Previous year questions: Linear Inequalities

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Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. The stems and options were checked against the official PDF pages and every final value against the official key. The JEE questions test the same skill as this chapter: turn a condition into an inequality, solve it, and read the solution set carefully.

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

The greatest integer function is written with square brackets. If the domain of f(x) equal to cos inverse of (4x plus 2 times bracket x) by 3 is the interval [alpha, beta], find 12(alpha plus beta).

Answer: The inverse cosine is defined only when its argument lies between minus 1 and 1. So minus 3 is at most 4x plus 2 bracket x is at most 3. Write x as n plus t where n is the integer part and t lies in [0, 1). Then the expression becomes 6n plus 4t. For n equal to 0 the expression runs from 0 up to 4, and the upper bound 3 cuts it at t equal to 3 by 4, giving the piece [0, 3 by 4]. For n equal to minus 1 the expression runs from minus 6 up to minus 2, and the lower bound minus 3 cuts it at t equal to 3 by 4, giving the piece [minus 1 by 4, 0). Other integers push the expression outside the band. The domain is [minus 1 by 4, 3 by 4], so alpha plus beta is 1 by 2 and 12 times that is 6. The official key is 6.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

If the set of all solutions of the equation |x squared plus x minus 9| equal to |x| plus |x squared minus 9| is [alpha, beta] union [gamma, infinity), find alpha squared plus beta squared plus gamma squared.

Answer: Use the triangle equality rule: |a plus b| equals |a| plus |b| exactly when the product ab is non-negative. Here a is x and b is x squared minus 9, so the condition is x times (x squared minus 9) at least 0, that is x(x minus 3)(x plus 3) at least 0. Mark the roots minus 3, 0 and 3 on a number line. The sign of the product is positive on the interval from minus 3 to 0 and again on the interval from 3 upward, because the number of negative factors is even there. So the solution is [minus 3, 0] union [3, infinity). Matching with the given form, alpha is minus 3, beta is 0 and gamma is 3. Their squares add to 9 plus 0 plus 9, that is 18, which is the official key.

Q3 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

The quadratic equation (lambda plus 2)x squared minus 3 lambda x plus 4 lambda equal to 0, with lambda not equal to minus 2, has two positive roots. Find the number of possible integral values of lambda.

Answer: Apply the three conditions for two positive roots. First, the roots must be real, so the discriminant is positive: 9 lambda squared minus 16 lambda(lambda plus 2) is positive, which simplifies to minus 7 lambda squared minus 32 lambda positive, that is lambda(7 lambda plus 32) negative. This gives lambda between minus 32 by 7 and 0, approximately minus 4.57 and 0.

Second, the sum of the roots is 3 lambda by (lambda plus 2), which must be positive. Third, the product is 4 lambda by (lambda plus 2), which must also be positive. Both are positive exactly when lambda by (lambda plus 2) is positive, that is lambda greater than 0 or lambda less than minus 2. Combining with the discriminant interval leaves lambda between minus 32 by 7 and minus 2. The integers inside are minus 4 and minus 3, so there are two integral values. The official key is 2.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

If the domain of f(x) equal to sine inverse of (x plus bracket x) by 3, with bracket standing for the greatest integer function, is [alpha, beta), find alpha squared plus beta squared.

Answer: The sine inverse needs its argument between minus 1 and 1, so minus 3 is at most x plus bracket x is at most 3. Write x equal to n plus t with t in [0, 1). The expression becomes 2n plus t. For n equal to 0, t runs from 0 up to 1, which stays inside the band, giving the piece [0, 1). For n equal to 1, the expression runs from 2 up to 3, which still fits, giving [1, 2). For n equal to 2 the expression starts at 4 and fails. For n equal to minus 1 the expression runs from minus 2 up to minus 1, which fits, giving [minus 1, 0). For n equal to minus 2 the expression ends below minus 3 and fails. The pieces join into the domain [minus 1, 2), so alpha is minus 1 and beta is 2. Then alpha squared plus beta squared is 1 plus 4, that is 5, matching the official key.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

The limit as x tends to 2 of the quotient, with numerator (tan(x minus 2)) times (rx squared plus (p minus 2)x minus 2p) and denominator (x minus 2) squared, equals 5 for some real r and p. Let the set of all possible values of q for which the roots of rx squared minus px plus q equal to 0 lie in the open interval (0, 2) be (alpha, beta]. Find 4(alpha plus beta).

