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6 months ago

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

If $\underset{x \to 0}{l i m} \frac{α e^{x} + β e^{- x} + γ s i n x}{x s i n^{2} x} = \frac{2}{3},$ where $α, β, γ \in R,$ then which of the following is NOT correct?

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Vishal Baghel

Contributor-Level 10

$\underset{x \to 0}{l i m} \frac{α e^{x} + β e^{- x} + γ s i n x}{x s i n^{2} x}$

$= \underset{x \to 0}{l i m} \frac{α e^{x} + β e^{- x} + γ s i n x}{x^{3}}$

$\Rightarrow α + β = 0, α - β + γ = 0, \frac{α + β}{2} = 0, \frac{α}{6} - \frac{β}{6} - \frac{γ}{6} = \frac{2}{3}$

$\Rightarrow β = - α γ = - 2 α α - β - γ = 4 \Rightarrow α + α + 2 α = 4 \Rightarrow γ = 1$

$α = 1, β = - 1, γ = - 2$

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6 months ago

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

$\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + . . . + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256},$ then the maximum value of a is:

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Vishal Baghel

Contributor-Level 10

$\frac{1}{(20 - a) (40 - a)} + \frac{1}{(40 - a) (60 - a)} + \frac{1}{(60 - a) (80 - a)} + . . . . . + \frac{1}{(180 - a) (200 - a)} = \frac{1}{256}$

LHS = $\frac{1}{20} (\frac{1}{20 - a} - \frac{1}{40 - a}) + \frac{1}{20} (\frac{1}{40 - a} - \frac{1}{60 - a}) + . . . + \frac{1}{20} (\frac{1}{180 - a} - \frac{1}{200 - a})$

$= \frac{1}{20} (\frac{1}{20 - a} - \frac{1}{200 - a}) = \frac{1}{20} . \frac{180}{(20 - a) (200 - a)}$

$\Rightarrow a^{2} - 220 a + 4000 - 2304 = 0 \Rightarrow a^{2} 220 a + 1696 = 0$

$a = \frac{220 \pm 204}{8} = 21 2^{2}$

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6 months ago

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

Let A and B be two 3 * 3 non-zero real matrices such that AB is a zero matrix. Then

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Vishal Baghel

Contributor-Level 10

$A_{3 * 3}, B_{3 * 3} & A B = 0$

$\neq 0 \neq 0 \Rightarrow | A | = 0 & | B | = 0$ only .

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6 months ago

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

The Q-value of a nuclear reaction and kinetic energy of the projectile particle, K_p are related as

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Payal Gupta

Contributor-Level 10

X + p → Y + b

$Q = K_{Y} + K_{b} - K_{P}$

$Q + K_{P} = K_{Y} + K_{b}$

$\Rightarrow Q + K_{P} > 0$

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6 months ago

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

If z = 2 + 3 i, then z5 + ${(\bar{z})}^{5}$ is equal to:

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Vishal Baghel

Contributor-Level 10

$z^{5} + {(\bar{z})}^{5}$

$= {(2 + 3 i)}^{5} + {(2 - 3 i)}^{5}$

$= 2 (32 - 720 + 810) = 244$

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6 months ago

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

THE de Broglie wavelength for an electron and a photon are $λ_{e} a n d λ_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{P_{e}} = \frac{h}{\sqrt[]{2 m_{e} K_{e}}} \Rightarrow λ_{e}^{2} = \frac{h^{2}}{2 m_{e} k_{e}} \Rightarrow K_{e} = \frac{h^{2}}{2 m_{e} λ_{e}^{2}}$ - (i)

$λ_{P} = \frac{h}{P_{P}} = \frac{h c}{K_{P}} \Rightarrow K_{P} = \frac{h c}{λ_{P}}$ - (ii)

K_e = K_P

$\Rightarrow λ_{P} α λ_{e}^{2}$

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6 months ago

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

Let R be a relation from the set ${1, 2, 3, . . . . ., 60}$ to itself such that R = ${(a, b) : b = p q}$ where p, q $\geq$ 3 are prime numbers. Then, the number of elements in R is:

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Vishal Baghel

Contributor-Level 10

R = { (a, b): b = pq, where p, q $\geq 3$ are prime}

$\Rightarrow 60 * 11 = 660$

p, q $\in$ {3, 5, 7, 11, 13, 17, 19, 23, 29, 3, 37, 41, 43, 47, 53, 59} total 16

p, q $\in$ {3, 5, 7, 11, 13, 17, 19}

{3, 5, 7, 11, 13, 17, 19}

7 + 3 + 1 = 11

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6 months ago

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

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a alight of wavelength 2440 $\overset{o}{A}$ is used to see the object will be

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Payal Gupta

Contributor-Level 10

Resolving power, = $\frac{d}{1.22 λ} = \frac{24.4 * 1 0^{- 2}}{1.22 * 2440 * 1 0^{- 10}}$

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New answer posted

6 months ago

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

The refracting angle of a prims is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be

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Payal Gupta

Contributor-Level 10

At minimum deviation –

$δ = 2 i - A \Rightarrow i = \frac{δ + A}{2}, r = \frac{A}{2}$

Using snell's low : - $s i n i = μ = s i n r$

$\Rightarrow \frac{δ + A}{2} = \frac{π - A}{2} \Rightarrow δ = π - 2 A$

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6 months ago

Class 12th Electromagnetic Waves

A radar sends an electromagnetic single of electric (E₀) = 2.25 V/m and magnetic field (B₀) = 1.5 * 10^-⁸ T which strikes a target on line of sight at a distance of 3km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?

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Payal Gupta

Contributor-Level 10

$E_{0} = 2.25 \frac{v}{m}$ Speed of declromagnetic signal, $v = \frac{E_{0}}{B_{0}} = \frac{2.25 v / m}{1.5 * 1 0^{- 8} T}$

$B_{0} = 1.5 * 1 0^{- 8} T \Rightarrow v = \frac{225}{150} * 1 0^{8} = 1.5 * 1 0^{8} m / s$

Time to reach each to the same

radar, $t = \frac{2 * 3 * 1 0^{3} m}{1.5 * 1 0^{8} m / s} = 4 * 1 0^{- 5} S$

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