11th Maths Important Questions 2026–2027 PDF | First Term, Quarterly, Half-Yearly & Public Exam | Kalvi Mini
This page covers Important 2-Mark, 3-Mark, and 5-Mark Questions and Answers for the First Midterm Exam 2026 covering Sets, Relations & Functions, Inequalities, Polynomials, Partial Fractions, Surds, and Trigonometric Identities — along with the Kalvi Mini Model Question Paper and Answer Key.
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| 11th Maths Important Questions 2026–2027 PDF | First Term, Quarterly, Half-Yearly & Public Exam | Kalvi Mini |
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Sets, Relations, Functions & Inequalities
Question No. 1: If n(A) = 10 and n(A∩B) = 3, find n((A∩B)′ ∩ A).
Solution: (A∩B)′ ∩ A = (A′ ∪ B′) ∩ A = (A′ ∩ A) ∪ (B′ ∩ A) = ∅ ∪ (B′ ∩ A) = (B′ ∩ A) = A − B.
So n((A∩B)′ ∩ A) = n(A − B) = n(A) − n(A∩B) = 7.
So n((A∩B)′ ∩ A) = n(A − B) = n(A) − n(A∩B) = 7.
Question No. 2: If A = {1,2,3,4} and B = {3,4,5,6}, find n((A∪B) × (A∩B) × (AΔB)).
Solution: We have n(A∪B) = 6, n(A∩B) = 2 and n(AΔB) = 4.
So, n((A∪B) × (A∩B) × (AΔB)) = n(A∪B) × n(A∩B) × n(AΔB) = 6 × 2 × 4 = 48.
So, n((A∪B) × (A∩B) × (AΔB)) = n(A∪B) × n(A∩B) × n(AΔB) = 6 × 2 × 4 = 48.
Question No. 3: If P(A) denotes the power set of A, then find n(P(P(P(∅)))).
Solution: Since P(∅) contains 1 element, P(P(∅)) contains 21 elements and hence P(P(P(∅))) contains 22 elements. That is, 4 elements.
Question No. 4: Find the largest possible domain for the real valued function f defined by f(x) = √(x² − 5x + 6).
Solution: As we are finding the square root of x² − 5x + 6, we must have x² − 5x + 6 ≥ 0 for all x in the domain.
Solving x² − 5x + 6 = 0, we get x = 2 and 3. Now we have three intervals: (−∞,2), (2,3) and (3,∞).
(i) Take any point in (−∞,2), say x = 1. Clearly x² − 5x + 6 is positive.
(ii) Take any point in (2,3), say x = 2.5. Clearly x² − 5x + 6 is negative.
(iii) Take any point in (3,∞), say x = 4. Clearly x² − 5x + 6 is positive.
For all x in the intervals (−∞,2) and (3,∞), x² − 5x + 6 is positive. At x = 2, 3 the value of x² − 5x + 6 is zero.
Thus, √(x² − 5x + 6) is defined for all x in (−∞,2] ∪ [3,∞). Hence the domain of √(x² − 5x + 6) is (−∞,2] ∪ [3,∞).
Solving x² − 5x + 6 = 0, we get x = 2 and 3. Now we have three intervals: (−∞,2), (2,3) and (3,∞).
(i) Take any point in (−∞,2), say x = 1. Clearly x² − 5x + 6 is positive.
(ii) Take any point in (2,3), say x = 2.5. Clearly x² − 5x + 6 is negative.
(iii) Take any point in (3,∞), say x = 4. Clearly x² − 5x + 6 is positive.
For all x in the intervals (−∞,2) and (3,∞), x² − 5x + 6 is positive. At x = 2, 3 the value of x² − 5x + 6 is zero.
Thus, √(x² − 5x + 6) is defined for all x in (−∞,2] ∪ [3,∞). Hence the domain of √(x² − 5x + 6) is (−∞,2] ∪ [3,∞).
Question No. 5: Solve |x − 9| < 2 for x.
Solution: |x − 9| < 2 implies −2 < x − 9 < 2. Thus, 7 < x < 11.
Question No. 6: Solve |2/(x−4)| > 1, x ≠ 4.
Solution: From the given inequality, we have that 2 > |x − 4|. That is, −2 < x − 4 < 2 and x ≠ 4.
Adding 4 throughout the inequality, we obtain 2 < x < 6 and x ≠ 4. So the solution set is (2,4) ∪ (4,6).
