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2 months ago

Class 11th Relations and Functions

Let R? and R? be two relations defined as follows:
R? = {(a, b) ∈ R²: a² + b² ∈ Q} and
R? = {(a, b) ∈ R²: a² + b² ∉ Q}, where Q is the set of all rational numbers. Then:

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alok kumar singh

Contributor-Level 10

For R? let a = 1+√2, b=1-√2, c = 8¹/?
aR? b => a² + b² = (1+√2)² + (1-√2)² = 6 ∈ Q
aR? c => b² + c² = (1-√2)² + (8¹/? )² = 3 ∈ Q
aR? c => a² + c² = (1+√2)² + (8¹/? )² = 3 + 4√2 ∉ Q
∴ R? is not transitive.
For R? let a = 1+√2, b=√2, c=1-√2
aR? B => a² + b² = (1+√2)² + (√2)² = 5+2√2 ∉ Q
bR? b => b² + c² = (√2)² + (1-√2)² = 5-2√2 ∉ Q
aR? c => a² + c² = (1+√2)² + (1-√2)² = 6 ∈ Q
∴ R? is not transitive.

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2 months ago

Class 11th Maths

Let the latus rectum of the parabola y² = 4x be the common chord to the circles C₁ and C₂ each of them having radius 2√5. Then, the distance between the centres of the circles C₁ and C₂ is:

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alok kumar singh

Contributor-Level 10

C? = 2C? S = 2√20-4 = 8

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Class 11th Maths

If $z_{1}, z_{2}$ are complex numbers such that $R e (z_{1}) = |z_{1} - 1|, R e (z_{2}) = |z_{2} - 1|$ and $a r g (z_{1} - z_{2}) = \frac{π}{6}$ , then $I m (z_{1} + z_{2})$ is equal to:

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alok kumar singh

Contributor-Level 10

$|z_{1} - 1| = R e ? (z_{1})$ Let $z_{1} x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$

${(x_{1} - 1)}^{2} + y_{1}^{2} = x_{1}^{2}$

$y_{1}^{2} - 2 x_{1} + 1 = 0$

$|z_{2} - 1| = R e ? (z_{2})$

${(x_{2} - 1)}^{2} + y_{2}^{2} = x_{2}^{2}$

$y_{2}^{2} - 2 x_{2} + 1 = 0$

$(y_{1} - y_{2}) (y_{1} + y_{2}) = 2 (x_{1} - x_{2})$

$y_{1} + y_{2} = 2 (\frac{x_{1} - x_{2}}{y_{1} - y_{2}})$

$a r g ? (z_{1} - z_{2}) = \frac{π}{6}$

${t a n}^{- 1} ? (\frac{y_{1} - y_{2}}{x_{1} - x_{2}}) = \frac{π}{6}$

$\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \frac{1}{\sqrt[]{3}}$

$∴ y_{1} + y_{2} = 2 \sqrt[]{3}$

$\Rightarrow I m ? (z_{1} + z_{2}) = 2 \sqrt[]{3}$

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2 months ago

Class 11th Maths

If the sum of the series $20 + 19 \frac{3}{5} + 19 \frac{1}{5} + 18 \frac{4}{5} + \dots$ upto $n^{th}$ term is 488 and the $n^{th}$ term is negative, then:

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alok kumar singh

Contributor-Level 10

$488 = \frac{n}{2} [2 (\frac{100}{5}) + (n - 1) (\frac{2}{5})]$

$488 = \frac{n}{2} (101 - n)$

$\Rightarrow n^{2} - 101 n + 2440 = 0$

$\Rightarrow n = 61$ or 40

For $n = 40 \Rightarrow T_{n} > 0$

For $n = 61 \Rightarrow T_{n} < 0$

$T_{n} = \frac{100}{5} + (61 - 1) (- \frac{2}{5}) = - 4$

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2 months ago

Class 11th Chemistry

$6.023 * 10^{22}$ molecules are present in $10 g$ of a substance ' $x$ '. The molarity of a solution containing $5 g$ of substance ' $x$ ' in $2 L$ solution is $* 10^{- 3}$ .