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Given, f(x) = 1 – |x − 3| 

Domain: We observe that f (x) is defined for all 

x ∈ R 

∴ Domain of f = R 

Range: Now, 

0 ≤ |x − 3| < ∞∀∈ R 

⇒ −∞ < −|x − 3| ≤ 0 ∀ x∈ R 

⇒ −∞ < −|∀ − 3| ≤ 1 ∀x ∈ R 

⇒ −∞ < f(x) ≤ 1 ∀ x ∈ R 

Hence, Range of f = (−∞, 1)

Given,f, (x) =\(\frac{1}{4-x^2}\) 

Let y = f (x) y =\(\frac{1}{4-x^2}\) 

x² = \(\frac{4y-1}{y}\) 

x = \(\sqrt{\frac{4y-1}{y}}\) 

Clearly, x will assume real values, if 

⇒ 4y – 1 ≥ 0 and y ≠ 0 

4y ≥ 1 and y ≠ 0 

⇒ y ≥ \(\frac{1}{4}\)and y ≠ 0 

⇒ y ∈ (– ∞, 0) ∪ [ \(\frac{1}{1}\) , ∞] 

∴ Range of y = (– ∞, 0) ∪ [ \(\frac{1}{1}\) , ∞]

Given, f (x) = \(\frac{4-x}{x-4}\) 

Domain: Clearly, f (x) is defined for all x ∈ R expect 

x = 4 

∴ Domain of f = R – {4 } = (−∞, 4),∪ (4, ∞) 

Range: Let y = f(x) 

⇒y = \(\frac{4-x}{x-4}\) 

⇒ y = \(\frac{-(x-4)}{x-4}\) = −1 

∴ Range of f = {– 1}

As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

∠X + ∠XYZ + ∠XZY = 180º

62º + 54º + ∠XZY = 180º

∠XZY = 180º − 116º

∠XZY = 64º

∠OZY = 64/2 = 32º (OZ is the angle bisector of ∠XZY)

Similarly, ∠OYZ = 54/2 = 27º

Using angle sum property for ΔOYZ, we obtain 

∠OYZ+ ∠YOZ + ∠OZY = 180º 

27º + ∠YOZ + 32º = 180º

∠YOZ = 180º − 59º

∠YOZ = 121º

From the figure,  \(\overrightarrow {MN}|| \overrightarrow {KL}\) and \(\overrightarrow {MK}\) is transversal.

∠M = 2x and ∠K = x + 30°

∠M + ∠K = 180° (Q co-interior angles are supplementary) 

⇒ 2x + x + 30° = 180° 

⇒ 3x + 30° = 180° 

⇒ 3x = 180° – 30° 

⇒ 3x = 150° 

⇒ x = 150° /3 

∴ x = 50°

Given that PQ ⊥ PS ; PQ // SR 

∠SQR = 28°, ∠QRT = 65° 

From the figure 

∠QSR = x° (∵ alt. int. angles for the lines PQ // SR) 

Also 65° = x + 28° 

(∵ ext. angles = sum of the opp. interior angles) 

∴ x° = 65° – 28° = 37° 

And x° + y° = 90° 

[ ∵ PQ ⊥ PS and PQ // SR. ⇒ ∠P = ∠S] 

37° + y = 90° 

∴ y = 90° – 37° = 53°

 In the figure \(\overline {AB}|| \overline {DE}\) and C is a point in between them. 

Draw a parallel line CF to \(\overline {AB}\) through C. 

\(\overline {AB}|| \overline {CF}\) and \(\overline {BC}\) is a transversal, 

x + 103° = 180° 

x = 180°- 103°

x = 77°

From the figure, 

\(\overline {DC}|| \overline {CF}\) and \(\overline {CD}\) is a transversal, 

y + 103° = 180° 

y = 180°- 103° 

y = 77° and ∠BCD = x + y = 77° + 77° = 154°

AB || DE and AE is a transversal

∠BAC = ∠CED (Alternate interior angles)

∠CED = 35º

In ΔCDE,

∠CDE + ∠CED + ∠DCE = 180º (Angle sum property of a triangle)

