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11 months ago

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

Let the mean and the variance of 5 observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5} b e \frac{24}{5} a n d \frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and a respectively, then (4a + x₅) is equal to:

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alok kumar singh

Contributor-Level 10

$\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{7}{2}$

$x_{1} + x_{2} + x_{3} + x_{4} = 14$

and $\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5} = \frac{24}{5}$

x₅= 10

Variance $\frac{\sum_{i = 1}^{4} x_{i}^{2}}{4} - {(\frac{Σ x i}{4})}^{2} = a$

$\frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}}{4} - \frac{49}{4} = a$

$x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = 4 a + 49$

and $\frac{x_{1}^{2} + x_{2}^{2} + x_{5}^{2} + x_{4}^{2} + x_{5}^{2}}{5} - {(\frac{24}{5})}^{2} = \frac{194}{25}$

$\frac{4 a + 49 + x_{5}^{2}}{5} = \frac{576 + 194}{25}$

49 + 49 + $x_{5}^{2} = \frac{770}{5}$

49 $+ x_{5}^{2} + 49 = 154$

4a + 149 = 154

4a = 5

now 4a + x₅= 15

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11 months ago

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

Let the tangent to the circle C₁ : x² + y² = 2 at the point M(1, 1) intersect the circle $C_{2} : {(x - 3)}^{2}$ + ${(y - 2)}^{2} = 5,$ at two distinct points A and B. If the tangents to C₂at the points A and B intersect at N, then the area of the triangle ANB is equal to:

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alok kumar singh

Contributor-Level 10

Tangent to C₁ at (-1, 1) is T = 0

x(-1) + 4(1) = 2

-x + y = 2

find OP by dropping from (3, 2) to centre

OP = $| \frac{3 - 2 + 2}{\sqrt[]{2}} | = \frac{3}{\sqrt[]{2}}$

AP = $\sqrt[]{r^{2} - O P^{2}}$

$= \sqrt[]{5 - \frac{9}{2}} = \frac{1}{\sqrt[]{2}}$

$t a n θ = \frac{O P}{A P} = \frac{3 / \sqrt[]{2}}{1 / \sqrt[]{2}} = 3$

area of $Δ A B N = \frac{1}{2} A N^{2} \cdot s i n 2 θ$

AN = $\frac{\sqrt[]{5}}{3}$

$= \frac{1}{2} \cdot \frac{5}{9} \cdot (\frac{2 t a n θ}{1 + t a n^{2} θ})$

$= \frac{5}{2.9} * \frac{2.3}{1 + 3^{2}} = \frac{1}{6}$

sin = $\frac{A P}{A N}$

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11 months ago

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

$C_{6} H_{12} O_{6} \overset{Z y m a s e}{\to}$ A $\to_{Δ}^{N a O l} B + C H l_{3}$

The number of carbon atoms present in the product B is______________.

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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11 months ago

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

Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $\frac{π}{2}$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ $a^{2} > b^{2} .$ If e is the eccentricity of the ellipse E, then the value of $\frac{1}{e^{2}}$ is equal to:

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alok kumar singh

Contributor-Level 10

$A P ⊥ B P$

M₁ M₂ = 1

$\frac{2 t}{t^{2} - 3} * \frac{2 / 6}{3 - \frac{1}{t^{2}}} = - 1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $= \frac{2 b^{2}}{a}$

$4 = \frac{2 b^{2}}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b²= a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e} (1 - e^{2})$

e²+ 2e – 1 = 0

$e = \frac{- 2 \pm \sqrt[]{4 + 4}}{2}$

$e = - 1 + \sqrt[]{2}$

now $\frac{1}{e^{2}}$ $=$ $3$ $+$ $2$

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11 months ago

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

Observe structure of the following compounds

The total number of structures / compounds which possess asymmetric atoms is________.

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Vishal Baghel

Contributor-Level 10

No. of compounds containing asymmetric carbon are

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11 months ago

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

Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55g of the compound neutralized 12.5 mL of 1M H₂SO₄ solution. The percentage of nitrogen in the compound is_________ (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

Millie q. of H₂SO₄ used by NH₃ = 12.5 * 1 * 2 = 25

So millimoles of N = 25

Moles of N = 25 * 10^-3

Wt of N = 14 * 25 * 10^-3

% of N = $\frac{14 * 25 * 1 0^{- 3}}{0.55} * 100 = 63.66 ≅ 64$

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11 months ago

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

$\int_{0}^{5} c o s (π (x - [\frac{x}{2}])) d x,$ where [t] denotes greatest integer less than or equal to t, is equal to:

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alok kumar singh

Contributor-Level 10

$I = \int_{0}^{5} c o s (π x - π [\frac{x}{2}]) d x$

$I = \int_{0}^{2} c o s (π x) d x + \int_{2}^{4} c o s (π x - π) d x + \int_{4}^{5} c o s (π x - 2 π) d x$

$I = {\frac{s i n π x}{π} |}_{0}^{2} + {\frac{s i n (π x - π)}{π} |}_{2}^{4} + {\frac{s i n (π x - 2 π)}{π} |}_{4}^{5} = 0$

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11 months ago

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

An acidified magnate solution undergoes disproportionation reaction. The spin- only magnetic moment value of the product having manganese in higher oxidation state is________ B.M. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$3 M n O_{4}^{2 -} + 4 H^{+} \to_{}^{} 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$

No. of unpaired electron in Mn⁷⁺ = 0

$μ = 0 B . M$

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11 months ago

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

The number of terminal oxygen atoms present in the product B obtained from the following reaction is_____________.

$F e C r_{2} O_{4} + N a_{2} C O_{3} + O_{2} \to A + F e_{2} O_{3} + C O_{2}$

$A + H^{+} \to B + H_{2} O + N a^{+}$

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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11 months ago

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

If the constant term in the expansion of ${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10} i s 2^{k} . l,$ where $l$ is an odd integer, then the value of k is equal to:

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alok kumar singh

Contributor-Level 10

${(3 x^{3} - 2 x^{2} + \frac{5}{x^{5}})}^{10}$

General term $10! \frac{{(3 x^{3})}^{α} \cdot {(- 2^{x})}^{β} \cdot {(5 x^{- 5})}^{γ}}{α! β! γ!}$

$= 10! \frac{3^{α} \cdot {(- 2)}^{β} \cdot 5^{γ} \cdot x^{3 α + 2 β - 5 γ}}{α! β! γ!}$

Now for constant term 3 + 2 $- 5 γ = 0$ ………….(i)

$α + β + γ = 10$ …………(ii)

From equation (i) & (ii)

$3 α + 2 (10 - α - γ) = 5 γ$

$α + 20 = 7 γ$

= 1, $γ$ = 3, = 6

Constant term $10! \cdot \frac{3^{1} \cdot {(- 2)}^{6} \cdot 5^{3}}{3! 6!} = 2^{9} \cdot 3^{2} \cdot 5^{4} \cdot 7^{1}$

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