 10 ≤ −2; −22 ≤ 5x + 13 ≤ 3; and −19 ≤ 2x − 9 ≤ −3.

1. (−5, −4]
2. [−5, −4]
3. [−5, −3)
4. (−5, −3)

Answer: [−5, −4]

From -17<3x+10<=-2: -9<x<=-4. From -22<=5x+13<=3: -7<=x<=-2. From -19<=2x-9<=-3: -5<=x<=3. The common solution is [-5,-4].

Q7. Find the point of intersection of the boundary lines of the inequalities 2x + y ≥ 2 and x - y ≤ 3, which gives the vertex of their common graph.

1. (0, 0)
2. (5/3, -4/3)
3. (5/3, 4/3)
4. (-4/3, 5/3)

Answer: (5/3, -4/3)

Solving 2x+y=2 and x-y=3 simultaneously: adding gives 3x=5 (after substitution x-y=3 -> y=x-3), so 2x+(x-3)=2 -> 3x=5 -> x=5/3, y=5/3-3=-4/3. Intersection (5/3, -4/3).

Q8. Let a and b be integers. If the set of positive integer values of x satisfying both inequalities 9x - a ≥ 0 and 8x - b < 0 is exactly {1, 2, 3}, then how many ordered pairs (a, b) are possible?

1. 72
2. 75
3. 81
4. None of these

Answer: 72

x>=a/9 and x<b/8 must give exactly {1,2,3}. Include x=1 and exclude lower bound: 0<a/9<=1 so a in {1,...,9} (9 values). Include 3, exclude 4: 3<b/8<=4 so 24<b<=32, b in {25,...,32} (8 values). Total ordered pairs = 9x8 = 72.

Q9. A manufacturer has cost and revenue functions C(x) = 2x + 80 and R(x) = 5x + 20, where x denotes the number of items produced and sold. For the manufacturer to make a profit, how many items must be sold?

1. more than 20
2. more than or equal to 20
3. more than 25
4. None of these

Answer: more than 20

To determine when the manufacturer makes a profit, we set the revenue function R(x) greater than the cost function C(x). Solving the inequality 5x + 20 > 2x + 80 leads to x > 20, indicating that selling more than 20 items is necessary for profit.

Q10. A person plans to cut three pieces from a single board measuring 91 cm in total length. Let the shortest piece have length x cm. The second piece is 3 cm longer than the shortest piece, and the third piece is twice the length of the shortest piece. If the third piece is at least 5 cm longer than the second piece, which of the following gives the possible length of the shortest piece?

1. less than 8 cm
2. greater than or equal to 8 cm but less than or equal to 22 cm
3. less than 22 cm
4. greater than 22 cm

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

The correct option is right because the conditions set by the problem require that the total length of the pieces, which are defined in terms of the shortest piece x, must not exceed 91 cm. Additionally, the requirement that the third piece is at least 5 cm longer than the second piece establishes a lower limit for x, while the total length constraint establishes an upper limit, resulting in the range of x being from 8 cm to 22 cm.

Q11. Which statement is correct for the pair of inequalities 3x + 2y ≤ 6 and 6x + 4y ≥ 20?

1. The two graphs do not intersect
2. Neither graph includes the origin
3. Both graphs pass through the point (1, 1)
4. None of the above

Answer: The two graphs do not intersect

The boundary 6x+4y=20 is 3x+2y=10, parallel to 3x+2y=6, so the two boundary lines never intersect. The region 3x+2y<=6 contains the origin (0<=6), so 'neither includes origin' is false; (1,1) gives 5<=6 and 10>=20 (false), so it is not on both. Correct: the graphs do not intersect.

Q12. Determine the set of all real values of x that satisfy the inequality |(|x+3|+x)/(x+2) |>1.

1. x∈(-5,-2)∪(-1,∞)
2. x∈(5,2)∪(-1,∞)
3. x∈(-5,2)
4. x∈(-1,∞)

Answer: x∈(-5,-2)∪(-1,∞)

The correct option is right because it accurately captures the intervals where the expression inside the absolute value exceeds 1, considering the behavior of the function across the critical points derived from the inequality.