53º + 35º + ∠DCE = 180º

∠DCE = 180º − 88º

∠DCE = 92º

Given that ∠BCD = 125° 

∠A : ∠B = 2 : 3 

Sum of the terms of the ratio

∠A : ∠B = 2 + 3 = 5 

We know that ∠A + ∠B = ∠BCD 

(∵ exterior angles of triangle is equal to sum of its opp. interior angles)

∴ ∠A = 2/5 x 125° = 50°

∠B = 3/5 x 125° = 75°

Given that AB // DE, ∠CDE = 53°; 

∠BAC = 35° 

Now ∠E = 35°

( ∵ alternate interior angles)

Now in ∆CDE 

∠C + ∠D + ∠E = 180° 

(∵angle sum property, ACDE) 

∴ ∠DCE + 53° + 35° = 180° 

⇒ ∠DCE = 180° – 88° = 92°

Given that BC // DE ; ∠BAC = 35°; ∠BCE = 102°

∠ADE + ∠CBD = 180° 

(∵ interior angles on the same side of the transversal) 

∠ADE + (78° + 35°) = 180° 

(∵ ∠CBD = ∠BAC + ∠BCA) 

∴ ∠ADE = 180° – 113° = 67°

Given that BC // DE ; ∠BAC = 35°; 

∠BCE = 102° 

From the figure 

102° + ∠BCA = 180° 

(∵ linear pair of angles) 

∴ ∠BCA = 180° – 102° = 78°

Given that BC // DE ; ∠BAC = 35°; 

∠BCE = 102°

From the figure . ∠CED = ∠BCA = 78° 

(∵ corresponding angles)

Since, given variable y has different values so, The pair of equations y = 0 and y = - 5 has no solution

Correct option is d) infinitely many solutions

C. Infinitely many solutions

Explanation:

Expressing y in terms of x in the equation 2x – 5y = 7, we get,

2x – 5y = 7

– 5y = 7 – 2x

y = ( 7 – 2x)/– 5

Hence, we can conclude that the value of y will be different for different values of x.

Hence, option C is the correct answer.

a₁x + b₁y + c₁ = 0

a₂x + b₂x + c₂ = 0

If \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\) infinite solution ,

If \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)  no solution

If  \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)  unique solution

2x + y = 5

4x + 2y = 10

Thus,

\(\frac{1}{2}\) = \(\frac{1}{2}\) = \(\frac{1}{2}\)

Infinitely many solution

a₁x + b₁y + c₁ = 0

a₂x + b₂x + c₂ = 0

If \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)= \(\frac{c_1}{c_2}\) infinite solution,

If  \(\frac{a_1}{a_2}\)= \(\frac{b_1}{b_2}\)≠ \(\frac{c_1}{c_2}\), no solution

If  \(\frac{a_1}{a_2}\)≠ \(\frac{b_1}{b_2}\)≠ \(\frac{c_1}{c_2}\), unique solution

x – 3y = 3

3x – 9y = 2

Thus,

\(\frac{1}{3}\) = \(\frac{1}{3}\) ≠  \(\frac{3}{2}\)

The system has no solution

Given linear equations are

x – 3y – 2 = 0 …(i)

-2x + 6y – 5 = 0 …(ii)

On comparing with ax + by c=0,

We get

a₁ =1, b₁ =-3, c₁ =- 2;

a₂ = -2, b₂ =6, c₂ =- 5;

a₁/a₂ = – ½

b₁/b₂ = – 3/6 = – ½

c₁/c₂ = 2/5

i.e., a₁/a₂ = b₁/b₂ ≠ c₁/c₂ [parallel lines]

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

 \(\frac{1}{2x}\)+\(\frac{1}{3y}\) = 2

\(\frac{1}{3x}\)+\(\frac{1}{2y}\) = \(\frac{13}{6}\)

Multiplying eq 1 by 1/2 and eq2 by 1/3 and subtracting

⇒ 1/4x – 1/9x = 1 – 13/18

⇒ 5/36x = 5/18

⇒ x = 1/2

Thus,

1 + 1/3y = 2

⇒ y = 1/3

Given:

  \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1

The given linear equation is in the slope intercept form. Intercept means the distance at which the given equation cuts or meets the coordinate axis. In this problem a & b are the intercepts on the x and y axis respectively.