Answer: Since tan(x minus 2) behaves like (x minus 2) near 2, one power cancels and the quotient tends to the limit of (rx squared plus (p minus 2)x minus 2p) divided by (x minus 2). For that limit to be finite the numerator must vanish at x equal to 2: 4r plus 2p minus 4 minus 2p is zero, so r equals 1. Then the limit is the derivative of x squared plus (p minus 2)x minus 2p at 2, which is 4 plus p minus 2, that is p plus 2. Setting this equal to 5 gives p equal to 3.

Now the equation is x squared minus 3x plus q equal to 0 with both roots in (0, 2). The sum of the roots is 3, which is already inside the allowed total. For both roots positive the product q must be positive. For both roots below 2, the quantity q minus 6 plus 4 must be positive, that is q greater than 2. Reality of the roots needs the discriminant 9 minus 4q to be non-negative, so q is at most 9 by 4. Hence q lies in (2, 9 by 4], alpha is 2 and beta is 9 by 4. Then 4 times the sum is 4 times 17 by 4, that is 17, matching the official key.

Q6 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The numbers a, b and c are chosen from the set 1, 2, 3 and 4. The probability that ax squared plus 2 root 2 b x plus c is positive for every real x is m by n in lowest terms. Find m plus n.

Answer: A quadratic with positive leading coefficient is positive for all real x exactly when its discriminant is negative. Here the leading coefficient a is at least 1, and the discriminant is 8b squared minus 4ac, so the condition is 2b squared less than ac. Count the ordered triples from 1, 2, 3 and 4. There are 64 in all. When b is 1 the condition is ac greater than 2, and only the three pairs (1, 1), (1, 2) and (2, 1) fail, leaving 13 pairs. When b is 2 the condition is ac greater than 8, and the pairs (3, 3), (3, 4), (4, 3) and (4, 4) work, giving 4. When b is 3 or 4 the product ac cannot exceed 16, so no triple works. The favourable count is 17 and the probability is 17 by 64. Then m plus n is 81, which is the official answer.

Q7 (NCERT Exemplar, 3 marks)

Solve the system 4x plus 3 at least 2x plus 17 and 3x minus 5 less than minus 2.

Answer: Solve each inequality separately. The first gives 2x at least 14, so x at least 7. The second gives 3x less than 3, so x less than 1. A value cannot be at least 7 and less than 1 at the same time, so the system has no solution and the solution set is empty. Testing x equal to 7 fails the second inequality and testing x equal to 0 fails the first, which confirms the empty intersection. Source: NCERT Exemplar Class 11 Mathematics, Chapter 6, Exercise 6.3, Short Answer Type, Q6.

Q8 (NCERT Exemplar, 3 marks)

A company has cost C(x) equal to 26,000 plus 30x and revenue R(x) equal to 43x for x cassettes made and sold in a week. How many cassettes must be sold to make a profit?

Answer: Profit means revenue exceeds cost, so solve 43x greater than 26,000 plus 30x. Subtracting 30x from both sides gives 13x greater than 26,000, so x is greater than 2,000. Since the number of cassettes is a whole number, the company must sell at least 2,001 cassettes in the week. The strict inequality is important: at exactly 2,000 cassettes the profit is zero, so that value is not enough. Source: NCERT Exemplar Class 11 Mathematics, Chapter 6, Exercise 6.3, Short Answer Type, Q7.

Q9 (NCERT Exemplar, 3 marks)

Water in a pool is normal when the average of three daily pH readings lies between 8.2 and 8.5. The first two readings are 8.48 and 8.35. Find the range of the third reading.

Answer: Let the third reading be x. The average condition is 8.2 less than (8.48 plus 8.35 plus x) by 3 less than 8.5. Multiply all parts by 3: 24.6 is less than 16.83 plus x, and 16.83 plus x is less than 25.5. The first gives x greater than 7.77, and the second gives x less than 8.67. Both bounds are strict because the word between is used without inclusive language, so the third reading must lie in (7.77, 8.67). Source: NCERT Exemplar Class 11 Mathematics, Chapter 6, Exercise 6.3, Short Answer Type, Q8.

Q10 (NCERT Exemplar, 3 marks)

A solution is kept between 40 and 45 degrees Celsius. Using F equal to 9C by 5 plus 32, find the range of the temperature in Fahrenheit.

Answer: Substitute the endpoints of the Celsius range into the linear conversion. At C equal to 40 the Fahrenheit value is 9 times 40 by 5 plus 32, that is 72 plus 32 equal to 104. At C equal to 45 the value is 81 plus 32 equal to 113. Because the conversion is increasing, the whole interval maps in order, so the temperature lies between 104 and 113 degrees Fahrenheit. The endpoints are included exactly when the Celsius endpoints are included. Source: NCERT Exemplar Class 11 Mathematics, Chapter 6, Exercise 6.3, Short Answer Type, Q10.