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6 months ago

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

Let $\vec{a} = α \hat{i} a + 3 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} - β \hat{j} + 4 \hat{k} a n d \vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k} w h e r e α, β \in R,$ be three vectors. If the projection of $\vec{a} o n \vec{c} i s \frac{10}{3} a n d \vec{b} * \vec{c} = - 6 \hat{i} + 10 \hat{j} + 7 \hat{k},$ then the value of + is equal to:

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alok kumar singh

Contributor-Level 10

$\vec{a} \cdot \hat{c} = \frac{α + 6 + 2}{\sqrt[]{1} + 4 + 4}$

$\frac{10}{3} = \frac{8 + α}{3} \Rightarrow α = 2$

$\vec{b} * \vec{c} = \hat{i} (2 β - 8) + \hat{j} (10) + \hat{k} (6 + β)$

$= - 6 \hat{i} + 10 \hat{j} + 7 \hat{k}$

So = 1

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6 months ago

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

Let f : R R be a function defined by:

$f (x) = [\begin{matrix} m a x (t^{3} - 3 t) & ; & t \leq x; x \leq 2 & x^{2} + 2 x - 6 \\ ; & 2 < x < 3 & [x - 3] + 9 & ; \\ 3 \leq x \leq 5 & 2 x + 1 & ; & x > 5 \end{matrix}$

where [t] is the greatest integer less than or equal to t. Let me be the number of points where f is not differentiable and $I = \int_{- 2}^{2} f (x) d x .$ Then the ordered pair (m, I) is equal to:

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alok kumar singh

Contributor-Level 10

Draw g(t) = t³ – 3t

g'(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³– 3t) = ${\begin{matrix} t^{3} - 3 t & ; - 2 < t < - 1 \\ 2 & ; - 1 < t < 2 \end{matrix}$

$I = \int_{- 2}^{2} f (x) d x$

= $\int_{- 2}^{- 1} (t^{3} - 3 t) d t + \int_{- 1}^{2} 2 d t$

I = $\frac{27}{4}$

again rewrite the f(x)

$f (x) = {\begin{matrix} \begin{matrix} x^{3} - 3 x & 2 \end{matrix} & ; & \begin{matrix} x \leq - 1 & - 1 < x \leq 2 \end{matrix} \\ \begin{matrix} x^{2} + 2 x - 6 & 9 \end{matrix} & ; & \begin{matrix} 2 < x < 3 & 3 \leq x < 4 \end{matrix} \\ \begin{matrix} 10 & 11 & 2 x + 1 \end{matrix} & ; & \begin{matrix} 4 \leq x < 5 & x = 5 & x > 5 \end{matrix} \end{matrix}}$

$f' (x) = {\begin{matrix} 3 x^{2} - 3 & ; & x < - 1 & 0 & ; & - 1 < x < 2 \\ 2 x + 2 & ; & 2 < x < 3 & 0 & ; & 3 < x < 4 \\ 0 & ; & 4 < x < 5 & 2 & ; & x > 5 \end{matrix}}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

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6 months ago

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

If the mirror image of the point (2, 4,7) in the plane 3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to:

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alok kumar singh

Contributor-Level 10

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

$\frac{x - 2}{3} = \frac{y - 4}{- 1} = \frac{z - 7}{4} = \frac{- 2 (6 - 4 + 28 - 2)}{3^{2} + {(- 1)}^{2} + 4^{2}}$

Let $\frac{x - 2}{3} = \frac{y - 4}{- 1} = \frac{z - 7}{4} = \frac{- 28}{13} = λ$

$x = 3 λ + 2 y = - λ + 4 z = 4 λ + 7$

Now according to the question

(a, b, c) = $(3 λ + 2, - λ + 4, 4 λ + 7)$

Now $2 a + b + 2 c = 6 λ + 4 - λ + 4 + 8 λ + 14$

= $13 λ + 22$

= 28 + 22 [Use $λ = \frac{- 28}{13}$ ]

= 6

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6 months ago

Class 12th Semiconductor Electronics

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x * 10⁴. The value of x is_____________.

