Class 12th

The domain of the function f(x) = $s i n^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) i s :$

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Contributor-Level 10

$- 1 \leq \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$

$\Rightarrow 0 \leq \frac{2 x^{2} - x + 9}{x^{2} + 2 x + 7} & \frac{- 5 x - 5}{x^{2} + 2 x + 7} \leq 0$

$x \in R & - 1 \leq x < \infty$

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Which of the following physical quantities have the same dimensions?

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Vishal Baghel

Contributor-Level 10

Electric Displacement

$\begin{array}{l} \vec{D} = \in o \vec{E} \\ [D] = [\in_{0} \vec{E}] = [\in_{0} \frac{σ}{\in_{0}}] \\ [D] = [σ] \end{array}$

Electric displacement $(\vec{D})$ and surface change density

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Class 12th Chemistry

250g solution of D-glucose in water contains 10.8% of carbon by weight. The molality of the solution is nearest to

(Given : Atomic weights are, H, 1u ; C, 12u ; O, 16u)

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Contributor-Level 10

$C_{6} H_{12} O_{6} \to G l u c o s e$

$\frac{M a s s o f C}{m a s s o f g l u c o s e} = \frac{72}{180}$

% of C = 10.8 = $\frac{M a s s o f C}{m a s s o f s o l u t i o n} * 100$

$\frac{10.8 * 250}{100} m a s s o f C$

$∴ M o l a r i t y = \frac{0.375}{0.1825} = 2.055 = 2.06$

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Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} * \vec{b}) . (\vec{b} * \vec{c}) + (\vec{b} * \vec{c}) . (\vec{c} * \vec{a}) + (\vec{c} * \vec{a}) . (\vec{a} * \vec{b})$ = 168, then $| \vec{a} | + | \vec{b} | + | \vec{c} |$ is equal to:

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Contributor-Level 10

Given : $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = c o s θ (s a y)$

$| \vec{a} | | \vec{b} | | \vec{c} | = 14$

$(\vec{a} * \vec{b}) \cdot (\vec{b} * \vec{c})$

$= \vec{a} \cdot [(\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}]$

$= (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - {| \vec{b} |}^{2} \vec{a} \cdot \vec{c}$

$= | \vec{a} | {| \vec{b} |}^{2} | \vec{c} | (c o s^{2} θ - c o s θ) = 14 | \vec{b} | (c o s^{2} θ - c o s θ)$

Similarly, $(\vec{b} * \vec{c}) \cdot (\vec{c} * \vec{a})$

= $| \vec{b} | {| \vec{c} |}^{2} | \vec{a} | (c o s^{2} θ - s i n θ) = 14 | \vec{c} | (c o s 2 θ - s i n θ) & (\vec{c} * \vec{a}) \cdot (\vec{a} * \vec{b})$

$= | \vec{c} | {| \vec{a} |}^{2} | \vec{b} | (c o s^{2} θ - s i n θ) = 14 | \vec{a} | (c o s^{2} θ - s i n θ)$

Given : 14 $(c o s^{2} θ - c o s θ) (| \vec{a} | + | \vec{b} | + | \vec{c} |) = 168$

$| \vec{a} | + | \vec{b} | + | \vec{c} | = \frac{12}{c o s^{2} θ - c o s θ} = \frac{12}{\frac{1}{4} - (- \frac{1}{2})} = \frac{12}{\frac{3}{4}} = 16$

Given : $\vec{a}, \vec{b}, \vec{c}$ are coplanar & pair wise equal angle.

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Let $S = {z = x + i y | z - 1 + i | \geq | z |, | z | < 2, | z + i | = | z - 1 |} .$ Then the set of all values of x, for which $w = 2 x + i y \in S$ for some $y \in R,$ is

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Contributor-Level 10

$| z - 1 + i | \geq | z |$

$| z + i | = | z - 1 |$

W = (2x, y) = (a, y)

Let S represent the line segment AB

For 'B'

x² + y² = 4

x = -y

-> x² = 2

$x = \pm \sqrt[]{2}$

-> $B (- \sqrt[]{2}, \sqrt[]{2})$

$A (\frac{1}{2}, - \frac{1}{2})$

W (2x, y) lies on AB

-> $- \sqrt[]{2} < 2 x \leq \frac{1}{2}$

$(| z | < 2)$

$- \frac{1}{\sqrt[]{2}} < x \leq \frac{1}{4}$

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Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The balls so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

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Contributor-Level 10

$\begin{matrix} 3 R & 2 R & 4 B \\ 5 B & 3 W & 2 W \end{matrix}$

I II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2^nd bag after a ball from I to II was transferred.

$P (\frac{E_{1}}{E}) = \frac{P (E_{1} \cap E)}{P (E)}$

$P (E) = P (E_{1} \cap E) + P (E_{2} \cap E) + P (E_{3} \cap E)$

= $P (E_{1}) \cdot P (\frac{E}{E_{1}}) + . . . . + . . . . .$

= $\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10} = \frac{54}{100}$

$P (E_{1} / E) = \frac{15 / 100}{54 / 100} = \frac{5}{18}$

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At a particular station, the TV transmission tower has a height of 100m. To triple its coverage range, height of the tower should be increased to

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Contributor-Level 10

Final range = 3 Initial Range

$\Rightarrow d_{f} = 3 d_{i}$ $\begin{array}{l} \Rightarrow \sqrt[]{2 h' R} = 3 \sqrt[]{2 h R} \\ \Rightarrow h' = 9 h = 900 m \end{array}$

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Class 12th Differential Equations

A logic gate circuit has two inputs A and B and output Y. The voltage waveforms of A, B and Y are shown below. The logic gate circuit is:

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5 months ago

Let a smooth curve y = f(x) be such that the slope of the tangent at any point (x, y) on it is directly proportional to $(\frac{- y}{x})$ . If the curve passes through the points (1, 2) and (8, 1), then $| y (\frac{1}{8}) |$ is equal to

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Payal Gupta

Contributor-Level 10

Slope of any point P (x, y) to y = f (x) is $\frac{d y}{d x} = - k \frac{y}{x}$

$\Rightarrow \frac{d y}{y} + k \frac{d x}{x} = 0$

Solving the equation the curve is xky = c

It passes (1, 2) c = 2 xky = 2 again it passes (8, 1) 8k = 2 k = $\frac{1}{3}$

$∴$ the equation of curve is $x^{1 / 3}$ y = 2 …… (i)

$∴ | y (\frac{1}{8}) | = | \frac{2}{{(\frac{1}{8})}^{1 / 3}} | = 4$

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What is the half – life period of a radioactive material if its activity drops to 1/16^th of its initial value in 30 years?

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