One question at a time. Build your Tamil Nadu HSE First Year Mathematics understanding.
First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
The primes less than 10 are 2, 3, 5 and 7. The number 1 is not prime, and 9 = 3 × 3 is composite, so neither of them belongs to the set.
A set with 5 elements has 2^5 = 32 subsets in total. Proper subsets exclude A itself, and non-empty proper subsets also exclude the empty set, giving 32 − 2 = 30.
A − B consists of the elements that belong to A but not to B. Removing from A the elements 4 and 5, which also lie in B, leaves {1, 2, 3}.
By the inclusion–exclusion principle, n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 20 + 15 − 6 = 29. Adding the two counts counts the 6 common students twice, so they are subtracted once.
De Morgan's law for sets states that the complement of a union is the intersection of the complements: (A ∪ B)′ = A′ ∩ B′. An element lies outside A ∪ B exactly when it lies outside both A and B.
A relation from A to B is any subset of A × B. Since n(A × B) = 3 × 4 = 12, the number of such subsets, and hence of relations, is 2^12 = 4096.
For each x in {1, 2, 3, 4}, the relation pairs x with x + 2, giving (1, 3), (2, 4), (3, 5) and (4, 6). The first coordinates form the domain and the second coordinates form the range.
Every element is related to itself, so R is reflexive. But (1, 2) ∈ R while (2, 1) ∉ R, so R is not symmetric; also (1, 2), (2, 3) ∈ R while (1, 3) ∉ R, so R is not transitive.
Let y = (3x + 2)/5. Solving for x gives 5y = 3x + 2, so x = (5y − 2)/3, and replacing y by x gives f⁻¹(x) = (5x − 2)/3. Checking with x = 1, both f and f⁻¹ return 1.
There are 2^4 = 16 functions from a 4-element set to a 2-element set. The two constant functions each miss one element of B, so the number of onto functions is 16 − 2 = 14.
In set-builder form the inequality −3 < x is strict, so −3 itself is not a member and the bracket at −3 must be round. The inequality x ≤ 4 is not strict, so 4 is a member and the bracket at 4 must be square, giving (−3, 4].
A singleton set has exactly one element. The only even prime number is 2, so {x : x is an even prime} = {2}. The remaining sets are respectively empty, a two-element set and a three-element set, so none of them is a singleton.
The ordered pair (a, b) belongs to R exactly when a divides b. Checking each first coordinate in {1, 2, 3, 4} gives (1,1), (1,2), (1,3), (1,4), (2,2), (2,4), (3,3) and (4,4), which is eight ordered pairs in all.
Since f(2) = f(−2) = 4, two distinct inputs give the same output, so f is not one-one. Also x² ≥ 0 for every real x, so no negative real number has a preimage and the range is [0, ∞), not R; hence f is not onto either.
By definition (g ∘ f)(1) = g(f(1)). First compute the inner value f(1) = 2(1) + 3 = 5, then apply g to that result: g(5) = 5² − 1 = 25 − 1 = 24.
Equivalence relations on a finite set correspond one-to-one with partitions of that set. The set {1, 2, 3} has the partitions {1}{2}{3}, {1,2}{3}, {1,3}{2}, {2,3}{1} and {1,2,3}, and each partition yields exactly one equivalence relation, giving five in total.
For |2x−3|=7, the expression inside the absolute value can equal 7 or −7. Solving 2x−3=7 gives x=5, and solving 2x−3=−7 gives x=−2. Both values satisfy the original equation, so the solution set is {−2, 5}.
Write 27 as 3^3, so 3^(x−1)=3^3. Since the bases are equal, the exponents must be equal: x−1=3. Hence x=4.
For x²−5x+6=0, the sum of roots is −(−5)/1=5 and the product is 6/1=6. Therefore α+β+αβ=5+6=11. The individual roots are 2 and 3, which also verify the result.
For 2x²−4x+3=0, the discriminant is b²−4ac=(−4)²−4(2)(3)=16−24=−8. A negative discriminant means the quadratic has no real roots. Therefore the roots are non-real complex conjugates.
Complete the square: x²−6x+11=(x−3)²+2. Since (x−3)²≥0 for every real x, the expression is at least 2. The minimum value 2 occurs at x=3.
Factor x²−x−6=(x−3)(x+2). The product is non-positive between the roots −2 and 3, including the endpoints. Hence the solution set is [−2, 3].
Squaring √(x+5)=x−1 gives x+5=(x−1)²=x²−2x+1, so x²−3x−4=0. Factoring gives (x−4)(x+1)=0, so x=4 or x=−1. Checking in the original equation, x=−1 fails because the right side is negative, while x=4 satisfies it.
Factor the numerator as (x−3)(x+3) and the denominator as (x−3)(x+2). Since x≠3, cancel the common factor x−3. The simplified expression is (x+3)/(x+2), with x≠−2 also excluded.
