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New answer posted

11 months ago

Maths Straight Lines

48. Two lines passing through the point (2, 3) intersects each other at an angle of 60^o. If slope of one line is 2, find equation of the other line.

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alok kumar singh

Contributor-Level 10

48.

Given, slope of line 1, m₁ = 2.

Let m₂ be the slope of line 2.

If θ is the angle between the two lines then we can write,

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11 months ago

Maths Relations and Functions

20. Show that the Modulus Function f : R → R, given by f(x) = |x| is neither one-one nor onto, where is |x| if x is positive or 0 and |-x| is -x if x is negative.

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Vishal Baghel

Contributor-Level 10

The $f x_{}^{n}$ $f : R \to R$ is given by $f (x) = | x |$

$\Rightarrow f (x) = (\begin{matrix} x_{}, i f & x \geq 0 \\ - x_{}, i f & x < 0 \end{matrix})$

For $x_{1} = - 1$ and $x_{2} = 1$

$f (x_{1}) = f (- 1) = | - 1 | = 1$

$f (x_{2}) = f (1) = | 1 | = 1$

So, $f (x_{1}) = f (x_{2})$ but $x_{1} \neq x_{2}$

i.e., $f$ is not one-one

For $x = - 1 \in R$

$f (x) = | x |$

i.e., $f (- 1) = | - 1 | = 1$

So, range of $f (x)$ is always a positive real number and is not equal to the co-domain $R$

i.e., $f$ is not onto

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11 months ago

Maths Relations and Functions

19.Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

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Vishal Baghel

Contributor-Level 10

The $f x_{}^{n}$ $f : R \to R$ is given by $f (x) = [x]$

Let $x_{1} = 1.5$ and $x_{2} = 1.2 \in R$ Then,

$f (x_{1}) = f (1.5) = [1.5] = 1$

$f (x_{2}) = f (1.2) = [1.2] = 1$

So, $f (x_{1}) = f (x_{2})$ but $x_{1} \neq x_{2}$

i.e., $f (1.5) = f (1.2)$ but $1.5 \neq 1.2$

So, $f$ is not one-one

The range of $f (x)$ is a set of all integers, $Z$ which is not a co-domain of $R$

$∴ f$ is not onto

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New answer posted

11 months ago

Maths Straight Lines

47. Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x –x₁) + B (y – y₁) = 0.

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alok kumar singh

Contributor-Level 10

47.

The slope of line 4x + by + c = 0 is

m = -A/B

As the required line is parallel to the line Ax + by + c = 0

They have the same slope ie, m = -A/B

So, equation of line with slope m and passing through (x₁, y₁)

is given by point-slope from as,

⇒ - A (x-x₁) B (y -y₁)

⇒ A (x-x₁) + B (y-y₁)= 0.

Hence proved.

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11 months ago

Maths Relations and Functions

2. Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: R → R given by f(x) = x²

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

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Vishal Baghel

Contributor-Level 10

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives a set of all positive real numbers

Hence, range of $f (x \neq)$ co-domain of $R$

So, $f$ is not subjective

(iv) $f : N \to N$ given by&n

...more

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives a set of all positive real numbers

Hence, range of $f (x \neq)$ co-domain of $R$

So, $f$ is not subjective

(iv) $f : N \to N$ given by $f (x) = x_{}^{3}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow x_{1}^{} = x_{2}^{} \in N$

So, $f$ is injective

For $x \in N,$

Range of $f (x) = {1_{}^{3}, 2_{}^{3}, 3_{}^{3} . . . .}$

${1_{}^{}, 8, 2 7_{}^{} . . . .} \neq$ co-domain of $N$

So, $f$ is not subjective

(v) $f : R \to R$ given by $f (x) = x_{}^{3}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow x_{1}^{} = x_{2}^{} \in N$

So, $f$ is injective

For $x \in R,$

Range of $f (x) = {1_{}^{3}, 2_{}^{3}, 3_{}^{3} . . . .}$

${1_{}^{}, 8, 2 7_{}^{} . . . .} \neq$ co-domain of R

So, $f$ is not subjective

less

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New answer posted

11 months ago

Maths Straight Lines

46. The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y −19 = 0 at right angle. Find the value of h.

0 Follower 26 Views

alok kumar singh

Contributor-Level 10

46. Slope of line 1 (passing) through (h. 3) and (4, 1) is

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New answer posted

11 months ago

Maths Straight Lines

45. Find angles between the lines √3x + y = 1and x + √3y = 1.

0 Follower 24 Views

alok kumar singh

Contributor-Level 10

Kindly go through the solution

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New answer posted

11 months ago

Maths Straight Lines

44. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and havingx intercept 3.

0 Follower 53 Views

alok kumar singh

Contributor-Level 10

44.

The slope of line x- 7y + 5 = 0 is

So, equation of line with slope m₂ and having x-intercept 3 is

y = m (x-d), d = x- intercept

⇒y = - 7 (x- 3)

⇒y = - 7x + 21

⇒ 7x + y- 21 = 0.

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New answer posted

11 months ago

Maths Straight Lines

43. Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing throughthe point (–2, 3).

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

43.

The slope of the line 3x - 4y + 2 = 0 is,

$m =- \frac{}{} \frac{3}{4}$
$= - \frac{3}{- 4} = \frac{3}{4}$

So, slope of line parallel to 3x - 4y + 2 = 0 is also m= $\frac{3}{4}$ .

Hence, this parallel line passes through ( -2, 3) we can write,

$m = \frac{y - y_{0}}{x - x_{0}}$

$\frac{3}{4} = \frac{y - 3}{x - (- 2)} = \frac{y - 3}{x + 2}$

3 (x + 2) = 4 (y - 3)

3x + 6 = 4y - 12

3x - 4y + 6 + 12 = 0.

3x - 4y + 18 = 0. Which is the required equation of line.

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New answer posted

11 months ago

Maths Straight Lines

42. Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0.

0 Follower 14 Views

alok kumar singh

Contributor-Level 10

42.

(i) Given, equation of lines are

15x + 8y- 34 = 0

15x + 8y + 31 = 0

So, c₁ = 34 and c₂ = 31, A = 15 and B = 8

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