Adding 4 throughout the inequality, we obtain 2 < x < 6 and x ≠ 4. So the solution set is (2,4) ∪ (4,6).
Question No. 7: Our monthly electricity bill contains a basic charge that is independent of units consumed and a charge that depends on the units consumed. Let us say the Electricity Board charges Rs. 110 as basic charge and charges Rs. 4 for each unit used. If a person wants to keep his electricity bill below Rs. 250, then what should be his electricity usage?
Solution: Let x denote the number of units used. Note that x ≥ 0. Then his electricity bill is Rs. 110 + 4x.
The person wants his bill to be below Rs. 250. Let us solve the inequality 110 + 4x < 250.
Thus, 4x < 140; which gives 0 ≤ x < 35. The person should keep his usage below 35 units in order to keep his bill below Rs. 250.
The person wants his bill to be below Rs. 250. Let us solve the inequality 110 + 4x < 250.
Thus, 4x < 140; which gives 0 ≤ x < 35. The person should keep his usage below 35 units in order to keep his bill below Rs. 250.
Question No. 8: Solve 3x − 5 ≤ x + 1 for x.
Solution: We have 3x − 5 ≤ x + 1; which is equivalent to 2x ≤ 6. Hence we have x ≤ 3; the solution set is (−∞,3].
Question No. 9: Solve the following system of linear inequalities: 3x − 9 ≥ 0, 4x − 10 ≤ 6.
Solution: Note that 3x − 9 ≥ 0 implies 3x ≥ 9, by multiplying both sides by 1/3 we get x ≥ 3.
Similarly, 4x − 10 ≤ 6 implies 4x ≤ 16 and hence x ≤ 4.
So the solution set of 3x − 9 ≥ 0, 4x − 10 ≤ 6 is the intersection of [3,∞) and (−∞,4]. Clearly, the intersection of these intervals gives [3,4].
Similarly, 4x − 10 ≤ 6 implies 4x ≤ 16 and hence x ≤ 4.
So the solution set of 3x − 9 ≥ 0, 4x − 10 ≤ 6 is the intersection of [3,∞) and (−∞,4]. Clearly, the intersection of these intervals gives [3,4].
Question No. 10: A girl A is reading a book having 446 pages and she has already finished reading 271 pages. She wants to finish reading this book within a week. What is the minimum number of pages she should read per day to complete reading the book within a week?
Solution: Let x denote the number of pages the girl should read per day. Then we need x to satisfy 7x + 271 ≥ 446.
Hence x ≥ 25; which implies that she should read at least 25 pages per day.
Hence x ≥ 25; which implies that she should read at least 25 pages per day.
Question No. 11: Solve 3x² + 5x − 2 ≤ 0.
Solution: On factorizing the quadratic polynomial we get 3(x + 2)(x − 1/3) ≤ 0. Draw the number line. Mark the critical points −2 and 1/3 where the factors vanish. On each sub-interval check the sign of (x + 2)(x − 1/3).
Interval (−∞,−2): Sign of (x+2) = −, Sign of (x−1/3) = −, Sign of 3x²+5x−2 = +
Interval (−2,1/3): Sign of (x+2) = +, Sign of (x−1/3) = −, Sign of 3x²+5x−2 = −
Interval (1/3,∞): Sign of (x+2) = +, Sign of (x−1/3) = +, Sign of 3x²+5x−2 = +
You can see the inequality is satisfied in [−2, 1/3].
Interval (−∞,−2): Sign of (x+2) = −, Sign of (x−1/3) = −, Sign of 3x²+5x−2 = +
Interval (−2,1/3): Sign of (x+2) = +, Sign of (x−1/3) = −, Sign of 3x²+5x−2 = −
Interval (1/3,∞): Sign of (x+2) = +, Sign of (x−1/3) = +, Sign of 3x²+5x−2 = +
You can see the inequality is satisfied in [−2, 1/3].
Question No. 12: Determine whether the following functions are even, odd or neither: (i) sin²x − 2cos²x − cosx (ii) sin(cos(x)) (iii) cos(sin(x)) (iv) sinx + cosx
Solution:
(i) Let f(x) = sin²x − 2cos²x − cosx. f(−x) = f(x) [since sin(−x) = −sinx and cos(−x) = cosx]. Thus, f(x) is even.
(ii) Let f(x) = sin(cos(x)). f(−x) = f(x), f(x) is an even function.