The triangle formed by a straight line with the coordinate axis is a right angled triangle where the angle subtended at origin is 90°. So the length of the x intercepts becomes the perpendicular and y intercept becomes the base of the triangle.

We know,

Area Of a triangle = \(\frac{1}{2}\) x (perpendicular length) x (base length)

So,

Area of the triangle becomes \(\frac{1}{2}\)ab

The Area of the triangle is \(\frac{1}{2}\)ab

 \(\frac{x}{3}\)+\(\frac{y}{4}\) = 11

⇒ 4x + 3y = 132

 \(\frac{5x}{6}\)- \(\frac{y}{3}\) = - 7

⇒ 5x – 2y = -42

Multiplying eq1 by 2 and eq2 by 3 and adding them

⇒ 8x + 6y + 15x – 6y = 264 – 126

⇒ 23x = 138

⇒ x = 6

Thus,

24 + 3y = 132

⇒ y = 36

 \(\frac{15}{u}\)+\(\frac{2}{v}\) = 17

\(\frac{1}{u}\)+\(\frac{1}{v}\) = \(\frac{36}{5}\)

Multiplying eq2 by 2 and subtracting

13/u = 17 – 72/5

⇒ u = 5

Thus,

3 + 2/v = 17

⇒ v = 1/7

0.5x + 0.7y = 0.74

0.3x + 0.5y = 0.5

Multiplying eq1 by 0.3 and eq2 by 0.5 and subtracting eq1 from eq2

⇒ 0.15x + 0.25y – 0.15x – 0.21y = 0.25 – 0.222

⇒ 0.04y = 0.028

⇒ y = 0.7

Thus,

x = 0.5

x + y = 5xy

x + 2y = 13xy, x ≠ 0, y ≠ 0

Subtracting the two eq.

⇒ - y = - 8xy

⇒ x = 1/8

Thus,

1/8 + y = 5y/8

⇒ y = 1/3

2(3u - v) = 5uv

2(u + 3v) = 5uv

Equating both equations

⇒ 6u – 2v = 2u + 6v

⇒ u = 2v

Substituting value of u

⇒ 2(6v – v) = 5 × 2v × v

⇒ v = 1

Thus,

2(3u – 1) = 5u

⇒ 6u – 2 = 5u

⇒ u = 2

 \(\frac{5}{x+1}\) - \(\frac{2}{y-1}\) = \(\frac{1}{2}\)

\(\frac{10}{x+1}\) + \(\frac{2}{y-1}\) = \(\frac{5}{2}\)

Adding eq1 and eq2

⇒ 15/(x + 1) = 3

⇒ x + 1 = 5

⇒ x = 4

Thus,

5/5 – 2/(y – 1) = 12

⇒ y – 1 = 4

⇒ y = 5

 \(\frac{3}{x}\) - \(\frac{1}{y}\) = - 9

\(\frac{2}{x}\)+\(\frac{3}{y}\) = 5

Multiplying eq1 by 3 and adding to eq1

⇒ 11/x = - 27 +5

⇒ x = - 1/2

Thus,

- 4 + 3/y = 5

⇒ y = 1/3

The interior angles of a triangle are the three angle elements inside the triangle.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

From the given figure, we have

∠ACB + x = 180° (Linear pair)

75°+ x = 180°

x = 105°

We know that the sum of all angles of a triangle is 180°

Therefore, for △ABC, we can say that:

