Ncert Solutions Maths class 12th

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New answer posted•a year 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

16 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

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

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year 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 $(\vec{a} \times \vec{b}) . (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) . (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) . (\vec{a} \times \vec{b})$ = 168, then $| \vec{a} | + | \vec{b} | + | \vec{c} |$  is equal to:

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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

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

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

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

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

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

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 : $\vec{a}, \vec{b}, \vec{c}$ &

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New answer posted•a year 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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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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New answer posted•a year 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:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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.

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

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New answer posted•a year 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:

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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New answer posted•a year 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:

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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New answer posted•a year 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:

13 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\frac{7 . {(α)}^{2 / 7}}{2} = 14$

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

=128

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New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{h - c o s θ}{2} = \frac{k - s i n θ}{3} = \frac{w - 0}{1}$

$= \frac{- 1 (2 c o s θ + 3 s i n θ - 6)}{14} \Rightarrow h = \frac{c o s - 2 (2 c o s θ + 3 s i n θ - 6)}{14}$

$k = \frac{5 s i n θ - 6 c o s θ + 18}{14}$

${(5 h + 6 k - 12)}^{2} + 4 {(3 h + 5 k - 9)}^{2} = 1$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

Let $S_{1} = {z_{1} \in C : | z_{1} - 3 | = \frac{1}{2}}$  and $S_{2} = {z_{2} \in C : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |} .$  Then, for $z_{1} \in S_{1}$ and $s_{2} \in S_{2},$  the least value of  $| z_{2} - z_{1} | i s :$

18 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$S_{1} = {z, \in c : ⌊ z_{1} - 3 ⌋ = \frac{1}{2}} a n d S_{2} = {z_{2} \in c : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |}$

Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2},$ the least value of |z₂ – z₁|

${| z_{2} + | z_{2} - 1 | |}^{2} = {| z_{2} - | z_{2} + 1 | |}^{2}$

$\Rightarrow (z_{2} + {\bar{z}}_{2}) (| z_{2} - 1 | + | z_{2} + 1 | - 2) = 0$

$∴ z_{2} + {\bar{z}}_{2} = 0 o r | z_{2} - 1 | + | z_{2} + 1 | - 2 = 0$

$∴$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $| z_{1} - 3 | = \frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$

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Ncert Solutions Maths class 12th

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