(iii) f(x) = cos(sin(x)). f(−x) = f(x). Thus, f(x) is an even function.
(iv) Let f(x) = sinx + cosx. f(−x) ≠ f(x) and f(−x) ≠ −f(x). Thus, f(x) = sinx + cosx is neither even nor odd.
(i) Let f(x) = sin²x − 2cos²x − cosx. f(−x) = f(x) [since sin(−x) = −sinx and cos(−x) = cosx]. Thus, f(x) is even.
(ii) Let f(x) = sin(cos(x)). f(−x) = f(x), f(x) is an even function.
(iii) f(x) = cos(sin(x)). f(−x) = f(x). Thus, f(x) is an even function.
(iv) Let f(x) = sinx + cosx. f(−x) ≠ f(x) and f(−x) ≠ −f(x). Thus, f(x) = sinx + cosx is neither even nor odd.
Polynomials & Partial Fractions
Question No. 13: Find a quadratic polynomial f(x) such that f(0) = 1, f(−2) = 0 and f(1) = 0.
Solution: Let f(x) = ax² + bx + c be the polynomial satisfying the given conditions.
f(0) = a(0)² + b(0) + c = 1, implies that c = 1.
Now the other two conditions f(−2) = 0, f(1) = 0 give 4a − 2b + c = 0 and a + b + c = 0.
Using c = 1, we get 4a − 2b = −1 and a + b = −1. Solving these two equations we get a = b = −1/2.
Thus, f(x) = −(1/2)x² − (1/2)x + 1.
f(0) = a(0)² + b(0) + c = 1, implies that c = 1.
Now the other two conditions f(−2) = 0, f(1) = 0 give 4a − 2b + c = 0 and a + b + c = 0.
Using c = 1, we get 4a − 2b = −1 and a + b = −1. Solving these two equations we get a = b = −1/2.
Thus, f(x) = −(1/2)x² − (1/2)x + 1.
Question No. 14: Construct a cubic polynomial function with rational coefficients having zeros at x = 2/5, 1 + √3 such that f(0) = −8.
Solution: Given that 2/5 and 1 + √3 are zeros of f(x). Thus, 1 − √3 is also a zero of f(x).
Let f(x) = a(x − 2/5)[(x − (1 + √3))][x − (1 − √3)] = a(x − 2/5)[(x − 1)² − 3].
Using f(0) = −8, we have (−2/5)a(−2) = −8, which gives a = −10.
Thus the required polynomial is f(x) = (−10)(x − 2/5)[x² − 2x − 2] = −10x³ + 24x² + 12x − 8.
Let f(x) = a(x − 2/5)[(x − (1 + √3))][x − (1 − √3)] = a(x − 2/5)[(x − 1)² − 3].
Using f(0) = −8, we have (−2/5)a(−2) = −8, which gives a = −10.
Thus the required polynomial is f(x) = (−10)(x − 2/5)[x² − 2x − 2] = −10x³ + 24x² + 12x − 8.
Question No. 15: Prove that ap + q = 0 if f(x) = x³ − 3px + 2q is divisible by g(x) = x² + 2ax + a².
Solution: Note that the degree of f(x) is 3 and the leading coefficient is 1. Since g(x) divides f(x), we have f(x) = (x + b)g(x), for some b ∈ R.
Thus, x³ − 3px + 2q = (x + b)(x² + 2ax + a²).
Equating like coefficients on both sides, we have 2a + b = 0, a² + 2ab = −3p and 2q = ba².
Thus, b = −2a, p = a², and q = −a³. Now, q = −a³ = −a(a²) = −ap, which gives ap + q = 0.
Thus, x³ − 3px + 2q = (x + b)(x² + 2ax + a²).
Equating like coefficients on both sides, we have 2a + b = 0, a² + 2ab = −3p and 2q = ba².
Thus, b = −2a, p = a², and q = −a³. Now, q = −a³ = −a(a²) = −ap, which gives ap + q = 0.
Question No. 16: The equations x² − 6x + a = 0 and x² − bx + 6 = 0 have one root in common. The other root of the first and the second equations are integers in the ratio 4 : 3. Find the common root.
Solution: Let α be the common root. Let α, 4β be the roots of x² − 6x + a = 0. Let α, 3β be the roots of x² − bx + 6 = 0.
Then, 4αβ = a and 3αβ = 6, which give αβ = 2 and a = 8.