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alok kumar singh

Contributor-Level 10

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} * \frac{R_{0}}{R_{i}}$

= ${(\frac{10 * 1 0^{- 3}}{100 * 1 0^{- 6}})}^{2} * \frac{2}{1}$

= 2 * 104 = x * 104

= 2

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6 months ago

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

Let the solution curve of the differential equation

$x \frac{d y}{d x} - y = \sqrt[]{y^{2} + 16 x^{2}}, y (1) = 3 b e y = y (x) .$ Then y(2) is equal to:

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alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} = \frac{y}{x} + \frac{\sqrt[]{y^{2} + 16 x^{2}}}{x}$

Put y = $V \cdot x$

differentiable worst = x

$\frac{d y}{d x} = V + x \cdot \frac{d V}{d x}$

V + $x \frac{d V}{d x} = V + \sqrt[]{V^{2} + 16}$

$x \frac{d V}{d x} = \sqrt[]{V^{2} + 16}$

apply variable separable method

$\int \frac{d V}{\sqrt[]{V^{2} + 16}} = \int \frac{d x}{x} + l n C$

$\Rightarrow l n | V + \sqrt[]{V^{2} + 16} | = l n C x$

$\frac{y}{x} + \frac{\sqrt[]{y^{2} + 16 x^{2}}}{x} = C x$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2} + \frac{\sqrt[]{y^{2} + 16.4}}{2} = 8.2$

y2 + 64 = (32 – y)2

y = 15

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6 months ago

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

The probability that a randomly chosen 2 * 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to:

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alok kumar singh

Contributor-Level 10

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$= \frac{10 + 4 *^{10} ? C_{2}}{1 0^{4}} = \frac{10 + \frac{4 * 10.9}{2}}{1 0^{4}} = \frac{19}{1000}$

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6 months ago

Class 12th Physics Nuclei

$\sqrt[]{d_{1}} a n d \sqrt[]{d_{2}}$ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an a particle is approaching a gold nucleus. For d₁ = x d₂, the value of x will be________.

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alok kumar singh

Contributor-Level 10

$\sqrt[]{d}$ $c o t \frac{θ}{2}$

cot² 30° = x cot² 45°

x = 3

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6 months ago

Class 12th Thermodynamics

300 cal. Of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be__________°C.

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alok kumar singh

Contributor-Level 10

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

...more

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

T₂ = 375k

T₂ = 102°C

less

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New answer posted

6 months ago

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

The variation of applied potential and flowing through a given wire is shown in figure. The length of wire is 31.4cm. The diameter of wire is measured as 2.4cm. The resistivity of the given wire is measured as x * 10^-3 $Ω c m .$ The value of x is_________.

[Take p = 3.14]

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alok kumar singh

Contributor-Level 10

R = tanq = tan 45° = $\frac{ρ L}{A}$

$? 1 = ρ * \frac{31.4}{π {(1.2)}^{2}}$

$ρ = \frac{3.14 * 1.2 * 1.2}{31.4}$

$ρ = 12 * 12 * 1 0^{- 3}$

$ρ = 144 * 1 0^{- 3} = x * 1 0^{- 3} Ω - c m$

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New answer posted

6 months ago

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

Let Q and R be two points on the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}$ at a distance $\sqrt[]{26}$ from the point P(4, 2, 7). Then the surface of the area of the triangle PQR is……………

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Vishal Baghel

Contributor-Level 10

Let PT perpendicular to QR

$\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2} = λ \Rightarrow T (2 λ - 1, 3 λ - 2, 2 λ + 1)$ therefore

$2 (2 λ - 5) + 3 (3 λ - 4) + 2 (2 λ - 6) = 0 \Rightarrow λ = 2$

$T (3, 4, 5) ∴ P T = \sqrt[]{1 + 4 + 4} = 3 ∴ Q T = \sqrt[]{26 - 9} = \sqrt[]{17}$

$∴ Δ P Q R = \frac{1}{2} * 2 \sqrt[]{17} * 3 = 3 \sqrt[]{17}$

Therefore square of $a r (Δ P Q R)$ = 153.

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