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11 months ago

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

Suppose y = y(x) be the solution curve to the differential equation $\frac{d y}{d x} - y = 2 - e^{- x}$ such that $\underset{x \to \infty}{l i m} y (x)$ is finite. If a and b are respectively the x and y intercepts of the tangent to the curve at x = 0, then the value of a 4b is equal to…………

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alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} - y = 2 - e^{- x}$

$I . F . = e^{- \int d x} = e^{- x}$

$∴$ soln

$y e^{- x} = \int (2 e^{- x} - e^{- 2 x}) d x$

$y = - 2 + \frac{e^{- x}}{2} + C e^{x}$

as for x $\infty$ y finite c = 0

$∴ y = \frac{e^{- x}}{2} - 2$

$\begin{array}{l} \Rightarrow x + 2 y = - 3 \\ \Rightarrow a = - 3 b = \frac{- 3}{2} \end{array}$

$∴ a = 4 b = - 3 + 6 = 3$

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11 months ago

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

The statement $(\sim (p \Leftrightarrow \sim q)) \land q i s :$

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Vishal Baghel

Contributor-Level 10

(D)

$\sim (p \Leftrightarrow \sim q) \land q = (p \Leftrightarrow \sim q) \land q i s :$

$∴ (\sim (P \Leftrightarrow \sim Q)) \land$ q is equivalent to $(p \Rightarrow q) \land$ p

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11 months ago

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

The plane passing through the line L : $l x - y + 3 (1 - l) z = 1, x + 2 y - z = 2$ and perpendicular to the plane 3x + 2y + z = 6 is 3x 8y + 7z = 4. If is the acute angle between the line L and the y-axis, then 415 cos2 is equal to……….

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alok kumar singh

Contributor-Level 10

$L_{1} : l x - y + 3 (1 - z) = 1, x + 2 y - z = 2$

plane containing the line P : 3x – 8y + 7z = 4

If $\vec{n}$ be vector parallel to L.

then $\vec{n} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ l & - 1 & 3 (1 - l) \\ 1 & 2 & - 1 \end{matrix} | = (6 l - 5) \hat{i} + (3 - 2 l) \hat{j} + (2 l + 1) \hat{k}$ as P containing the line

$∴ 3 (6 l - 5) - 8 (3 - 2 l) + 7 (2 l + 1) = 0$

$\Rightarrow l = \frac{2}{3}$

If be the acute angle between line L & Y axis then cos = $\frac{5 / 3}{\sqrt[]{1 + \frac{25}{9} + \frac{49}{9}}} = \frac{5}{\sqrt[]{83}}$

$∴ 415 c o s^{2} θ = 125$

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11 months ago

Class 12th Electromagnetic Waves

The intensity of the light from a bulb incident on a surface is 0.22 W/m². The amplitude of the magnetic field in this light – wave is__________ * 10^-9 T.

(Given : Permittivity of vacuum $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} - m^{- 2},$ speed of light in vacuum c = 3 * 10⁸ ms^-1)

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alok kumar singh

Contributor-Level 10

$\frac{B_{0}^{2}}{2 μ_{0}} C$

$B_{0}^{2} = \frac{I 2 μ_{0}}{C}$

$B_{0}^{2} = \frac{0.22 * 2 * 4 * 1 0^{- 7}}{9 * 108}$

B₀ = 42.9 * 10^-9

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11 months ago

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

$t a n (2 t a n^{- 1} \frac{1}{5} + s e c^{- 1} \frac{\sqrt[]{5}}{2} + 2 t a n^{- 1} \frac{1}{8})$ is equal to:

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Vishal Baghel

Contributor-Level 10

$t a n (2 t a n^{- 1} \frac{1}{5} + s e c^{- 1} \frac{\sqrt[]{5}}{2} + 2 t a n^{- 1} \frac{1}{8}) t a n (2 t a n^{- 1} \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} * \frac{1}{8}} + s e c^{- 1} \frac{\sqrt[]{5}}{2})$

$= t a n (t a n^{- 1} \frac{3}{4} + t a n^{- 1} \frac{1}{2}) = t a n (t a n^{- 1} \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}})$

$= t a n (t a n^{- 1} \frac{\frac{5}{4}}{\frac{5}{8}}) = 2$

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11 months ago

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

The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $σ$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 σ^{2}$ is equal to………

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alok kumar singh

Contributor-Level 10

$\bar{x} = \frac{\sum x i}{40} = 30 \Rightarrow \sum x i = 1200$ ………. (i)

$α^{2} = \frac{1}{40} \sum x i_{i}^{2} - {(30)}^{2} = 25$

$\Rightarrow \sum x_{i}^{2} = 37000$

after omitting two wrong observations

$\sum y_{i}^{2} = 37000 - 144 - 100 = 36756$

$∴ 38 a^{2} = 36756 - 36158 = 238$

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11 months ago

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

Let S = ${θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16} .$ Then $n (S) + \sum_{θ \in S}^{} (s e c (\frac{π}{4} + 2 θ) c o s e c (\frac{π}{4} + 2 θ))$ is equal to:

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Vishal Baghel

Contributor-Level 10

$S = {θ \in [0, 2 π] : 8^{2 s i n^{2} x} + 8^{2 c o s^{2} x} = 16}$

Now apply AM $\geq G M$ for $\frac{8^{2 s i n^{2} x} + 8^{2 c o s^{2} x}}{2} \leq {(8^{2 s i n^{2} x + 2 c o s^{2} x})}^{\frac{1}{2}} \Rightarrow 8^{2 s i n^{2} x} = 8^{2 c o s^{2} x}$

$s i n^{2} θ = c o s^{2} θ$ $∴ θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

$= 4 + [c o s e c (\frac{π}{2} + π) + c o s e c (\frac{π}{2} + 3 π) + c o s e c (\frac{π}{2} + 5 π) + c o s e c (\frac{π}{2} + 7 π)]$

$= 4 - 2 (4) = - 4$

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11 months ago

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

If the sum of solutions of system of equations 2sin²q - cos2q = 0 and 2 cos²q + 3sinq = 0 in the interval [0, 2p] is kp, then k is equal to…….

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alok kumar singh

Contributor-Level 10

$2 s i n^{2} θ = c o s 2 θ = 1 - 2 s i n^{2} θ$

$\Rightarrow 4 s i n^{2} θ = 1 \Rightarrow s i n θ = \pm \frac{1}{2}$

$\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}$

$S u m \frac{7 π}{6} + \frac{11 π}{6} = 3 π \Rightarrow k = 3$

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11 months ago

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

$\sum_{k = 1}^{10} \frac{k}{k^{4} + k^{2} + 1} = \frac{m}{n};$ where m and n are co-prime, then m + n is equal to…….

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alok kumar singh

Contributor-Level 10

$\sum_{k = 1}^{10} \frac{k}{k^{4} + k^{2} + 1}$

$= \frac{1}{2} \sum_{k = 1}^{10} [\frac{1}{k^{2} - k + 1} - \frac{1}{k^{2} + k + 1}]$

$= \frac{1}{2} [1 - \frac{1}{111}] = \frac{110}{222} = \frac{55}{111} = \frac{m}{n}$

$∴ m + n = 166$

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11 months ago

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

Let S = ${θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16} .$ Then $n (S) + \sum_{θ \in S}^{} (s e c (\frac{π}{4} + 2 θ) c o s e c (\frac{π}{4} + 2 θ))$ is equal to:

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