The roots of x² − 6x + 8 = 0 are 2, 4. If α = 2, then β = 1. If α = 4, then β = 1/2, which is not an integer.
Hence, the common root is 2.
Then, 4αβ = a and 3αβ = 6, which give αβ = 2 and a = 8.
The roots of x² − 6x + 8 = 0 are 2, 4. If α = 2, then β = 1. If α = 4, then β = 1/2, which is not an integer.
Hence, the common root is 2.
Question No. 17: Find the values of p for which the difference between the roots of the equation x² + px + 8 = 0 is 2.
Solution: Let α and β be the roots of the equation x² + px + 8 = 0. Then, α + β = −p, αβ = 8 and |α − β| = 2.
Now, (α + β)² − 4αβ = (α − β)², which gives p² − 32 = 4. Thus, p = ±6.
Now, (α + β)² − 4αβ = (α − β)², which gives p² − 32 = 4. Thus, p = ±6.
Question No. 18: Resolve into partial fractions: x / [(x+3)(x−4)]
Solution: Let x / [(x+3)(x−4)] = A/(x+3) + B/(x−4), where A and B are constants.
Then, x = A(x − 4) + B(x + 3). When x = 4, we have B = 4/7. When x = −3, we have A = 3/7.
Hence, x / [(x+3)(x−4)] = 3 / [7(x+3)] + 4 / [7(x−4)].
Then, x = A(x − 4) + B(x + 3). When x = 4, we have B = 4/7. When x = −3, we have A = 3/7.
Hence, x / [(x+3)(x−4)] = 3 / [7(x+3)] + 4 / [7(x−4)].
Question No. 19: Resolve into partial fractions: 2x / [(x²+1)(x−1)]
Solution: In this case, note that the denominator has a factor x² + 1 which does not have real zeros.
Let 2x / [(x²+1)(x−1)] = A/(x−1) + (Bx+C)/(x²+1), where A, B, C are constants.
We have, 2x = A(x² + 1) + (Bx + C)(x − 1). When x = 1, we get A = 1. When x = 0, we have A − C = 0 and hence A = C = 1.
When x = −1, we have 2A − 2(C − B) = −2, which gives B = −1.
Thus, 2x / [(x²+1)(x−1)] = 1/(x−1) + (1−x)/(x²+1).
Let 2x / [(x²+1)(x−1)] = A/(x−1) + (Bx+C)/(x²+1), where A, B, C are constants.
We have, 2x = A(x² + 1) + (Bx + C)(x − 1). When x = 1, we get A = 1. When x = 0, we have A − C = 0 and hence A = C = 1.
When x = −1, we have 2A − 2(C − B) = −2, which gives B = −1.
Thus, 2x / [(x²+1)(x−1)] = 1/(x−1) + (1−x)/(x²+1).
Question No. 20: Resolve into partial fractions: (x+1) / [x²(x−1)]
Solution: Let (x+1) / [x²(x−1)] = A/x + B/x² + C/(x−1).
Then, x + 1 = Ax(x − 1) + B(x − 1) + Cx². When x = 0, we have B = −1 and when x = 1, we get C = 2.
When x = −1, we have 2A − 2B + C = 0, which gives A = −2.
Thus, (x+1) / [x²(x−1)] = −2/x − 1/x² + 2/(x−1).
Then, x + 1 = Ax(x − 1) + B(x − 1) + Cx². When x = 0, we have B = −1 and when x = 1, we get C = 2.
When x = −1, we have 2A − 2B + C = 0, which gives A = −2.
Thus, (x+1) / [x²(x−1)] = −2/x − 1/x² + 2/(x−1).
Surds & Trigonometry
Question No. 21: Simplify: (i) (x1/2y−3)1/2; where x, y ≥ 0. (ii) √(x² − 10x + 25).
Solution:
(i) Since x, y ≥ 0, we have (x1/2y−3)1/2 = x1/4/y3/2.
(ii) Observe that √(x² − 10x + 25) = √((x−5)²) = |x − 5|.
(i) Since x, y ≥ 0, we have (x1/2y−3)1/2 = x1/4/y3/2.
(ii) Observe that √(x² − 10x + 25) = √((x−5)²) = |x − 5|.
Question No. 22: Rationalize the denominator of √5 / (√6 + √2).
Solution: Multiplying both numerator and denominator by (√6 − √2), we get:
√5 / (√6 + √2) = √5(√6 − √2) / [(√6 + √2)(√6 − √2)] = (√30 − √10) / 4.
