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11 months ago

Class 12th Statistics

Let the mean and the variance of 20 observations x1, x2,…….x20 be 15 and 9, respectively. For $\in R$ , if the mean of ${(x_{1} + α)}^{2}, {(x_{2} + α)}^{2}, . . . ., {(x_{20} + α)}^{2}$ is 178, then the square of the maximum value of is equal to………….

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Vishal Baghel

Contributor-Level 10

$\bar{x} = 15 \Rightarrow \sum_{i = 1}^{20} x_{i} = 300$

$\frac{\sum_{i = 1}^{20} {(x_{i} - 15)}^{2}}{20} = 9 \Rightarrow \sum_{i = 1}^{20} {(x_{i} - 15)}^{2} = 180$

$\sum_{i = 1}^{20} {(x_{i} + α)}^{2} = 178 * 20 = 3560$

$4680 + 2 α (300) + 20 α^{2} = 3560$

$α^{2} + 30 α + 234 - 178 = 0$

= 2, 28

$α_{m a x} 2 = {(- 2)}^{2} = 4$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let $S = {θ \in (0, 2 π) : 7 c o s^{2} θ - 3 s i n^{2} θ - 2 c o s^{2} 2 θ = 2} .$ Then, the sum of roots of all the equations $x^{2} - 2 (t a n^{2} θ + c o t^{2} θ) x + 6 s i n^{2} θ = 0, θ \in S, i s . . . . . . . . .$

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Vishal Baghel

Contributor-Level 10

$\frac{7}{2} (1 + c o s 2 θ) - \frac{3}{2} (1 - c o s 2 θ) - 2 c o s^{2} 2 θ = 2$

Put cos 2 $θ$ = t

Equation 2t²– 5t = 0, t (2t – 5) = 0

$t = 0, \frac{5}{2}$

cos 20 = 0, 0 < 2 < 4

$x_{1} + x_{2} = 2 (t a n^{2} θ + c o t^{2} θ) \Rightarrow ?$

$= 2 (1 + 1) + 2 (1 + 1) + 2 (1 + 1) + 2 (1 + 1) = 16$

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

The position vector of 1kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) m s^{- 1} .$ The magnitude of is angular momentum is $\sqrt[]{x} N m$ where x is________.

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Payal Gupta

Contributor-Level 10

$\vec{L} = \vec{r} * m \vec{v}$

= $(3 \hat{i} - \hat{j}) * (3 \hat{j} + \hat{k})$

$= 9 \hat{k} + 3 (- \hat{j}) - \hat{i}$

$= - \hat{i} - 3 \hat{j} + 9 \hat{k}$

$| \vec{L} |$ $=$ =√ $1 + 9 + 81$ $=$ =√91

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

Let f(x) = $3^{{(x^{2} - 2)}^{3} + 4}, x \in R .$ Then which of the following statements are true?

P : x = 0 is a point of local minima of f

Q : x = $\sqrt[]{2}$ is a point of inflection of f

R : $f'$ is increasing for x > $\sqrt[]{2}$

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Vishal Baghel

Contributor-Level 10

$f (x) = 3^{{(x^{2} - 2)}^{3} + 4} = 81 . 3^{{(x^{2} - 2)}^{3}}$

$f' (x) = 81 . 3^{{(x^{2} - 2)}^{3}} . l n 3.3 {(x^{2} - 2)}^{2} . 2 x$

From graph : P, Q, R

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

An AC source is connected to an inductance of 100 mH, a capacitance of 100 $μ F$ and a resistance of 120 $Ω$ as shown in figure. The time in which the resistance having a thermal capacity 2 J/°C will get heated by 16°C is________ s.

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Payal Gupta

Contributor-Level 10

$x_{R} = 120 ?$

$x_{L} = ? L = 100 * 100 * 1 0^{? 3} 10 ?$

$x_{C} = \frac{1}{? c} = \frac{1}{100 * 100 * 1 0^{? 6}} = 100 ?$

$z = \sqrt[]{12 0^{2} + {(100 ? 10)}^{2}} = \sqrt[]{12 0^{2} + 9 0^{2}}$

= 150 $?$

$? 32 = \frac{160}{75} * t ? t = \frac{32 * 75}{160} = 15$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let S = {1, 2, 3,…., 2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022) = 1, is:

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Vishal Baghel

Contributor-Level 10

S = {1,2, 3, …., n, 2022}

HCF (n, 2022) = 1

2022 = 2 * 1011 ->3 * 337

2022 = 2 * 3 * 337 (prime factorization)

Let n (A) = no members divisible by 2 = 1011

Let n (B) = no members divisible by 3 = 674

Let n (C) = no members divisible by 337 = 6

$n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (C \cap A) + n (A \cap B \cap C)$

= 1011 + 674 + 6 – 337 – 2 – 3 + 1

= 1350

$n (A \cap B) = 337$

$n ((A \cup B \cup C)') = 2022 - 1350 = 672$

Prob. $= \frac{672}{2022} = \frac{336}{1011} = \frac{112}{337}$

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant $h = 6.62 * 10^{- 34} J . s$ )

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alok kumar singh

Contributor-Level 10

$e V s = h v - ?$

$\begin{array}{r} At V_{s} = 0 \Rightarrow h v = ? \\ \Rightarrow ? = [6.62 * 10^{- 34}] [10^{14}] [5.5] \\ \Rightarrow ? = \frac{ [6.62 * 10^{- 34}] [10^{14}] [5.5] e V}{(1.6 * 10^{- 19}]} \end{array}$

$= 2.27$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

The number of points, where the function f : R R,

$f (x) = | x - 1 | c o s | x - 2 | s i n | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |,$ is NOT differentiable, is:

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Vishal Baghel

Contributor-Level 10

$f (x) = | x - 1 | c o s | x - 2 | s i n | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$

$f (x) = | x - 1 | c o s (x - 2) s i n | x - 1 | + (x - 3) | x - 1 | | x - 4 |$

x = 1, 4 (doubtful points)

Diff. at x = 1

$\underset{x \to 1}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1}{l i m} \frac{| x - 1 | (s i n | x - 1 | c o s (x - 2) + (x - 3) | x - 4 |)}{x - 1}$

$\begin{array}{l} R H D = \underset{x \to 4^{+}}{l i m} \frac{3 (s i n 3 c o s 2)}{0^{+}} = + \infty \\ L H D = \underset{x \to 4^{-}}{l i m} \frac{3 (s i n 3 c o s 2)}{0^{-}} = - \infty \end{array}} \Rightarrow$ Not diff. at x = 4

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

Let the focal chord of the parabola P : y² = 4x along the line L : y = mx + c, m > 0 meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola H : x² – y² = 4. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is:

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Vishal Baghel

Contributor-Level 10

L : y = mx + c, m > 0

y = m (x – 1)

$\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$

$y = m x \pm \sqrt[]{4 m^{2} - 4}$

$\pm \sqrt[]{4 m^{2} - 4} = - m, m > 0$

$\Rightarrow \sqrt[]{4 m^{2} - 4} = m$

$M (\frac{5 + \sqrt[]{21}}{2}, \sqrt[]{3} + \sqrt[]{7}), N (\frac{5 - \sqrt[]{21}}{2}, \sqrt[]{3} - \sqrt[]{7})$

$=$ $2$ $(27)$ $=$ $2$

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11 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ in the $x$ -direction. The magnitude of gravitational potential on the $x$ -axis at a distance $x$ , taking its value to be zero at infinity is:

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alok kumar singh

Contributor-Level 10

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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