Class 12th

Contributor-Level 10

$A^{2} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}]$

$a \leftrightarrow$ R2

$= - [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$

$R_{2} \leftrightarrow R_{3}$

$= [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ =1

$\begin{array}{l} B_{0} = A^{49} + 2 A^{98} = A + 2 I \\ B_{n} = A d j (B_{n} - 1) \\ B_{4} = A d j (A d j (A d j (A d j B_{0}))) \end{array}$

$= {| B_{0} |}^{{(n - 1)}^{4}} = {| B_{0} |}^{16}$

$B_{0} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] = [\begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{matrix}]$

$= 2 (4 - 0) - 1 (0 - 1) = 9$

$B_{4} {(9)}^{16} = 3^{32}$

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New answer posted

11 months ago

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to:

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Class 12th Electromagnetic Waves

Contributor-Level 10

$\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$

IF = $e^{\int p d x} = \frac{{(x + 1)}^{2} (x + 2)}{x + 3}$

$\int P d x = \int \frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6} d n = \int (\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}) d x$

$= l n ({(x + 1)}^{2} \cdot (x + 2) / (x + 3))$

$\frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

$2 x^{2} + 11 x + 13 = A (x + 2) (x + 3) + B (x + 1) (x + 3) + C (x + 1) (x + 2)$

x = -1

->4 = 2A Þ A = 2

x = -2

-> -1 = -B Þ B = 1

x₂ – 3 Þ -2 = 2c

c = -1

$y \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} = \int \frac{x + 3}{x + 1} \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} d x$

= $\int (x + 1) (x + 2) d x$

= $\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c$

$(0, 1) \Rightarrow 1 \cdot \frac{2}{3} = c$

x = 1 $y \cdot (3) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = \frac{3}{2} + 3 = \frac{9}{2}$

y = $\frac{3}{2}$

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New answer posted

11 months ago

A beam of light travelling along X-axis is described by the electric field E_y = 900sin (t – x/c). The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3 * 10⁷ ms^-1 will be:

(Given speed of light = 3 * 10⁸ ms^-1)

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Class 12th Complex Numbers and Quadratic Equations

Contributor-Level 10

$E_{y} = 900 s i n ω (t - \frac{x}{c})$

$\frac{F_{E}}{F_{B}} = \frac{q E}{| q \bar{V} * \bar{B} |} = \frac{C}{V} = \frac{3 * 1 0^{8}}{3 * 1 0^{7}} = 10$

$\Rightarrow \frac{F_{E}}{F_{B}} = \frac{10}{1}$

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New answer posted

11 months ago

The remainder when 7²⁰²² + 3²⁰²² is divided by 5 is:

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Class 12th Physics Alternating Current

Contributor-Level 10

$7^{2022} + 3^{2022}$

$= {(49)}^{1011} + {(9)}^{1011}$

$= {(50 - 1)}^{1011} + {(10 - 1)}^{1011}$

$= 5 λ - 1 + 5 k - 1$

= 5m – 2

Remainder = 5 – 2 = 3

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New answer posted

11 months ago

A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of $3 Ω a n d 2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:

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Class 12th Differential Equations

Contributor-Level 10

In DC :- In AC:-

i₁ = 4A i₂ = 4 sin $ω t$

R₁ = 3 $Ω$ R₂ = $2 Ω$

In same time internal of 't' - $\frac{H_{1}}{H_{2}} = \frac{i_{1}^{2} R_{1} t}{{(l_{2^{- r m s}})}^{2} R_{2} t} = \frac{16 * 3 t}{{(\frac{4}{\sqrt[]{2}})}^{2} * 2 * t} = \frac{48}{(\frac{32}{2})} = 3 : 1$

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New answer posted

11 months ago

If y = y(x) $\in$ (0, (π)/2) be the solution curve of the differential equation $(s i n^{2} 2 x) \frac{d y}{d x} + (3 s i n^{2} 2 x + 2 s i n 4 x) y = 2 e^{- 4 x} (2 s i n 2 x + c o s 2 x),$ $^{} \frac{}{}^{}^{}$ with y((π)/4) = e, then y((π)/6) is equal to:

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Contributor-Level 10

If y = y(x), $x \in (0, \frac{π}{2})$ $\frac{}{}$ be the solution curve of the different equation

$∴ \frac{d y}{d x} + (8 + 4 c o t 2 x) y = \frac{2 e^{- 4 x}}{s i n^{2} 2 x} (2 s i n x + c o s 2 x)$ which is a linear different equation