√5 / (√6 + √2) = √5(√6 − √2) / [(√6 + √2)(√6 − √2)] = (√30 − √10) / 4.
Question No. 23: Find the square root of 7 − 4√3.
Solution: Let √(7 − 4√3) = a + b√3, where a, b are rationals.
Squaring on both sides, we get 7 − 4√3 = a² + 3b² + 2ab√3.
So, a² + 3b² = 7 and 2ab = −4. Therefore a = −2/b.
From a² + 3b² = 7, we get (−2/b)² + 3b² = 7, which gives 4/b² + 3b² = 7, or 3b⁴ − 7b² + 4 = 0.
Solving for b², we get b² = [7 ± √(49−48)] / 6. Thus b² = 1 or b² = 4/3. Since b is rational, b² = 1 and hence b = ±1.
If b = 1, then a = −2. If b = −1, then a = 2.
√(7 − 4√3) = |2 − √3| = 2 − √3.
Squaring on both sides, we get 7 − 4√3 = a² + 3b² + 2ab√3.
So, a² + 3b² = 7 and 2ab = −4. Therefore a = −2/b.
From a² + 3b² = 7, we get (−2/b)² + 3b² = 7, which gives 4/b² + 3b² = 7, or 3b⁴ − 7b² + 4 = 0.
Solving for b², we get b² = [7 ± √(49−48)] / 6. Thus b² = 1 or b² = 4/3. Since b is rational, b² = 1 and hence b = ±1.
If b = 1, then a = −2. If b = −1, then a = 2.
√(7 − 4√3) = |2 − √3| = 2 − √3.
Question No. 24: Prove that (tanθ − secθ + 1) / (tanθ + secθ − 1) = (1 + sinθ) / cosθ.
Solution:
(tanθ − secθ + 1) / (tanθ + secθ − 1) = [tanθ − secθ + 1] / [tanθ + secθ − (sec²θ − tan²θ)]
= [tanθ − secθ + 1] / {(tanθ + secθ)[1 − (secθ − tanθ)]}
= tanθ + secθ = (1 + sinθ) / cosθ.
(tanθ − secθ + 1) / (tanθ + secθ − 1) = [tanθ − secθ + 1] / [tanθ + secθ − (sec²θ − tan²θ)]
= [tanθ − secθ + 1] / {(tanθ + secθ)[1 − (secθ − tanθ)]}
= tanθ + secθ = (1 + sinθ) / cosθ.
Question No. 25: Prove that (secA − cosecA)(1 + tanA + cotA) = tanA secA − cotA cosecA.
Solution:
L.H.S. = (1/cosA − 1/sinA)[1 + sinA/cosA + cosA/sinA]
= [(sinA − cosA)/(sinA cosA)] × [(sinA cosA + sin²A + cos²A)/(sinA cosA)]
= [(sinA − cosA)(1 + sinA cosA)] / (sin²A cos²A)
= [sinA + sin²A cosA − cosA − sinA cos²A] / (sin²A cos²A)
= sinA/(sin²A cos²A) + sin²A cosA/(sin²A cos²A) − cosA/(sin²A cos²A) − sinA cos²A/(sin²A cos²A)
= 1/(sinA cos²A) + 1/cosA − 1/(sin²A cosA) − 1/sinA
= cosecA sec²A + secA − cosec²A secA − cosecA
= secA(secA cosecA + 1) − cosecA(cosecA secA + 1)
= (secA cosecA + 1)(secA − cosecA)
= sec²A cosecA + secA − cosec²A secA − cosecA
= tanA/cosA − cotA/sinA
= tanA secA − cotA cosecA = R.H.S.
L.H.S. = (1/cosA − 1/sinA)[1 + sinA/cosA + cosA/sinA]
= [(sinA − cosA)/(sinA cosA)] × [(sinA cosA + sin²A + cos²A)/(sinA cosA)]
= [(sinA − cosA)(1 + sinA cosA)] / (sin²A cos²A)
= [sinA + sin²A cosA − cosA − sinA cos²A] / (sin²A cos²A)
= sinA/(sin²A cos²A) + sin²A cosA/(sin²A cos²A) − cosA/(sin²A cos²A) − sinA cos²A/(sin²A cos²A)
= 1/(sinA cos²A) + 1/cosA − 1/(sin²A cosA) − 1/sinA
= cosecA sec²A + secA − cosec²A secA − cosecA
= secA(secA cosecA + 1) − cosecA(cosecA secA + 1)
= (secA cosecA + 1)(secA − cosecA)
= sec²A cosecA + secA − cosec²A secA − cosecA
= tanA/cosA − cotA/sinA
= tanA secA − cotA cosecA = R.H.S.
