Algebra
- For what value of k, the system of equations kx + 2y = 2 and 3x+ y = 1 will be coincident?
-
View Hint View Answer Discuss in Forum
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 will be coincident if
a1 = b1 = c1 a2 b2 c2 ⇒ k = 2 = 2 3 1 1
⇒ k = 3 × 2 = 6
The system of equations has infinite solutions.Correct Option: D
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 will be coincident if
a1 = b1 = c1 a2 b2 c2 ⇒ k = 2 = 2 3 1 1
⇒ k = 3 × 2 = 6
The system of equations has infinite solutions.
-
If 
2x + 1 
- x - 1 - 7 = 0. then two values of x are x² x
-
View Hint View Answer Discuss in Forum
⇒ 2 x² + 1 - x - 1 - 7 = 0. x² x ⇒ 2 
x - 1 ² + 2 
- x - 1 - 7 = 0. x x ⇒ 2 x - 1 ² + 4 - x - 1 - 7 = 0. x x ⇒ 2 x - 1 ² - x - 1 - 3 = 0. x x if - x = 1 = y then, x
2 y² – y – 3 = 0
⇒ 2 y² – 3 y + 2y – 3 = 0
⇒ y (2y – 3) + 1 (2y – 3) = 0
⇒ (y + 1) (2y – 3) = 0
⇒ y = – 1 or 3/2
when y = –1⇒ x – 1 = -1 x
⇒ x² + x + = 0
The value of x will not be real.
Again,⇒ x – 1 = 3 x 2 ⇒ x² - 1 = 3 x 2
⇒ 2x² – 2 = 3x
⇒ 2x² – 3x – 2 = 0
⇒ 2x² – 4x + x – 2 = 0
⇒ 2x (x – 2) +1 (x – 2) = 0
⇒ (2x + 1 ) (x – 2) = 0⇒ x = - 1 or 2 2 Correct Option: B
⇒ 2 x² + 1 - x - 1 - 7 = 0. x² x ⇒ 2 x - 1 ² + 2 - x - 1 - 7 = 0. x x ⇒ 2 x - 1 ² + 4 - x - 1 - 7 = 0. x x ⇒ 2 x - 1 ² - x - 1 - 3 = 0. x x if - x = 1 = y then, x
2 y² – y – 3 = 0
⇒ 2 y² – 3 y + 2y – 3 = 0
⇒ y (2y – 3) + 1 (2y – 3) = 0
⇒ (y + 1) (2y – 3) = 0
⇒ y = – 1 or 3/2
when y = –1⇒ x – 1 = -1 x
⇒ x² + x + = 0
The value of x will not be real.
Again,⇒ x – 1 = 3 x 2 ⇒ x² - 1 = 3 x 2
⇒ 2x² – 2 = 3x
⇒ 2x² – 3x – 2 = 0
⇒ 2x² – 4x + x – 2 = 0
⇒ 2x (x – 2) +1 (x – 2) = 0
⇒ (2x + 1 ) (x – 2) = 0⇒ x = - 1 or 2 2
- The total area (in sq. unit) of the triangles formed by the graph of 4x + 5y = 40, x - axis, y - axis and x = 5 and y = 4 is
-
View Hint View Answer Discuss in Forum
Putting x = 0 in 4x + 5y = 40,
4 × 0 + 5y = 40 ⇒ 5y = 40⇒ y = 40 = 8 5
∴ Point of intersection on y-axis = (0, 8)
Again, putting y
= 0 in 4x + 5y = 40,
4x + 5 × 0 = 40 ⇒ 4x = 40⇒ x = 40 = 10 4
∴ Point of intersection on x-axis = (10, 0)
OA = 10 units
OD = 5 units = EC
∴ DA = 10 – 5 = 5 units
Again, OB = 8 units
OE = 4 units
BE = 8 – 4 = 4 units∴ Area of ∆ADC = 1 × DA × DC = 1 × 5 × 4 = 10 sq. units 2 2 ∴ Area of ∆BEC = 1 × EC × BE = 1 × 5 × 4 = 10 sq. units 2 2
∴ Required area = 10 + 10 = 20 sq. units.Correct Option: B
Putting x = 0 in 4x + 5y = 40,
4 × 0 + 5y = 40 ⇒ 5y = 40⇒ y = 40 = 8 5
∴ Point of intersection on y-axis = (0, 8)
Again, putting y
= 0 in 4x + 5y = 40,
4x + 5 × 0 = 40 ⇒ 4x = 40⇒ x = 40 = 10 4
∴ Point of intersection on x-axis = (10, 0)
OA = 10 units
OD = 5 units = EC
∴ DA = 10 – 5 = 5 units
Again, OB = 8 units
OE = 4 units
BE = 8 – 4 = 4 units∴ Area of ∆ADC = 1 × DA × DC = 1 × 5 × 4 = 10 sq. units 2 2 ∴ Area of ∆BEC = 1 × EC × BE = 1 × 5 × 4 = 10 sq. units 2 2
∴ Required area = 10 + 10 = 20 sq. units.
- Area of the triangle formed by the graph of the straight lines x – y = 0, x + y = 2 and the x– axis is
-
View Hint View Answer Discuss in Forum
On putting x = 0 in x + y = 2,
0 + y = 2 Þ y = 2
∴ Point of intersection on y-axis = (0, 2)
Again,
putting y = 0 in x + y = 2, x = 2
∴ Point of intersection on x-axis = (2, 0)
x – y = 0 will pass through origin and be equally inclined to axes.
On putting x = y in x + y = 2, 2y = 2 ⇒ y = 1
∴ CD = 1 OA = 2Area of ∆ OAC = 1 × OA × CD = 1 × 2 × 1 = 1 sq. units 2 2 Correct Option: A
On putting x = 0 in x + y = 2,
0 + y = 2 Þ y = 2
∴ Point of intersection on y-axis = (0, 2)
Again,
putting y = 0 in x + y = 2, x = 2
∴ Point of intersection on x-axis = (2, 0)
x – y = 0 will pass through origin and be equally inclined to axes.
On putting x = y in x + y = 2, 2y = 2 ⇒ y = 1
∴ CD = 1 OA = 2Area of ∆ OAC = 1 × OA × CD = 1 × 2 × 1 = 1 sq. units 2 2
- If the ordinate and abscissa of the point (k, 2k –1) be equal, then the value of k is
-
View Hint View Answer Discuss in Forum
Abscissa = k,
Ordinate = 2k –1
According to the question,
k = 2k – 1 ⇒ 2k – k = 1 ⇒ k = 1Correct Option: C
Abscissa = k,
Ordinate = 2k –1
According to the question,
k = 2k – 1 ⇒ 2k – k = 1 ⇒ k = 1