For equal real roots, the discriminant must be zero. Here a=1, b=k and c=9, so k²−4(1)(9)=k²−36=0. Thus k=6 or k=−6, and both values give a perfect square (x±3)².
For the equation to be quadratic, m−1≠0, so m≠1. Its discriminant is [2(m−1)]²−4(m−1)(1)=4(m−1)(m−2). Equal real roots require this discriminant to be zero, giving m=2, since m=1 is not allowed.
log₂ 8 = 3 because 2³ = 8, and log₃ 81 = 4 because 3⁴ = 81. Adding these values gives x + y = 3 + 4 = 7.
27^(2/3) = (3³)^(2/3) = 3² = 9, while 9^(1/2) = 3. Dividing gives 9 ÷ 3 = 3.
The inequality |x − 4| < 2 means the distance from x to 4 is less than 2. Therefore 2 < x < 6, which is the open interval (2, 6).
The critical points are x = 1, where the numerator is zero, and x = −2, where the denominator is zero and the expression is undefined. Testing intervals gives positive values on (−∞, −2) and (1, ∞), with x = 1 included because the inequality is non-strict.
Combining the logarithms gives log₃((x − 1)(x + 1)) = 1, so x² − 1 = 3 and x² = 4. The domain requires x > 1, so only x = 2 is admissible.
By the factor theorem, substituting x = 2 into the polynomial must give zero. Thus 8 − 12 + 2k − 2 = 0, which simplifies to 2k − 6 = 0, so k = 3.
For the given equation, α + β = 3/2 and αβ = 1/2. The new roots have sum (α + β)/(αβ) = 3 and product 1/(αβ) = 2, so the required equation is x² − 3x + 2 = 0.
Since x² + 6x + 9 = (x + 3)², its square root is the absolute value of x + 3, not simply x + 3. For example, at x = −5 the expression equals 2, while x + 3 equals −2.
Let r be the common root. Subtracting the equations gives (a − b)r + (b − a) = 0, so (a − b)(r − 1) = 0; since a ≠ b, r = 1. Substituting r = 1 into either equation gives 1 + a + b = 0, hence a + b = −1.
Factoring gives (x − 2)(x + 2)/[(x − 1)(x + 1)]. The sign changes at −2, −1, 1 and 2; testing intervals shows the expression is non-positive on [−2, −1) and (1, 2], with x = −2 and x = 2 included because the numerator is zero, while x = ±1 are excluded.
To convert degrees to radians, multiply the degree measure by π/180. Therefore 210° = 210 × π/180 = 7π/6 radians, after dividing numerator and denominator by 30.
Write 240° as 180° + 60°, so sin 240° = −sin 60°. Since 240° lies in the third quadrant where sine is negative, the value is −√3/2.
For a function of the form sin(kx), the period is 2π divided by |k|. Here k = 3, so the period is 2π/3; the coefficient 5 changes only the amplitude.
In the third quadrant both sine and cosine are negative, while tangent is positive. Taking opposite side 5 and adjacent side 12 gives hypotenuse 13 by the Pythagorean theorem, so sin θ = −5/13.
For sin θ = sin α, the general solution is θ = nπ + (−1)^n α. Since 1/2 = sin(π/6), taking α = π/6 gives θ = nπ + (−1)^n π/6.
Use cos(A + B) = cos A cos B − sin A sin B with A = 45° and B = 30°. This gives cos 75° = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
Using sin²θ + cos²θ = 1, the numerator becomes sin²θ and the denominator becomes cos²θ. Their quotient is sin²θ/cos²θ = tan²θ.
Use the sum-to-product identity sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2). With C = 5x and D = 3x, this becomes 2 sin 4x cos x.
Let P = cos 20° cos 40° cos 80°. Multiplying by 8 sin 20° and repeatedly using 2 sin A cos A = sin 2A gives 8 sin 20° P = sin 160° = sin 20°, so P = 1/8.
Let t = tan θ. Then sec θ = 2 − t, and squaring sec²θ = 1 + tan²θ gives (2 − t)² = 1 + t². Expanding yields 4 − 4t + t² = 1 + t², so 4t = 3 and t = 3/4.
Sine is negative in the third and fourth quadrants, since the y-coordinate of the terminal point is negative there. Tangent is positive exactly when sine and cosine have the same sign, which happens only in the first and third quadrants. The quadrant satisfying both conditions at once is therefore the third quadrant.
Using the allied-angle formula tan(180° − θ) = −tan θ with θ = 45°, we get tan 135° = −tan 45° = −1. The same conclusion follows from the terminal point in the second quadrant, where the y-value is positive and the x-value is negative, so their quotient is negative.
Because A and B are acute, cos A = 4/5 and sin B = 12/13. Applying sin(A + B) = sin A cos B + cos A sin B gives (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65.
The triple-angle identity cos 3θ = 4cos³θ − 3cos θ gives 4(1/27) − 3(1/3) = 4/27 − 1 = −23/27. The negative sign is essential because 3θ lies in a region where the cosine is negative when cos θ = 1/3.