Kalvi Mini First Midterm Exam 2026 - Model Question Paper
PART-I: Choose the Correct Answer (5 x 1 = 5 Marks)
Question No. 1: If A = {(x,y) : y = ex, x∈R} and B = {(x,y) : y = e−x, x∈R} then n(A∩B) is
(1) Infinity (2) 0 (3) 1 (4) 2
(1) Infinity (2) 0 (3) 1 (4) 2
Question No. 2: If kx / [(x+2)(x−1)] = 2/(x+2) + 1/(x−1), then the value of k is
(1) 1 (2) 2 (3) 3 (4) 4
(1) 1 (2) 2 (3) 3 (4) 4
Question No. 3: The value of sin(480°) is
(1) √3/2 (2) 1/2 (3) 1/√2 (4) √3
(1) √3/2 (2) 1/2 (3) 1/√2 (4) √3
Question No. 4: The relation R defined on a set A = {0,−1,1,2} by xRy if |x² + y²| ≤ 2, then range of R is
(1) {(0,0),(0,−1),(0,1),(−1,0),(−1,1),(1,2),(1,0)}
(2) R⁻¹ = {(0,0),(0,−1),(0,1),(−1,0),(1,0)}
(3) {0,−1,1,2}
(4) {0,−1,1}
(1) {(0,0),(0,−1),(0,1),(−1,0),(−1,1),(1,2),(1,0)}
(2) R⁻¹ = {(0,0),(0,−1),(0,1),(−1,0),(1,0)}
(3) {0,−1,1,2}
(4) {0,−1,1}
Question No. 5: The number of roots of (x+3)⁴ + (x+5)⁴ = 16 is
(1) 4 (2) 2 (3) 3 (4) 0
(1) 4 (2) 2 (3) 3 (4) 0
PART-II: Short Answer Problems (5 x 2 = 10 Marks)
Question No. 6: If n(A) = 10 and n(A∩B) = 3, find n((A∩B)′ ∩ A).
Question No. 7: If n(P(A)) = 1024, n(A∪B) = 15 and n(P(B)) = 32, then find n(A∩B).
Question No. 8: Solve |x − 9| < 2 for x.
Question No. 9: Solve 3x − 5 ≤ x + 1 for x.
Question No. 10: Simplify: (x1/2y−3)1/2; where x, y ≥ 0.
PART-III: Medium Application Problems / Theorems (5 x 3 = 15 Marks)
Question No. 11: Find the largest possible domain for the real valued function f defined by f(x) = √(x² − 5x + 6).
Question No. 12: Solve the following system of linear inequalities: 3x − 9 ≥ 0, 4x − 10 ≤ 6.
Question No. 13: Solve 3x² + 5x − 2 ≤ 0.
Question No. 14: Prove that (tanθ − secθ + 1) / (tanθ + secθ − 1) = (1 + sinθ) / cosθ.
Question No. 15: Determine whether f(x) = sin²x − 2cos²x − cosx is even, odd or neither.
PART-IV: Long Answers / Proofs (4 x 5 = 20 Marks)
Question No. 16: Construct a cubic polynomial function with rational coefficients having zeros at x = 2/5, 1 + √3 such that f(0) = −8.
Question No. 17: Prove that ap + q = 0 if f(x) = x³ − 3px + 2q is divisible by g(x) = x² + 2ax + a².
Question No. 18: Resolve into partial fractions: 2x / [(x²+1)(x−1)]
Question No. 19: Find the square root of 7 − 4√3.
Kalvi Mini First Midterm Exam 2026 - Answer Key
PART-I
Question No. 1: (3) 1
Question No. 2: (3) 3
Question No. 3: (1) √3/2
Question No. 4: (4) {0,−1,1}
Question No. 5: (4) 0
PART-II to PART-IV
Question No. 6: (A∩B)′ ∩ A = (A′ ∪ B′) ∩ A = (A′ ∩ A) ∪ (B′ ∩ A) = ∅ ∪ (B′ ∩ A) = (B′ ∩ A) = A − B.
So n((A∩B)′ ∩ A) = n(A − B) = n(A) − n(A∩B) = 10 − 3 = 7.
