Full Syllabus Mock Test

For sin⁻¹(y), −1 ≤ y ≤ 1. Solve −1 ≤ x² − 3x + 2 ≤ 1. x² − 3x + 2 ≤ 1 → x² − 3x + 1 ≤ 0 → x ∈ [(3−√5)/2, (3+√5)/2]. x² − 3x + 2 ≥ −1 → x² − 3x + 3 ≥ 0, discriminant negative so always true. Domain is [(3−√5)/2, (3+√5)/2].

The sum 7 can be formed by (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That is 6 outcomes out of 36. 6/36 = 1/6.

dy/dx = cos x − sin x. d²y/dx² = −sin x − cos x = −(sin x + cos x) = −y. Therefore, d²y/dx² + y = 0.

An involutory matrix is a matrix that is its own inverse, meaning that when multiplied by itself, it yields the identity matrix, A² = I.

Directly from intercept form.

y₁ = 2 tan⁻¹ x / (1+x²). So (1+x²)y₁ = 2 tan⁻¹ x. Differentiating: 2x y₁ + (1+x²)y₂ = 2/(1+x²). Multiply by (1+x²): 2x(1+x²)y₁ + (1+x²)² y₂ = 2.

Slope m = dy/dx = -3x² + 6x + 9. To maximize m, dm/dx = -6x + 6 = 0 → x = 1. Max slope = -3(1)² + 6(1) + 9 = 12.

Volume = ∫[0 to 5] 10e^(0.2t) dt = [10/0.2 e^(0.2t)] from 0 to 5 = 50[e¹ - e⁰] = 50(e - 1) liters.

The rise (opposite) is 1m and run (adjacent) is 5m. tan(θ) = 1/5. Hence, θ = tan⁻¹(1/5).

TATANAGAR has 8 pairs, 'TA' appears twice (prob 2/8 = 1/4). CALCUTTA has 7 pairs, 'TA' appears once (prob 1/7). P(TATA|TA) = (1/2 × 1/4) / [(1/2 × 1/4) + (1/2 × 1/7)] = (1/4) / (1/4 + 1/7) = 7/11.

Area = |det[2 1; 1 3]| = |6 − 1| = 5.

Marginal cost = dC/dx = 6x + 5. At x = 3, MC = 18 + 5 = 23.

ILATE stands for Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential — this is the priority order for choosing the first function u.

Since every real ₹500 note has a completely unique serial number, a note can only share a serial number with itself. Thus R = {(a, a) for all notes a}, which is the identity relation.

There are two independent arbitrary constants (a and b) in the given family of curves. Hence, the differential equation will be of order 2.

Commutative: (a+b)/2 = (b+a)/2. Associative: (a∗b)∗c = ((a+b)/2 + c)/2 = (a+b+2c)/4. a∗(b∗c) = (a + (b+c)/2)/2 = (2a+b+c)/4. Not equal in general, so not associative.

2ˣ is strictly increasing → one-one. Range is (0, ∞), not all real numbers → not onto.

Let f(x) = sin⁻¹x and g(x) = πx/2. Plotting both on their valid domain [−1, 1], they intersect at the origin (x = 0) and at the boundaries x = 1 (where both equal π/2) and x = −1 (where both equal −π/2). Thus, there are exactly three real solutions.

The transpose of a matrix is formed by turning all its rows into columns and all its columns into rows.

g(x) is always continuous as the maximum function of a continuous function. It may not be differentiable where f stops being the maximum, but it remains continuous.

LHS = 2x² − 15. RHS = (1)(3) − (4)(−2) = 3 + 8 = 11. So 2x² − 15 = 11 → 2x² = 26 → x² = 13 → x = ±√13. Option 0 is ±√13. We adjust.

Transpose of an m×n matrix is n×m. A row matrix is 1×n, so its transpose is n×1, which is a column matrix.

Let a = sin⁻¹(x). Then (cos⁻¹(x))² = (π/2 − a)². The sum is 2a² − πa + π²/4. The maximum occurs at the domain boundary a = −π/2 (when x = −1), yielding 5π²/4.

k×i = j.

Area = ∫[1 to e] log x dx = [x log x − x]₁ᵉ = (e·1 − e) − (1·0 − 1) = 0 − (−1) = 1 sq unit.

Trace(A) = 1 + 2 + 3 = 6.

a * b = a/b, but b * a = b/a, so not commutative. Also, (a * b) * c = (a/b)/c = a/(bc), whereas a * (b * c) = a/(b/c) = (ac)/b. Hence, not associative.

General point is (2λ+1, −3λ−1, λ). Distance squared from (1, −1, 0) is (2λ)² + (−3λ)² + λ² = 14λ². Given distance is √14, so 14λ² = 14 → λ² = 1 → λ = ±1. If λ = 1, point is (3, −4, 1). If λ = −1, point is (−1, 2, −1).

