Class 12th

The minimum value of the twice differentiable function f(x) = $\int_{0}^{x} e^{x - t} f' (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ is:

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Contributor-Level 10

$f (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t$

$f' (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t + e^{x} . \frac{f' (x)}{e^{x}} - [(2 x - 1) . e^{x} + (x^{2} - x + 1) . e^{x}]$

$\begin{array}{l} f (x) = (2 x + 1) . e^{x} - 2 e^{x} + c \\ f (x) = e^{x} (2 x - 1) c = 0 \end{array}$

$f (- \frac{1}{2}) = \frac{- 2}{\sqrt[]{e}}$

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Class 12th Physics

An electron (mass m) with an initial velocity $\vec{v} = v_{0} \hat{i} (v_{0} > 0)$ is moving in an electric field $\vec{E} = - E_{0} \hat{i} (E_{0} > 0)$ where E₀ is constant. If at t = 0 de Broglie wavelength is $λ_{0} = \frac{h}{m v_{0}}$ , then its de Broglie wavelength after time t is given by

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Contributor-Level 10

Given that $\vec{v} = v_{0} i a n d \vec{E} = - E_{0} \hat{i}$

$\vec{v} (t) = v_{0} \hat{i} + \frac{e E t}{m} \hat{i}$

So, de Broglie wave length, $λ = \frac{h}{m v}$

$\Rightarrow λ = \frac{h}{m v_{0} (1 + \frac{q E_{0}}{m v_{0}} t)} = \frac{λ_{0}}{(1 + \frac{q E_{0}}{m v_{0}} t)}$

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Class 12th Functions

Let a, b and $γ$ be three positive real number. Let f(x) = ax⁵ + bx³ + $γ x, x \in R$ and g : R ® R be such that $g (f (x)) = x$ for all x $\in R .$ If $a_{1}, a_{2}, a_{3} . . . . ., a_{n}$ be in arithmetic progression with mean zero then the value of $f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i})))$ is equal to:

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Contributor-Level 10

Case I

= = 0 then

f (x) = yx

$g (x) = \frac{x}{y}$

$\frac{1}{x} \sum_{i = 1}^{n} f (a_{i}) \Rightarrow \frac{y}{x} (a_{1} + a_{2} + . . . . + a_{x}) = 0$

$\Rightarrow f (g (0)) \Rightarrow f (0)$

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If the minimum value of f(x) = $\frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0,$ $\frac{^{}}{} \frac{}{^{}}$ is 14, then the value of is equal to:

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Contributor-Level 10

$\frac{5 x^{2}}{2} + \frac{α}{2 x^{5}} + \frac{α}{2 x^{5}} \geq 7 {(\frac{α^{2}}{2^{7}})}^{\frac{1}{7}}$

$\frac{7 . {(α)}^{2 / 7}}{2} = 14$

${(α^{2})}^{\frac{1}{7}} = 2^{2} \Rightarrow α = {(2^{2})}^{\frac{7}{2}} = 2^{7}$

=128

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Class 12th Physics

A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $λ,$ calculate the change of microscope's resolving power due to oil and choose the correct option

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Contributor-Level 10

Resolving power

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The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

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Contributor-Level 10

$\frac{h - c o s θ}{2} = \frac{k - s i n θ}{3} = \frac{w - 0}{1}$

$= \frac{- 1 (2 c o s θ + 3 s i n θ - 6)}{14} \Rightarrow h = \frac{c o s - 2 (2 c o s θ + 3 s i n θ - 6)}{14}$

$k = \frac{5 s i n θ - 6 c o s θ + 18}{14}$

${(5 h + 6 k - 12)}^{2} + 4 {(3 h + 5 k - 9)}^{2} = 1$

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Let $S_{1} = {z_{1} \in C : | z_{1} - 3 | = \frac{1}{2}}$ and $S_{2} = {z_{2} \in C : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |} .$ Then, for $z_{1} \in S_{1}$ and $s_{2} \in S_{2},$ the least value of $| z_{2} - z_{1} | i s :$

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Contributor-Level 10

$S_{1} = {z, \in c : ⌊ z_{1} - 3 ⌋ = \frac{1}{2}} a n d S_{2} = {z_{2} \in c : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |}$

Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2},$ the least value of |z₂ – z₁|

${| z_{2} + | z_{2} - 1 | |}^{2} = {| z_{2} - | z_{2} + 1 | |}^{2}$

$\Rightarrow (z_{2} + {\bar{z}}_{2}) (| z_{2} - 1 | + | z_{2} + 1 | - 2) = 0$

$∴ z_{2} + {\bar{z}}_{2} = 0 o r | z_{2} - 1 | + | z_{2} + 1 | - 2 = 0$

$∴$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $| z_{1} - 3 | = \frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$

Clearly $| z_{1} - z_{2} | m i n = \frac{5}{2} - 1 = \frac{3}{2}$

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Class 12th Matrices

Let the matrix $A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$ and the matrix B₀ = A²⁹ + 2A⁹⁸. If B_n = Adj(B_n=1) for all $n \geq 1,$ then det

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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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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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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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