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Ncert Solutions Maths class 12th

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Ncert Solutions Maths class 12th Class 12th

Let $\vec{a} = \hat{i} - \hat{j} + 2 \hat{k} a n d l e t \hat{b}$ be a vector such that $\vec{a} * \vec{b} = 2 \hat{i} - \hat{k} a n d \vec{a}, \vec{b} = 3 .$ Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b} i s :$

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Payal Gupta

Contributor-Level 10

$\vec{a} = \hat{i} - \hat{j} + 2 \hat{k}$

$\vec{a} * \vec{b} = 2 \hat{i} - \hat{k}$

$\vec{a} . \vec{b} = 3$

${| \vec{a} * \vec{b} |}^{2} + {| \vec{a} . \vec{b} |}^{2} = {| \vec{a} |}^{2} . {| \vec{b} |}^{2}$

$= \frac{\vec{b} . \vec{a} - {| \vec{b} |}^{2}}{| \vec{a} - \vec{b} |} = \frac{3 - \frac{7}{3}}{\sqrt[]{\frac{7}{3}}} = \frac{2}{\sqrt[]{21}}$

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Ncert Solutions Maths class 12th Class 12th

The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = Z a n d \frac{7 - x}{2} = y - 2 = z - 6$ is

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Payal Gupta

Contributor-Level 10

$A (- 7 \hat{i} + 6 \hat{j}) C (7 \hat{i} + 2 \hat{j} + 6 \hat{k})$

$\vec{b} = - 6 \hat{i} + 7 \hat{j} + \hat{k}$ $\vec{d} = - 2 \hat{i} + \hat{j} + \hat{k}$

$\vec{b} * \vec{d} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 6 & 7 & 1 \\ - 2 & 1 & 1 \end{matrix} | = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$

the shortest distance between the lines

$=$ $| \frac{42 - 8 + 24}{29} |$ $=$ $\frac{58}{29}$ $=$ $2$

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2 months ago

Ncert Solutions Maths class 12th Class 12th

A plane E is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4, and passes through the point P(1, -1, 1). If the distance of the plane E from the point Q(a, a, 2) is $3 \sqrt[]{2},$ then (PQ)² is equal to

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Payal Gupta

Contributor-Level 10

First plane, P₁ = 2x – 2y + z = 0,

normal vector $\equiv n_{1} = (2, - 2, 1)$

Second plane, P₂ = x – y + 2z = 4,

normal vector $\equiv n_{2} = (1, - 1, 2)$

Plane perpendicular to P₁ and P₂ will have normal vector n₃ where n₃ = (n₁* n₂)

Hence,

n₃ = (3, 0)

Distance PQ

$= \sqrt[]{21} \Rightarrow {(P Q)}^{2} = 21$

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2 months ago

Ncert Solutions Maths class 12th Class 12th

The domain of the function f(x) = $s i n^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) i s :$

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alok kumar singh

Contributor-Level 10

$- 1 \leq \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$

$\Rightarrow 0 \leq \frac{2 x^{2} - x + 9}{x^{2} + 2 x + 7} & \frac{- 5 x - 5}{x^{2} + 2 x + 7} \leq 0$

$x \in R & - 1 \leq x < \infty$

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2 months ago

Ncert Solutions Maths class 12th Class 12th

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} * \vec{b}) . (\vec{b} * \vec{c}) + (\vec{b} * \vec{c}) . (\vec{c} * \vec{a}) + (\vec{c} * \vec{a}) . (\vec{a} * \vec{b})$ = 168, then $| \vec{a} | + | \vec{b} | + | \vec{c} |$ is equal to:

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alok kumar singh

Contributor-Level 10

Given : $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = c o s θ (s a y)$

$| \vec{a} | | \vec{b} | | \vec{c} | = 14$

$(\vec{a} * \vec{b}) \cdot (\vec{b} * \vec{c})$

$= \vec{a} \cdot [(\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}]$

$= (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - {| \vec{b} |}^{2} \vec{a} \cdot \vec{c}$

$= | \vec{a} | {| \vec{b} |}^{2} | \vec{c} | (c o s^{2} θ - c o s θ) = 14 | \vec{b} | (c o s^{2} θ - c o s θ)$

Similarly, $(\vec{b} * \vec{c}) \cdot (\vec{c} * \vec{a})$

= $| \vec{b} | {| \vec{c} |}^{2} | \vec{a} | (c o s^{2} θ - s i n θ) = 14 | \vec{c} | (c o s 2 θ - s i n θ) & (\vec{c} * \vec{a}) \cdot (\vec{a} * \vec{b})$

$= | \vec{c} | {| \vec{a} |}^{2} | \vec{b} | (c o s^{2} θ - s i n θ) = 14 | \vec{a} | (c o s^{2} θ - s i n θ)$

Given : 14 $(c o s^{2} θ - c o s θ) (| \vec{a} | + | \vec{b} | + | \vec{c} |) = 168$

$| \vec{a} | + | \vec{b} | + | \vec{c} | = \frac{12}{c o s^{2} θ - c o s θ} = \frac{12}{\frac{1}{4} - (- \frac{1}{2})} = \frac{12}{\frac{3}{4}} = 16$

Given : $\vec{a}, \vec{b}, \vec{c}$ are coplanar & pair wise equal angle.