Differentiating x² = 4y yields 2x = 4(dy/dx), so dy/dx = x/2. At (2,1), tangent slope = 1. Normal slope = -1. Equation: y - 1 = -1(x - 2) → x + y = 3.

p = (100 − x)/2. Revenue R = p x = x(100 − x)/2 = 50x − x²/2. MR = dR/dx = 50 − x. When p = 20, x = 100 − 40 = 60. MR = 50 − 60 = −10? That's negative. We check: if x = 100 − 2p, then p = 50 − x/2. R = 50x − x²/2. MR = 50 − x. At p = 20, x = 100 − 40 = 60, MR = −10. Not in options. Maybe marginal revenue with respect to p? No. Perhaps the question means p = 20 is the selling price, and demand is x = 100 − 2p. Then x = 60. R = 20 × 60 = 1200. Marginal revenue usually means derivative of R with respect to quantity. dR/dx = 50 − x = −10. To get a positive option, we change to x = 200 − 2p. Then p = 100 − x/2. R = 100x − x²/2. MR = 100 − x. At p = 20, x = 200 − 40 = 160, MR = 100 − 160 = −60. Still negative. Maybe it's a linear demand and marginal revenue is always half? Actually, if p = a − bx, R = ax − bx², MR = a − 2bx. With demand x = 100 − 2p → p = 50 − 0.5x. R = 50x − 0.5x², MR = 50 − x. At p = 20, x = 60, MR = −10. Not matching. We'll adjust options to include −10 or change numbers. We set x = 100 − p. Then p = 100 − x. R = 100x − x². MR = 100 − 2x. At p = 20, x = 80, MR = 100 − 160 = −60. No. We use p = 50, then x = 0, MR = 100. We'll just make the correct answer something else and ensure it's in options. We do: demand x = 200 − 10p, then p = 20 − x/10. R = 20x − x²/10. MR = 20 − x/5. At p = 10, x = 100, MR = 20 − 20 = 0. Not good. We'll scrap and write a simpler one: R(x) = 5x − 0.01x², find MR at x = 100. MR = 5 − 0.02x = 5 − 2 = 3. Options: ₹3, etc.

a × b − a × c = 0 → a × (b − c) = 0. This means that either b − c = 0 (so b = c), or vector a is parallel to vector (b − c). Thus, b=c is not 'definitely' true, but a being parallel to (b−c) encompasses the general relationship (including the zero vector case trivially).

A line cannot be perpendicular to itself (not reflexive). If L₁ ⊥ L₂, then L₂ ⊥ L₁ (symmetric). If L₁ ⊥ L₂ and L₂ ⊥ L₃, then L₁ is parallel to L₃, not perpendicular (not transitive).

When the equation is expressed as dx/dy = f(x/y), it is easier to substitute x = vy, which gives dx/dy = v + y(dv/dy).

Complete square: 3 + 2x - x² = 4 - (x-1)². Integral is ∫ dx / (2² - (x-1)²). Formula gives (1/(2*2)) log|(2+(x-1)) / (2-(x-1))| = (1/4) log|(x+1)/(3-x)| + C.

Substitute x = 0: sin⁻¹(1) − 0 = π/2, which is true. If x = 1/2: sin⁻¹(1/2) − 2sin⁻¹(1/2) = π/6 − 2(π/6) = −π/6 ≠ π/2. Thus x=0 is the only solution.

Multiplication by a zero matrix always yields a zero matrix. So AB = 0.

Sum = k(1²) + k(2²) + k(3²) = k(1 + 4 + 9) = 14k = 1. Therefore, k = 1/14.

Locality 2 corresponds to row 2: [150, 160]. Total = 150 + 160 = 310.

Collinear vectors are parallel but can have opposite directions (anti-parallel). Thus, they don't must have the same direction.

∫dx/√(x² + a²) = ln|x + √(x² + a²)| + C or sinh⁻¹(x/a) + C. Both forms are equivalent.

Let I = ∫₀¹ xˣ dx. Put x = 1 − t, dx = −dt. Limits: 0→1, 1→0. I = ∫₁⁰ (1−t)ˣ (−dt) = ∫₀¹ (1−t)ˣ dt. So equality holds for any α. The variable of integration is dummy; the integral of xˣ from 0 to 1 equals integral of (1−x)ˣ from 0 to 1 by substitution, regardless of α.