I.F = $e^{\int (8 + 4 c o t 2 x) d x} = e^{8 x + 2 c o s (s i n 2 x)} = e^{8 x} . s i n^{2} 2 x$ $^{}^{}^{}^{}$

Given, $y (\frac{π}{4}) = e^{- π} \Rightarrow c = 0$ $\frac{}{}^{}$

$∴ y = \frac{e^{- 4 x}}{s i n 2 x}$

$∴ y (\frac{π}{6}) = \frac{e^{- 4 \frac{π}{6}}}{s i n (2 . \frac{π}{6})} = \frac{2}{\sqrt[]{3}} e^{\frac{2 π}{3}}$

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New answer posted

11 months ago

If the solution curve of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passes through the points (2, 1) and (k + 1, 2), k > 0, then

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Contributor-Level 10

$\frac{d y}{d x} = \frac{x + y - 2}{x - y}$

Let x – 1 = X, y – 1 = Y

then DE: $\frac{d Y}{d X} = \frac{X + Y}{X - Y} = \frac{1 + \frac{Y}{X}}{1 - \frac{Y}{X}}$

Put y = vx

then $\frac{d Y}{d X} = V + X \frac{d V}{d X}$

V + $X \frac{d V}{d X} = \frac{1 + V}{1 - V}$

$X \frac{d V}{d X} = \frac{1 + V}{1 - V} - V$

$= \frac{1 + V^{2}}{1 - V}$

$\frac{1 - V}{1 + V^{2}} d V = \frac{d X}{X}$

$\frac{V - 1}{V^{2} + 1} d V + \frac{d X}{X} = 0$

$\frac{1}{2} l n | V^{2} + 1 | - t a n^{- 1} V + l n | X | = c$

$l n (\sqrt[]{1 + {(\frac{Y - 1}{X -})}^{2}} \cdot | X - 1 |) t a n^{- 1} \frac{Y - 1}{X - 1} = c$

$(2, 1) \Rightarrow l n (\sqrt[]{1 + 0} \cdot 1) - 0 = c$

c = 0

$l n \sqrt[]{{(X - 1)}^{2} + {(Y - 1)}^{2}} = t a n^{- 1} \frac{Y - 1}{X - 1}$

point (k + 1, 2) $l n \sqrt[]{k^{2} + 1} = t a n^{- 1} \frac{1}{k}$

$\Rightarrow \frac{1}{2} l n (k^{2} + 1) = t a n^{- 1} \frac{1}{k}$

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New answer posted

11 months ago

Out of 60% female and 40% male candidates appearing in an exam, 60% candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is:

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Contributor-Level 10

Probability that chosen candidate is female $= \frac{40}{60} = \frac{2}{3}$

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New answer posted

11 months ago

For I(x) = $\int \frac{s e c^{2} x - 2022}{s i n^{2022} x} d x, i f I (\frac{π}{4}) = 2^{1011}, t h e n$

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Contributor-Level 10

I(x) = $\int \frac{s e c^{2} x - 2022}{s i n^{2022} x} d x$

$= \int s i n^{- 2022} x \cdot s e c^{2} x d x - \int 2022 s i n^{- 2022} x d x$

$= s i n^{- 2022} x \cdot t a n x - \int (- 2022) s i n^{- 2023} x \cdot c o s x \cdot t a n x d x - \int 2022 s i n^{- 2022} x d x$

$= t a n x \cdot s i n^{- 2022} x + 2022 \int s i n^{- 2022} - 2022 \int s i n^{- 2022} x d x$

I(x) = $t a n x \cdot s i n^{- 2022} x + c$

Given, $I (\frac{π}{4}) = 2^{1011}$

-> $2^{1011} = \frac{1}{{(\frac{1}{\sqrt[]{2}})}^{2022}} + c \Rightarrow c = 0$

$I (x) = \frac{t a n x}{s i n^{2022} x}, I (\frac{π}{3}) = \frac{\sqrt[]{3}}{(\frac{\sqrt[]{3}}{2})} = \frac{2^{2022}}{{(\sqrt[]{3})}^{2021}}$

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New answer posted

11 months ago

Let a vector $\vec{a}$ has magnitude 9. Let a vector $\vec{b}$ be such that for every (x, y) $\in R * R - {(0, 0)},$ the vector $(x \vec{a} + y \vec{b})$ is perpendicular to the vector $(6 y \vec{a} - 18 x \vec{b}) .$ Then the value of $| \vec{a} * \vec{b} |$ is equal to:

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