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2 months ago

Ncert Solutions Maths class 12th Class 12th

Let $S = {z = x + i y | z - 1 + i | \geq | z |, | z | < 2, | z + i | = | z - 1 |} .$ Then the set of all values of x, for which $w = 2 x + i y \in S$ for some $y \in R,$ is

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alok kumar singh

Contributor-Level 10

$| z - 1 + i | \geq | z |$

$| z + i | = | z - 1 |$

W = (2x, y) = (a, y)

Let S represent the line segment AB

For 'B'

x² + y² = 4

x = -y

-> x² = 2

$x = \pm \sqrt[]{2}$

-> $B (- \sqrt[]{2}, \sqrt[]{2})$

$A (\frac{1}{2}, - \frac{1}{2})$

W (2x, y) lies on AB

-> $- \sqrt[]{2} < 2 x \leq \frac{1}{2}$

$(| z | < 2)$

$- \frac{1}{\sqrt[]{2}} < x \leq \frac{1}{4}$

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2 months ago

Ncert Solutions Maths class 12th Class 12th

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The balls so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

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alok kumar singh

Contributor-Level 10

$\begin{matrix} 3 R & 2 R & 4 B \\ 5 B & 3 W & 2 W \end{matrix}$

I II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2^nd bag after a ball from I to II was transferred.

$P (\frac{E_{1}}{E}) = \frac{P (E_{1} \cap E)}{P (E)}$

$P (E) = P (E_{1} \cap E) + P (E_{2} \cap E) + P (E_{3} \cap E)$

= $P (E_{1}) \cdot P (\frac{E}{E_{1}}) + . . . . + . . . . .$

= $\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10} = \frac{54}{100}$

$P (E_{1} / E) = \frac{15 / 100}{54 / 100} = \frac{5}{18}$

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New answer posted

2 months ago

Ncert Solutions Maths class 12th Class 12th

The minimum value of the twice differentiable function f(x) = $\int_{0}^{x} e^{x - t} f' (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ is:

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Vishal Baghel

Contributor-Level 10

$f (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t$

$f' (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t + e^{x} . \frac{f' (x)}{e^{x}} - [(2 x - 1) . e^{x} + (x^{2} - x + 1) . e^{x}]$

$\begin{array}{l} f (x) = (2 x + 1) . e^{x} - 2 e^{x} + c \\ f (x) = e^{x} (2 x - 1) c = 0 \end{array}$

$f (- \frac{1}{2}) = \frac{- 2}{\sqrt[]{e}}$

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New answer posted

2 months ago

Ncert Solutions Maths class 12th Functions

Let a, b and $γ$ be three positive real number. Let f(x) = ax⁵ + bx³ + $γ x, x \in R$ and g : R ® R be such that $g (f (x)) = x$ for all x $\in R .$ If $a_{1}, a_{2}, a_{3} . . . . ., a_{n}$ be in arithmetic progression with mean zero then the value of $f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i})))$ is equal to:

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Vishal Baghel

Contributor-Level 10

Case I

= = 0 then

f (x) = yx

$g (x) = \frac{x}{y}$

$\frac{1}{x} \sum_{i = 1}^{n} f (a_{i}) \Rightarrow \frac{y}{x} (a_{1} + a_{2} + . . . . + a_{x}) = 0$

$\Rightarrow f (g (0)) \Rightarrow f (0)$

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2 months ago

Ncert Solutions Maths class 12th Class 12th

If the minimum value of f(x) = $\frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0,$ $\frac{^{}}{} \frac{}{^{}}$ is 14, then the value of is equal to:

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Vishal Baghel

Contributor-Level 10

$\frac{5 x^{2}}{2} + \frac{α}{2 x^{5}} + \frac{α}{2 x^{5}} \geq 7 {(\frac{α^{2}}{2^{7}})}^{\frac{1}{7}}$

$\frac{7 . {(α)}^{2 / 7}}{2} = 14$

${(α^{2})}^{\frac{1}{7}} = 2^{2} \Rightarrow α = {(2^{2})}^{\frac{7}{2}} = 2^{7}$

=128

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