For a given x, there could be multiple y satisfying the equation, or none. The relation is not a function because one x may correspond to more than one y (or no y). For example, x=5 gives 2y=70 → y=35; x=15 gives 2y=40 → y=20. But each x gives exactly one y actually: y = (85−3x)/2. Since it's a relation from N to N, it associates exactly one y to each x where 85−3x is even and positive. This defines a function from a subset of N. Since y = (85−3x)/2 gives a unique y for each x, it is a function. However, the domain is not all N, only those x where (85−3x) is a positive even integer. But the question says relation 'from N to N'. A function can have domain a proper subset. Actually, a relation R from A to B is a function if for every a ∈ A there is at most one b such that (a,b) ∈ R. Here, for each x ∈ N, there is at most one y. So it is a function (partial function). But the question might intend that not all x give a natural y, so it's not a function 'from N to N'? Usually 'function from A to B' requires domain = A. Since some x ∈ N do not yield y ∈ N, it's not a function from N to N. We go with 'not a function' because domain is not all of N. Yes, a function must be defined on the whole domain. So answer is 'not a function'.

Roots are x=0, 4. Area = ∫(0 to 4) (4x − x²) dx = [2x² − x³/3] = 32 − 64/3 = 32/3 sq units.

LHS = √(2cos²x) = √2|cos x| = −√2 cos x (since cos x ≤ 0 in [π/2, π]). RHS = √2 cos⁻¹(cos x). For x ∈ [π/2, π], cos⁻¹(cos x) = 2π−x? No, in [0,π], cos⁻¹(cos x)=x. So RHS = √2 x. Equation: −√2 cos x = √2 x → cos x = −x. Drawing graphs of y=cos x and y=−x in [π/2, π] shows 1 intersection.

Total upward force = T₁ + T₂ = (2î + 3ĵ) + (−2î + 4ĵ) = 0î + 7ĵ = 7ĵ.

Differentiating gives 2y(dy/dx)=2x.

Smaller part is to the right of x = a/√2. Area = 2 ∫[a/√2 to a] √(a² − x²) dx = 2[(x/2)√(a²−x²) + (a²/2)sin⁻¹(x/a)]_(a/√2)^a = 2[0 + (a²/2)(π/2) − (a/(2√2))(a/√2) − (a²/2)(π/4)] = 2[a²π/4 − a²/4 − a²π/8] = 2[a²π/8 − a²/4] = a²π/4 − a²/2 = (a²/2)(π/2 − 1) sq units.

Surface area S = 2πr(r + h) = 2πr(r + 3) = 2π(r² + 3r). dS/dr = 2π(2r + 3). At r = 7, dS/dr = 2π(17) = 34π ≈ 106.76. dr = 0.1. dS ≈ 10.676 m². Cost change ≈ 20 × 10.676 ≈ 213.5. We compute exactly: dS = 2π(2r + h) dr = 2π(14 + 3)(0.1) = 2π(17)(0.1) = 3.4π ≈ 10.68. Cost = 20 × 3.4π = 68π ≈ 213.6. Not in options. We use formula: S = 2πr² + 2πrh. dS = (4πr + 2πh) dr = 2π(2r + h) dr = 2π(14 + 3)(0.1) = 3.4π. Cost = 68π ≈ 213.6. Maybe h = 3, r = 7, so S = 2π(49 + 21) = 2π(70) = 140π. dS = 4πr dr + 2πh dr = 4π(7)(0.1) + 2π(3)(0.1) = 2.8π + 0.6π = 3.4π. Cost = 68π. If ₹20, then 20 × 3.4π = 68π ≈ 213.6. To get one of the options, maybe the tank is open? Or different values. We assume open cylindrical tank (no top). S = πr² + 2πrh. dS = 2πr dr + 2πh dr = 2π(7 + 3)(0.1) = 2π. Cost = 40π ≈ 125.6, not matching. We try: if sheet cost ₹20 per m², and dS = 3.4π, cost = 68π. If π ≈ 22/7, 68 × 22/7 ≈ 213.7. Still no match. We'll adjust options to include 68π or change numbers. We change cost to ₹10 per m²: cost = 34π ≈ 106.76. Not in options. We make radius 3.5 m (7/2), h = 3. Then dS = 2π(2(3.5) + 3)(0.1) = 2π(7 + 3)(0.1) = 2π. Cost = 40π ≈ 125.6. Still no. To get ₹88, cost change = 88, so dS = 88/20 = 4.4 m². Then 2π(2r + h) dr = 4.4. If dr = 0.1, 2π(2r + 3) = 44 → 2r + 3 = 44/(2π) = 22/π ≈ 7, so 2r ≈ 4, r ≈ 2. We'll just set r = 2, h = 3. Then S = 2π(2)(5) = 20π. dS = 2π(4 + 3)(0.1) = 1.4π ≈ 4.4. Cost = 20 × 4.4 = 88. So We'll change question to radius 2 m, height 3 m.

Let numbers be x and 15-x. Sum of squares S = x² + (15-x)². S' = 2x - 2(15-x) = 4x - 30. S' = 0 gives x = 7.5. The numbers are 7.5 and 7.5.

Cyclic permutations of the vectors in a scalar triple product leave its value unchanged. Hence, [b c a] = [a b c].