So n((A∩B)′ ∩ A) = n(A − B) = n(A) − n(A∩B) = 10 − 3 = 7.
Question No. 7: n(P(A)) = 1024 ⇒ 2n(A) = 210 ⇒ n(A) = 10.
n(P(B)) = 32 ⇒ 2n(B) = 25 ⇒ n(B) = 5.
n(A∩B) = n(A) + n(B) − n(A∪B) = 10 + 5 − 15 = 0.
n(P(B)) = 32 ⇒ 2n(B) = 25 ⇒ n(B) = 5.
n(A∩B) = n(A) + n(B) − n(A∪B) = 10 + 5 − 15 = 0.
Question No. 8: |x − 9| < 2 implies −2 < x − 9 < 2. Thus, 7 < x < 11.
Question No. 9: 3x − 5 ≤ x + 1; which is equivalent to 2x ≤ 6. Hence x ≤ 3; the solution set is (−∞,3].
Question No. 10: Since x, y ≥ 0, we have (x1/2y−3)1/2 = x1/4/y3/2.
Question No. 11: x² − 5x + 6 ≥ 0 ⇒ (x−2)(x−3) ≥ 0. Critical points are 2, 3. The intervals are (−∞,2], [3,∞). Domain is (−∞,2] ∪ [3,∞).
Question No. 12: 3x − 9 ≥ 0 ⇒ x ≥ 3. 4x − 10 ≤ 6 ⇒ x ≤ 4. Intersection of [3,∞) and (−∞,4] is [3,4].
Question No. 13: 3(x+2)(x−1/3) ≤ 0. Testing intervals: sign is negative in (−2,1/3). At endpoints, value is zero. Solution is [−2, 1/3].
Question No. 14: (tanθ − secθ + 1)/(tanθ + secθ − 1) = [tanθ − secθ + 1] / [tanθ + secθ − (sec²θ − tan²θ)] = [tanθ − secθ + 1] / {(tanθ + secθ)[1 − (secθ − tanθ)]} = tanθ + secθ = (1 + sinθ)/cosθ.
Question No. 15: f(−x) = sin²(−x) − 2cos²(−x) − cos(−x) = (−sinx)² − 2(cosx)² − cosx = sin²x − 2cos²x − cosx = f(x). Thus, f(x) is even.
Question No. 16: Zeros are 2/5, 1+√3, 1−√3. f(x) = a(x−2/5)(x−(1+√3))(x−(1−√3)) = a(x−2/5)(x²−2x−2).
f(0) = a(−2/5)(−2) = 4a/5 = −8 ⇒ a = −10. f(x) = −10(x−2/5)(x²−2x−2) = −10x³ + 24x² + 12x − 8.
f(0) = a(−2/5)(−2) = 4a/5 = −8 ⇒ a = −10. f(x) = −10(x−2/5)(x²−2x−2) = −10x³ + 24x² + 12x − 8.
Question No. 17: x³ − 3px + 2q = (x+b)(x²+2ax+a²). Equating coefficients: 2a + b = 0 ⇒ b = −2a; a² + 2ab = −3p ⇒ a² + 2a(−2a) = −3p ⇒ −3a² = −3p ⇒ p = a²; 2q = ba² ⇒ 2q = (−2a)a² = −2a³ ⇒ q = −a³. q = −a(a²) = −ap ⇒ ap + q = 0.
Question No. 18: 2x/[(x²+1)(x−1)] = A/(x−1) + (Bx+C)/(x²+1). 2x = A(x²+1) + (Bx+C)(x−1). x=1 ⇒ 2=2A ⇒ A=1. x=0 ⇒ 0=A−C ⇒ C=1. x=−1 ⇒ −2=2A−2(C−B) ⇒ −2=2−2(1−B) ⇒ B=−1.
Answer: 1/(x−1) + (1−x)/(x²+1).
Answer: 1/(x−1) + (1−x)/(x²+1).
Question No. 19: Let √(7−4√3) = a+b√3. a²+3b²=7, 2ab=−4. a=−2/b ⇒ 4/b²+3b²=7 ⇒ 3b⁴−7b²+4=0. (3b²−4)(b²−1)=0. b²=1 ⇒ b=±1. b=1, a=−2 or b=−1, a=2.
√(7−4√3) = |2−√3| = 2−√3.
√(7−4√3) = |2−√3| = 2−√3.
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