Mathematics Test - 5

We have, p.d.f of a random variable

X is f (x) = 3(1- 2x² ), 0

= 0, otherwise

= 179/864

We have f(x) = x + (1/x) , x ≠ 0

On differentiating w.r.t x. we get

For maxima or minima, we put f'(x)

⇒

⇒ x² -1 = 0 ⇒ x = -11

Now, since f'(x) > 0, x ∈ (-1- h, -1) and f'(x) <0,= "" x="" ∈="" (-1,="" -1="" +=""/>

⇒ x = -1 is the point of local maxima.

and local maxima value

and f'(x) <0,= "" when="" x="" ∈="" (1="" -="" h,=""/>

and f"(x) > 0 when x ∈(1, 1 + h)

⇒ x = 1 is point of minima

and local minima value,

When a player tosses 2 fair coins, then S = [HT, TH, TT, HH)

Let X be a random variable that denotes the amount received by the player.

Then.X can take values 5, 2, and 1.

Now, P(X = 5) = 1/4, P(X = 2) = 2/4 = 1/2 and P(X = 1) = 1/4

Thus, the probability distribution of X is

We have , A = {x ∈R /x² + 5|x| + 6 = 0|

Now, x² + 5|x| + 6 = 0 …..(i)

⇒x² + 3 |x| + 2 |x| + 6 = 0

⇒|x|(|x| + 3) + 2(|x| + 3|) = 0

⇒(|x| + 2)(|x| + 3) = 0

⇒|x| = - 2 or |x| = -3

⇒x ≠2 or |x| ≠3

Here, no value of x satisfy the Eq. (i).

∴n(A) = 0

We have, differential equation

On me integrating both sides, we get

⇒ tan^-1 x = -tan^-1 y + C

⇒ tan^-1 x + tan ^-1 y = C

Key Idea: If lines (x - x₁ )/a₁ = (y-y₁ )/b₁ = (z -z₁ )/c₁

and (x -x₂ )/a₂ = (y - y₂ )/b₂ = (z - z₂ )/c₂

are perpendicular, then

a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

Given lines are (2x -4)/λ = (y - 1)/2 = (z - 3)/1

⇒ (x - 2)/λ = (y - 1)/2 = (z - 3)/1 ….(i)

Since, ines (i) and (ii) are perpendicular.

⇒ λ/2 + 2λ/3 + 1 = 0

⇒ 3λ + 4λ + 6 = 0

⇒ 7λ = -6

⇒ λ = -(6/7)

We have

= k, x = 0 is continuous at x = 0

⇒ e^2x1x1 = k

⇒ k = e²

We know that, I² + m² + n² = 1

On adding Eqs. (i) and (ii). we get

⇒I = 7 π/18

Let y =

Put t = tan θ ⇒ θ = tan^-1 t

sin^-1 (sin θ) = θ = tan^-1 t

= cos^-1 (cos θ)

= θ = tan^-1 t

Key Idea: Use sine rule. I.e.

a/Sin A = b/sinB = c/sin C

Given, sin B sin C = bc/a²

⇒ a² = bc/sinBsinC

⇒ a² = (b/sinB) (c/sinC)

⇒ a² = (a/sinA)²

⇒ sin² 4 = 1

⇒ sinA = 1 sin90º

⇒ A = 90º

Put tan x = t

⇒ 1 = A(t + 2) + B(t + 1)

Put t = -2

∴ 1 = B(-1) ⇒ B = -1

∴ Put t + 1 = 0 ⇒ t = -1

∴ 1 = A(1) ⇒ A = 1

Let E = sin(3sin^-1 (0.4)

Put sin^-1 (0.4) = θ ⇒ sin θ = 0.4

∴ E = sin(3θ) = 3 sinθ - 4sin³ θ

= 3(0.4) - 4(0.4)³

= 1.2 -4(0.064)

= 1.2 - 0.256

= 0.944

We have, S_n = 5(2ⁿ - 1)

We know that, a_n = S_n - S_n-1

= 5(2ⁿ - 1) - 5(2^n-1 - 1)

= 5(2ⁿ - 2^n-1 )

= 5(2^n-1 )

Key Idea: Use period of cos kθ is 2π/k

∴ Period of cos (πx) is 2π/π = 2

Let the first term and common ratio of a GP are m and n, respectively.

Then, A = m (n^{p - 1} ) ...(i)

B = m(n^{q - 1} ) ...(ii)

and c = m(n^{r - 1} ) ...(ii)

Now, A^{q - r} . B^{r - p} . C^{p - q}

= m^{(q - r)} , n^{(q - r)(p -1)} , m^{(r - p)} , n^{(r - p)(q - 1)} , m^{(p - q)} , n^{(p - q)(r - 1)}

We have, Z = 5x + 4y

Subject to constraints y ≤ 2x, x ≤ 2y, x + y ≤ 3, x ≥ 0, y ≥ 0 On taking given constraints as equations, we get the following graph.

Intersecting point of line y = 2x and x + y = 3 is A(1, 2) and intersecting point of line x = 2y and x + y = 3 is B(2, 1)

Here, OABO is the required feasible region whose components are O(0, 0), 4(1, 2) and B(2, 1).