Class 12th

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

8 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

8 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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Contributor-Level 10

$| \vec{a} | = 9 & (x \vec{a} + y \vec{b}) . (6 y \vec{a} - 18 x \vec{b}) = 0$

$\Rightarrow 6 x y {| \vec{a} |}^{2} - 18 x^{2} (\vec{a} . \vec{b}) + 6 y^{2} (\vec{a} . \vec{b}) - 18 x y {| \vec{b} |}^{2} = 0$

This should hold $\forall x, x, y \in R * R$

$∴ {| \vec{a} |}^{2} = 3 {| \vec{b} |}^{2} & (\vec{a} . \vec{b}) = 0$

$∴ | \vec{a} * \vec{b} | = \frac{{| \vec{a} |}^{2}}{\sqrt[]{3}} = \frac{81}{\sqrt[]{3}} = 27 \sqrt[]{3}$

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New answer posted

8 months ago

Class 12th Physics

A magnet hung at 45° with magnetic meridian makes an angle of 60° with the horizontal. The actual value of the angle of dip is-

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Contributor-Level 10

Let angle of dip =

Then B cos 45° = B' cos 60°- (i)

B' sin 60 = B sin - (ii)

$(i) \div (i i) \Rightarrow c o s t 60 = c o t α c o s 45 °$

$c o t α = \frac{c o t 60 °}{c o s 45 °} = \frac{1 \sqrt[]{2}}{\sqrt[]{3} x} = \sqrt[]{\frac{2}{3}}$

$t a n α = \sqrt[]{\frac{3}{2}}$

$α = t a n^{- 1} \sqrt[]{\frac{3}{2}}$

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New answer posted

8 months ago

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a₀ = a₁ = 0 and a_n+2 = 3a_n+1 – 2a_n + 1, $\forall n \geq 0 .$

Then $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ is equal to

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Class 12th Inverse Trigonometric Functions

Contributor-Level 10

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2 = 3a_n+1 – 2a_n + 1

$(+) \Rightarrow (a_{2} + a_{3} + a_{4} + . . . . + a_{n + 1} + a_{n + 2})$

-> a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

-> a_n+2 – 2a_n+1 = n + 1

-> a_n+1 -2a_n = n

-> 24 * 22 = 528

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New answer posted

8 months ago

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos^-(x) – 2sin^-¹(x) = cos^-¹(2x) is equal to:

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Contributor-Level 10

cos^-1 x = 2 sin^-1 x = cos^-1 2x

$c o s^{- 1} x - 2 (\frac{π}{2} - c o s^{- 1} x) = c o s^{- 1} 2 x$

$c o s (3 c o s^{- 1} x) = c o s (π + c o s^{- 1} 2 x)$

$4 x^{3} - 3 x = - 2 x$

$4 x^{3} = x \Rightarrow x = 0 \pm \frac{1}{2}$

All satisfy the original equation

Sum = $- \frac{1}{2} + 0 + \frac{1}{2} = 0$

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New answer posted

8 months ago

Let the vectors $\vec{a} = (1 + t) \hat{i} + (1 - t) \hat{j} + \hat{k}, \vec{b} = (1 - t) \hat{i} + (1 + t) \hat{j} + 2 \hat{k}$ and $\vec{c} = t \hat{i} - t \hat{j} + \hat{k}, t \in R$ be such that for $α, β, γ \in R, α \vec{a} + β \vec{b} + γ \vec{c} = \vec{0} \Rightarrow α = β = γ = 0 .$ Then, the set of all values of t is:

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Class 12th Physics Moving Charges and Magnetism

Contributor-Level 10

By its given condition

$∴ \vec{a}, \vec{b}, \vec{c}$ are linearly independent vectors is $[\vec{a} \vec{b} \vec{c}] \neq 0$

Now, $[\vec{a} \vec{b} \vec{c}]$ = $| \begin{matrix} 1 + t & 1 - t & 1 \\ 1 - t & 1 + t & 2 \\ t & - t & 1 \end{matrix} | \neq 0$ $\begin{matrix} \end{matrix}$

$R_{1} - R_{2}, R_{2} + R_{3}$

$R_{1} - R_{2} \Rightarrow | \begin{matrix} 0 & 0 & - 3 \\ 1 & 1 & 3 \\ t & - t & l \end{matrix} | \neq 0 \Rightarrow t \neq 0$

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New answer posted

8 months ago

Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3s and 4s respectively. If their moments of inertia are in the ratio of 3 : 2, then the ratio of their magnetic moments will be:

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Class 12th Inverse Trigonometric Functions

Contributor-Level 10

T₁ = 3 sec T₂ = 4sec

$T = 2 π \sqrt[]{\frac{I}{μ B}}$

$\frac{I_{1}}{I_{2}} = \frac{3}{2} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt[]{\frac{l_{1} μ_{2}}{μ_{1} l_{2}}}$

$\Rightarrow {(\frac{3}{4})}^{2} = \frac{I_{1}}{I_{2}} (\frac{μ_{2}}{μ_{1}}) \Rightarrow \frac{μ_{2}}{μ_{1}} = \frac{l_{2}}{l_{1}} * \frac{9}{16} = \frac{2}{3} * \frac{9}{16} = \frac{3}{8}$ $\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{8}{3}$

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New answer posted

8 months ago

Considering only the principal values of the inverse trigonometric functions, the domain of the function f(x) = $c o s^{- 1} (\frac{x^{2} - 4 x + 2}{x^{2} + 3})$ $^{} \frac{^{}}{^{}}$ is:

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Contributor-Level 10

$| \frac{x^{2} + 4 x + 2}{x^{2} + 3} | \leq 1$

${(x^{2} - 4 x + 2)}^{2} \leq {(x^{2} + 3)}^{2}$

$\Rightarrow (2 x^{2} - 4 x + 5) (- 4 x - 1) \leq 0$

$\Rightarrow - 4 x - 1 \leq 0 \to x \geq - \frac{1}{4}$

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New answer posted

8 months ago

Let the function $f (x) = {\frac{l o g_{e} (1 + 5 x) - l o g_{e} (1 + α x)}{\begin{matrix} x & 10 \end{matrix}} \begin{matrix} ; i f x \pm 0 & ; i f x = 0 \end{matrix} b e c o n t i n u o u s a t x = 0 .$

Then a is equal to

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