Q13. One corner point of the region enclosed by the inequalities x ≥ 0, x + 2y ≥ 0, and 2x + y ≤ 4 is

1. (1, 1)
2. (0, 1)
3. (3, 0)
4. (0, 0)

Answer: (0, 0)

Region: x>=0, x+2y>=0, 2x+y<=4. Vertices are intersections of the boundaries: x=0 & x+2y=0 -> (0,0); x=0 & 2x+y=4 -> (0,4); x+2y=0 & 2x+y=4 -> (8/3,-4/3). Among the options only (0,0) is a corner point. (3,0) gives 2x+y=6>4, outside the region.

Q14. Which pairs of consecutive even positive integers, each greater than 5, have a sum less than 23?

1. (4, 6), (6, 8), (8, 10), (10, 12)
2. (6, 8), (8, 10), (10, 12)
3. (6, 8), (8, 10), (10, 12), (12, 14)
4. (8, 10), (10, 12)

Answer: (6, 8), (8, 10), (10, 12)

The correct option includes pairs of consecutive even integers starting from 6, where each pair's sum is less than 23. The pairs (6, 8), (8, 10), and (10, 12) all meet the criteria, while the other options either include integers less than 6 or exceed the sum limit.

Q15. The region enclosed by the inequalities x + y ≤ 1 and x − y ≤ 1 lies in which quadrants?

1. I and II
2. I and III
3. II and III
4. All four quadrants

Answer: All four quadrants

The region x+y<=1 and x-y<=1 is unbounded. Sample points (0.4,0.4), (-1,1), (-1,-1), (0.5,-0.4) all satisfy both inequalities, so the region lies in all four quadrants.

Q16. If the inequality (2m + x)/3 ≤ (4mx - 1)/2 has solution set x ≥ 3/4, what is the value of the parameter m?

1. 7/10
2. 9/10
3. 9/11
4. None of these

Answer: 9/10

(2m+x)/3 <= (4mx-1)/2 -> 4m+2x <= 12mx-3 -> x(12m-2) >= 4m+3 -> x >= (4m+3)/(12m-2). Setting (4m+3)/(12m-2) = 3/4 gives 16m+12 = 36m-6, so 20m = 18 and m = 9/10.

Q17. A natural number x is selected uniformly at random from the first 100 natural numbers. The probability that the inequality x + 100/x > 50 holds is

1. 1/20
2. 11/20
3. 1/3
4. 3/20

Answer: 11/20

The inequality can be rearranged and analyzed to find the values of x that satisfy it. By determining the range of x that meets the condition, we find that 11 out of the 20 possible values fulfill the inequality, leading to a probability of 11/20.

Q18. The number of values of k for which the system of linear equations, (k + 2)x + 10y = k, kx + (k + 3)y = k − 1 has no solution, is

1. Infinitely many
2. 3
3. 1
4. 2

Answer: 1

For no solution, determinant (k+2)(k+3)-10k = k^2-5k+6 = 0 gives k=2 or k=3. At k=2 all coefficient ratios equal (consistent -> infinitely many solutions). At k=3 the coefficient ratios match (5/3) but differ from the constant ratio (3/2), giving no solution. Hence exactly 1 value of k.

Q19. The least integral value α of x such that (x−5)/(x²+5x−14) > 0, satisfies:

1. α² + 3α − 4 = 0
2. α² − 5α + 4 = 0
3. α² − 7α + 6 = 0
4. α² + 5α − 6 = 0

Answer: α² + 5α − 6 = 0

(x-5)/((x+7)(x-2)) > 0 holds on (-7,2) U (5, infinity). The least integer in this set is x=-6 (since -7 is excluded). Then alpha=-6 satisfies alpha^2+5*alpha-6 = 36-30-6 = 0, option 3, not the stored option 1.

Q20. If a R and the equation -3(x - [x])² + 2(x - [x]) + a² = 0 (where [x] denotes the greatest integer x) has no integral solution, then all possible values of a lie in the interval

1. (-2, -1)
2. (-, -2) (2,)
3. (-1, 0) (0, 1)
4. (1, 2) 2

Answer: (-1, 0) (0, 1)

The equation has no integral solutions when the quadratic expression in terms of the fractional part of x, which is bounded between 0 and 1, does not yield any real roots. The values of a must be such that the discriminant of the quadratic is negative, which occurs in the intervals (-1, 0) and (0, 1), ensuring that the equation cannot equal zero for any integer value.