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6 months ago

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

Let $P_{1} : \vec{r} . (2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let P₂be another plane which passes through the points $(2, - 3, 2), (2, - 2, - 3) a n d (1, - 4, 2)$ . If the direction ratios of the line of intersection of P₁ and P₂ be 16, , then the value of + is equal to……………

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alok kumar singh

Contributor-Level 10

2x + y – 3z = 4

$π_{2} = | \begin{matrix} x - 2 & y + 3 & z - 2 \\ 0 & 1 & - 5 \\ - 1 & - 1 & 0 \end{matrix} | = 0$

$π_{2} : - 5 x + 5 y + z + 23 = 0$

now line lying in both the planes have DR.

$\frac{a}{1 + 15} = \frac{b}{15 - 2} = \frac{c}{10 + 5}$

$\frac{a}{16} = \frac{b}{13} = \frac{c}{15}$

So direction ratio's a : b : c = 16 : 13 : 15

$x + f' (x) = f' (0)$

$f' (0) = 1 - c^{2} f r o m e q u a t i o n (i)$

$x + f' (x) = 1 - c^{2}$

$\frac{x^{2}}{2} + f (x) = (1 - c^{2}) x + d$

f(0) = 0

$f (x) = \frac{- x^{2}}{α} + (1 - c^{2}) x$

c = 3/2

$f " (x) = - 1 = 2 (c + 1)$

$f (x) = \frac{- x^{2}}{2} + (\frac{- 5}{4}) x$

now $(f (1) + f (2) + . . . . . . . f (20))$

$= \frac{- 1}{2} (1^{2} + 2^{2} + . . . . . . + 2 0^{2}) - \frac{5}{4} (1 + 2 + . . . . + 20)$

$= \frac{20 \cdot 21}{2} \cdot \frac{(- 82 - 15)}{12}$

now $2 | f (1) + f (2) + . . . . + f (20) |$

$= \frac{20 \cdot 21 \cdot 97}{12} = 3395$

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6 months ago

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

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > 0, b > 0,$ be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (2 \sqrt[]{2} + \sqrt[]{14}) .$ If the eccentricity H is $\frac{\sqrt[]{11}}{2},$ then the value of a² + b² is equal to………….

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alok kumar singh

Contributor-Level 10

Given 2a + 2b = $4 (2 \sqrt[]{2} + \sqrt[]{14})$

$= 4 \sqrt[]{2} (2 + \sqrt[]{7})$

$b^{2} = a^{2} (\frac{11}{4} - 1)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2} a$

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6 months ago

Class 12th Laws of Motion

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

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Payal Gupta

Contributor-Level 10

By conservation of Angular momentum

L_i = L_f

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

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6 months ago

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

Let $c, k \in R . I f f (x) = (c + 1) x^{2} + (1 - c^{2}) x + 2 k$ and $f (x + y) = f (x) + f (y) - x y,$ for all x, y $\in$ R, then the value of $| 2 (f (1) + f (2) + f (3) + . . . . + f (20)) |$ is equal to………….

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alok kumar singh

Contributor-Level 10

f (x) = (c + 1)x² + (1 – c²)x + 2k

$f' (x) = ((c + 1)) 2 x + (1 - c^{2}) . . . . . . . . . . . . . (i)$

$f' (x) = \underset{y \to 0}{l i m} I \frac{f (x + y) - f (x)}{x + y - x}$

$= \underset{y \to 0}{l i m} \frac{f (y) - x y}{y}$

$f' (x) = \underset{y \to 0}{l i m} \frac{f (y)}{y} - x$

$x + f' (x) = \underset{y \to 0}{l i m} \frac{f (y) - f (0)}{y - 0} a s f (c) = 0$

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6 months ago

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

$50 t a n (3 t a n^{- 1} (\frac{1}{2}) + 2 c o s^{- 1} (\frac{1}{\sqrt[]{5}})) + 4 \sqrt[]{2} t a n (\frac{1}{2} t a n (2 \sqrt[]{2}))$ is equal to…………….

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alok kumar singh

Contributor-Level 10

$c o s^{- 1} \frac{1}{\sqrt[]{5}} = c o t^{- 1} \frac{1}{2}$

$t a n (2 (t a n^{- 1} \frac{1}{2} + c o t^{- 1} \frac{1}{2})) + t a n^{- 1} (\frac{1}{2}) = t a n (π + t a n^{- 1} \frac{1}{2}) = \frac{1}{2}$

$t a n 2 α = 2 \sqrt[]{2} \Rightarrow \frac{2 t a n α}{1 - t a n^{2} α} = 2 \sqrt[]{2}$

$\sqrt[]{2} t a n^{2} α + t a n α - \sqrt[]{2} = 0$

$t a n α = \frac{1}{\sqrt[]{2}}, (- \sqrt[]{2}) \to r e j e c t e d$

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6 months ago

Class 12th System of Particles and Rotational Motion

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

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Payal Gupta

Contributor-Level 10

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

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6 months ago

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

$50 t a n (3 t a n^{- 1} (\frac{1}{2}) + 2 c o s^{- 1} (\frac{1}{\sqrt[]{5}})) + 4 \sqrt[]{2} t a n (\frac{1}{2} t a n (2 \sqrt[]{2}))$ is equal to…………….

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alok kumar singh

Contributor-Level 10

$c o s^{- 1} \frac{1}{\sqrt[]{5}} = c o t^{- 1} \frac{1}{2}$

$t a n (2 (t a n^{- 1} \frac{1}{2} + c o t^{- 1} \frac{1}{2})) + t a n^{- 1} (\frac{1}{2}) = t a n (π + t a n^{- 1} \frac{1}{2}) = \frac{1}{2}$

$t a n 2 α = 2 \sqrt[]{2} \Rightarrow \frac{2 t a n α}{1 - t a n^{2} α} = 2 \sqrt[]{2}$

$\sqrt[]{2} t a n^{2} α + t a n α - \sqrt[]{2} = 0$

$t a n α = \frac{1}{\sqrt[]{2}}, (- \sqrt[]{2}) \to r e j e c t e d$

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6 months ago

Class 12th Electrostatic Potential and Capacitance

A person is standing in an elevator. In which situation, he experiences weight loss?

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Payal Gupta

Contributor-Level 10

F.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss.

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6 months ago

Class 12th Electromagnetic Waves

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} l o g_{e} (\frac{k t}{β t});$ where α and β are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of will be

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Payal Gupta

Contributor-Level 10

$P = \frac{α}{β} l o g_{e} (\frac{k t}{β x})$

$\frac{k t}{β x}$ = Dimensionless

$β = \frac{k t}{x} = \frac{[M L^{2} T^{- 2} k^{- 1}] [k]}{[L]}$

$\frac{α}{β} =$ dimensionless

= dimensionless of

= MLT²

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6 months ago

Class 12th Biomolecules

Match List-I with List-II.

List – I List – II

(A) Sucrose

...more

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Payal Gupta

Contributor-Level 10

(a) Sucrose → α-D glucose and β-D fructose

(b) Lactose → β-D- galactose and β-D glucose

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