Mathematics Test - 1

This is equal to zero only if n + 2 is odd i.e. n is odd integer.

13, a₁ , a₂ , a₃ , ....a₂₀ , 67 are in AP

∴

= 800

Now, AM ≥ GM

∵

∴∴The maximum value of a₁ ⋅a₂ ⋅a₃ ⋅....a₂₀ is (40)²⁰

Let p(h,k) be the mid point of the line segment joining the focus (a,0) and a general point Q(x, y) on the parabola. Then,

⇒ x = 2h − a, y = 2k

Put these values of x and y in y² = 4ax to get

4k² = 4a(2h − a)

⇒ 4k² = 8ah − 4a²

⇒ k² = 2ah − a²

So, the locus of P(h, k) is y² = 2ax − a²

Its directrix is x − a/2 = −a/2

⇒ x = 0.

Let, the centre of the required circle is (1, k) and radius |k|

So, the equation of the required circles is

(x−1)² +(y − k)² = k²

Given, it passes through (2, 3)

Thus diameter = 10/3

We have p.d.f of a random variable x is given by

f(x)= 1/4a, 0 0) and also,

⇒k = 1

ydx - xdy = xydx

=>ydx - xydx = xdy

=>y(1 - x)dx = xdy

=>∫dy/y = ∫(1 - x)/x dx

=>log y = ∫(1/x - 1)dx

=>log y = log x - x + c

=>log y - log x = -x + c

=>log(y/x) = -x + c

=>log(y/x) = -x + c ...... Taking negative sign common

=>x/y = e^{(x + c)}

=>x/y = c e^x

We have,

⇒ 3 + 5 + 7 … + (2n + 1) = 440

⇒ n/2[2 x 3 + (n-1)(2) = 440

⇒ n(3 + n - 1) = 440

⇒ n(n + 2) = 440

⇒ n = 20

We have standard deviation of X =

⇒Vai(X) = 3pq

and mean, E(X) = 3p

Now, 3pq = E(x² )-(3p² )

(∵ Var(X) = E(x² ) - (E(x)² ))

⇒ E(x² ) = 3pq + 9p²

= 3p(1-p) + 9p² (∵p+q = 1)

= 3p - 3p² + 9p²

⇒ 3p + 6p² ⇒ 3p(1 + 2p)

Key Idea Given, ax² + 2hxy + by² = 0 represents perpendicular pair of straight lines then a + b = 0

we have, pair of Snes represented by (1+sin² θ)x² + 2hxy + 2sin θy² =0, θ∈(0, 2 π)

∴(1 + sin² θ) + 2Sin θ= 0

⇒(1+ sin θ)² = 0

⇒1 + sin θ= 0

⇒

⇒

In any ΔABC we know that Area of Δ = ½(bc sinA)

⇒ sin A = 2Δ/bc …(i)

also, R = a/2sinA …(ii)

From Eqs. (i) and (ii), we have

⇒ R = abc/4Δ

⇒ Δ = A(ΔABC) = abc/4R

We have rcos θ = -√2 and r sinθ = √2

∴ r² cos² θ + r² sin² θ = 2 + 2 = 4

⇒ r² (cos² θ + sin² θ) = 4

⇒ r = ±2 and θ = tan^-1

θ = tan^-1 (-1) ⇒ θ = 3π/4

∴ Required polar co-ordinate is (r, θ) = (2, 3π/4)

We have, A is non-singular matrix

∴|A| = 0

and (A + l)(A - l) = 0

⇒A²  - I²  = 0

⇒A²  = I²  = I

⇒A ·A = I

⇒A^-1  = A

∴A + A^-1  = A + A = 2A

Slope of the given line, x - 2y = 4 is ½

Equation of a line passing through 4(6, 1) and perpendicular to given line is y - 1 = (-2)(x - 6)

⇒ y - 1 = -2x + 12

⇒ 2x + y = 13

⇒

∴ the y-intercept of slope (i) is 13.

We have, tanA + tanB + tanC = 6

⇒ tanA tanB tanC = 6 …(i)

and tanA.tanB = 2 …(ii)

From Eqs. (i) and (ii) we get, tanC = 3

We have, maximise z = 4x₁ + 2x₂

Subject to contracts, 3x₁ + 2x₂ ≥ 9, x₁ - x₂ ≤ 3, x₁ ≥ 0, x₂ ≥ 0

On taking given constraints as equation, we get the following graphs

Here, we get feasible region is unbounded.