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6 months ago

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

Different A.P's are constructed with the first term 100, the last term 199 and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is…….

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alok kumar singh

Contributor-Level 10

$d_{1} = \frac{199 - 100}{2} \in l$

$d_{2} = \frac{199 - 100}{3} = 33$

$d_{3} = \frac{199 - 100}{4} \in l$

$d_{n} = \frac{199 - 100}{i + 1} \in l$

$\Rightarrow d i = 33 + 11 o r 9$

$∴$ sum of common differences = 33 + 11 + 9 = 53

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6 months ago

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

If for some q, q, r $\in$ R, not at all have same sign, one of the roots of the equation $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$ is also a root of the equation x² + 2x – 8 = 0, then $\frac{q^{2} + r^{2}}{p^{2}}$ is equal to………….

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Vishal Baghel

Contributor-Level 10

Let and are the roots of $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + q^{2} + r^{2} = 0$

$∴ α + β > 0 a n d α β > 0$ Also, it has a common root with x² + 2x – 8 = 0

$∴$ The common root between above two equations is 4.

$\Rightarrow 16 (p^{2} + q^{2}) - 8 q (p + r) + q^{2} + r^{2} = 0 \Rightarrow (16 p^{2} - 8 p q + q^{2}) + (16 q^{2} - 8 q r + r^{2}) = 0$

$\Rightarrow {(4 p - q)}^{2} + {(4 q - r)}^{2} = 0 \Rightarrow q = 4 p a n d r = 16 p ∴ \frac{q^{2} + r^{2}}{p^{2}} = \frac{16 p^{2} + 256 p^{2}}{p^{2}} = 272$

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6 months ago

Class 12th Physics Thermal Properties of Matter

As per the given figure, two plates A and B of thermal conductivity K and 2K are joined together of form a compound plate. The thickness of plates are 4.0cm and 2.5cm respectively and the area the area of cross-section is 120 cm² for each plate. The equivalent thermal is 120 cm² for each plate. The equivalent thermal conductivity of the compound plat is $(1 + \frac{5}{α}) K,$ then the value of a will be________.

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alok kumar singh

Contributor-Level 10

k_{eq=
$\frac{L_{1} + L_{2}}{\frac{L_{1}}{k_{1}} + \frac{L_{2}}{k_{2}}} = \frac{4 + 2.5}{\frac{4}{k} + \frac{2.5}{2 k}}$}

_{keq=
$\frac{6.5}{\frac{8 + 2.5}{2 k}} = \frac{6.5}{10.5} * 2 k = \frac{13 * 2}{21} k$}

_{=
$(1 + \frac{5}{51}) k$}

_{=
$(1 + \frac{5}{α})$
=21}

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6 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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6 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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6 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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6 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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6 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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6 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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6 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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