∠BAC+ ∠ABC +∠ACB = 180°

40°+ y +75° = 180°

y = 65°

2x - \(\frac{3}{y}\) = 9

3x + \(\frac{7}{y}\) = 2, y ≠ 0

Multiplying eq1 by 3 and eq2 by 2 and subtracting eq1 from eq2

⇒ 6x + 21/y – 6x + 6/y = 4 – 27

⇒ 23/y = - 23

⇒ y = -1

Thus,

2x + 3 = 9

⇒ x = 3

x + y = 2xy

x – y = 6xy

Adding the two equation

⇒ 2x = 8xy

⇒ y = 1/4

Thus,

x + 1/4 = 2 × x × 1/4

⇒ x/2 = - 1/4

⇒ x = - 1/2

 \(\frac{2}{x}\)+ \(\frac{5}{y}\) = 1

\(\frac{60}{x}\)+\(\frac{40}{y}\) = 19, x ≠ 0, y ≠ 0

Multiplying eq1 by 8 and subtracting from eq2

⇒ 44/x = 19 – 8

⇒ x = 4

Thus,

2/4 + 5/y = 1

⇒ 5/y = 1/2

⇒ y = 10

Let us assume that A and B are the two interior opposite angles.

We know that ∠A is equal to ∠B.

We also know that the sum of interior opposite angles is equal to the exterior angle.

Therefore from the figure we have,

∠A + ∠B = 80°

∠A +∠A = 80° (because ∠A = ∠B)

2∠A = 80°

∠A = 40/2 =40°

∠A= ∠B = 40°

Thus, each of the required angles is of 40°.

The interior angles of a triangle are the three angle elements inside the triangle.

The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

We know that the sum of all angles of a triangle is 180^o.

Therefore, for △DBC, we have

30^o + 50^o + ∠DBC = 180^o

∠DBC = 100^o

From the figure we can say that

x + ∠DBC = 180^o is a Linear pair

x = 80^o

From the exterior angle property we have

y = 30^o + 80^o = 110^o

There are 3 median in a triangle. 

(A) Right angle triangle

 \(\frac{1}{5x}\)+\(\frac{1}{6y}\) = 12

\(\frac{1}{3x}\) - \(\frac{3}{7y}\) = 8, x ≠ 0, y ≠ 0

Multiplying eq1 by 1/3 and eq2 by 1/5 and subtracting

⇒ 1/18y + 3/35y = 4 - 8/5

⇒ 89/630y = 12/5

⇒ y = 89/1512

Thus,

1/5x + 1512/534 = 12

⇒ 1/5x = 816/89

⇒ x = 89/4080

∵ Sum of three angles of a triangle = 180°
∴ 50° + 50° + third angle = 180°
⇒ 100° + third angle = 180°
⇒ third angle = 180° – 100°
⇒ = 80°

There are 3 sides in a triangle.

A triangle has three vertices and three sides. Therefore each vertex makes a median after joining the mid point of its opposite sides. Therefore there may be three medians.

The correct option is (B) 110°.

There is no triangle is possible because if each angle is greater than 60° then the sum of all angles will be greater then 180°.

Let in ∆ABC, = 80° and ∠B = ∠C

∵∠A + ∠B +∠c = 180°

⇒ 80° +∠B + ∠B = 180°

[∵ ∠A= 80° and 2C = ∠B]

⇒ 2∠B = 180° – 80°

⇒ ∠B = ( 100/2)= 50°

Measure of each of the equal angles = 50°

If two angles in a triangle are 60° and 40° then the third angle will be 80°.

Yes, a median lies inside the triangle completely.

The sum of all three angles of a triangle will be 180°.

Let ∆ABC is a triangle where ∠B = 30° and ∠C = 80°.

Now, ∠B + ∠C = 30° + 80° = 110°

∵∠A + ∠B + ∠C = 180° or ∠A + 110° = 180° or ∠A = 180° – 110° = 70°

Thus, third angle is 70°.

The sum of two sides of a triangle should be more than third side.

Here 4 + 5 ⊁ 9

∴ Such triangle is not possible.

The equilateral triangle has equal angles.

∴ x° + x° + c°= 180°

⇒ 3x° = 180°

⇒ x° = 180/3 = 60°

∴ All angles of an equilateral triangle are 60°, 60°, 60°.

No, ∴ Sum of three angles = 180

⇒ 70° + 21 + x° = 180°
⇒ x = 180 – 90°
⇒ x = 89°
≠ 9

∵ Third angle is not a right angle.

Hence. Such right angle triangle is not possible.