Let C and D are two mid-points of line segment AB. According to question 4, we have, AC = 1/2 AB and AD = 1/2 AB ⇒ AC = AD [Things which are equal to the same thing are equal to one another] It is possible only if C and D coincide with each other. Hence, the mid-point C is unique.
Let C and D are two mid-points of line segment AB. According to question 4, we have,
AC = 1/2 AB and AD = 1/2 AB
⇒ AC = AD [Things which are equal to the same thing are equal to one another]
It is possible only if C and D coincide with each other.
Hence, the mid-point C is unique.
Let, the work done = y and let the distance = x Given: the constant force = 5 units Work done by the body is directly proportional to is distance travelled. Therefore, y α x ⇒ y = kx Here, k is proportionality constant and givenk = 5 units. Hence, y = 5x For the graph: Putting x = 0, we have, y = 5Read more
Let, the work done = y and let the distance = x
Given: the constant force = 5 units
Work done by the body is directly proportional to is distance travelled. Therefore, y α x
⇒ y = kx
Here, k is proportionality constant and givenk = 5 units.
Hence, y = 5x
For the graph:
Putting x = 0, we have, y = 5 × 0 = 0
Putting x = 1, we have, y = 5 × 1 = 5
Putting x = 2, we have, y = 5 × 2 = 10
Hence, A(0, 0), B(1, 5) and C(2, 10) are the three solutions of the
given equation.
(i) If the distance travelled is 2 units, the work done is 10 units.
(ii) If the distance travelled is 0 units, the work done is 0 units.
(ii) p(x) = x³ + 3x² + 3x + 1 and g(x) = x + 2 Putting x + 2 = 0, we get, x = 2 Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by (g(x) = x + 2, remainder is given by p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 Since, remainder p(-2) ≠ 0, hence g(x) is not a factor oRead more
(ii) p(x) = x³ + 3x² + 3x + 1 and g(x) = x + 2
Putting x + 2 = 0, we get, x = 2
Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by (g(x) = x + 2, remainder is given by p(-2)
= (-2)³ + 3(-2)² + 3(-2) + 1
= -8 + 12 – 6 + 1
= -1
Since, remainder p(-2) ≠ 0, hence g(x) is not a factor of p(x).
(iii) p(x) = x³ - 4x² + x + 6 and g(x) = x - 3 Putting x - 3 = 0, we get, x = 3 Using remainder theorm, when p(x) = x³ - 4x² + x + 6 is divided by g(x) = x -3, remainder is given by p(3) = (3)³ - 4(3)² + (3) + 6 = 27 - 36 + 3 + 6 = 0 Since, remainder p(3) = 0, hence g(x) is a factor of p(x).
(iii) p(x) = x³ – 4x² + x + 6 and g(x) = x – 3
Putting x – 3 = 0, we get, x = 3
Using remainder theorm, when p(x) = x³ – 4x² + x + 6 is divided by g(x) = x -3,
remainder is given by p(3)
= (3)³ – 4(3)² + (3) + 6
= 27 – 36 + 3 + 6 = 0
Since, remainder p(3) = 0, hence g(x) is a factor of p(x).
(i) p(x) = x² + x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = x² + x + k is divided by x - 1, remainder is given by p(1) = (1)² + (1) + k = 2 + k Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ 2 + k = 0 ⇒ k = -2
(i) p(x) = x² + x + k
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = x² + x + k is divided by x – 1, remainder is given by p(1)
= (1)² + (1) + k
= 2 + k
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ 2 + k = 0
⇒ k = -2
(ii) p(x) = 2x² + kx + √2 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x - 1, remainder is given by p(1) = 2(1)² + k(1) + √2 = 2 + k + √2 Since x - 1 is a fctor of p(x), hence remainder p(1) = 0 ⇒ 2 + k + √2 = 0 ⇒ k = -2 - √2
(ii) p(x) = 2x² + kx + √2
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x – 1, remainder is given by p(1)
= 2(1)² + k(1) + √2
= 2 + k + √2
Since x – 1 is a fctor of p(x), hence remainder p(1) = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 – √2
(iii) p(x) = kx² – √2 x + 1 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x - 1, remainder is given by p(1) = k(1)² – √2(1) + 1 = k – √2 + 1 Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ k - √2 + 1 = 0 ⇒ k = √2 - 1
(iii) p(x) = kx² – √2 x + 1
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x – 1, remainder is given by p(1)
= k(1)² – √2(1) + 1
= k – √2 + 1
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ k – √2 + 1 = 0
⇒ k = √2 – 1
(iv) p(x) = kx² - 3x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² - 3x + k is divided by x - 1, remainder is given by p(1) = k(1)² - 3(1) + k = 2k - 3 Since x - 1 is a factor p(x), hence remainder p(1) = 0 ⇒ 2k - 3 = 0 ⇒ k = 3/2
(iv) p(x) = kx² – 3x + k
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – 3x + k is divided by x – 1, remainder is given by p(1)
= k(1)² – 3(1) + k
= 2k – 3
Since x – 1 is a factor p(x), hence remainder p(1) = 0
⇒ 2k – 3 = 0
⇒ k = 3/2
In NCERT Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Let C and D are two mid-points of line segment AB. According to question 4, we have, AC = 1/2 AB and AD = 1/2 AB ⇒ AC = AD [Things which are equal to the same thing are equal to one another] It is possible only if C and D coincide with each other. Hence, the mid-point C is unique.
Let C and D are two mid-points of line segment AB. According to question 4, we have,
See lessAC = 1/2 AB and AD = 1/2 AB
⇒ AC = AD [Things which are equal to the same thing are equal to one another]
It is possible only if C and D coincide with each other.
Hence, the mid-point C is unique.
In Fig. 5.10, if AC = BD, then prove that AB = CD.
Given that: AC = BD ⇒ AC - BC = BD - BC [∵ If equals are subtracted from equals, the remainders are equal] ⇒ AB = CD
Given that: AC = BD
See less⇒ AC – BC = BD – BC [∵ If equals are subtracted from equals, the remainders are equal]
⇒ AB = CD
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is (i) 2 units (ii) 0 unit
Let, the work done = y and let the distance = x Given: the constant force = 5 units Work done by the body is directly proportional to is distance travelled. Therefore, y α x ⇒ y = kx Here, k is proportionality constant and givenk = 5 units. Hence, y = 5x For the graph: Putting x = 0, we have, y = 5Read more
Let, the work done = y and let the distance = x
See lessGiven: the constant force = 5 units
Work done by the body is directly proportional to is distance travelled. Therefore, y α x
⇒ y = kx
Here, k is proportionality constant and givenk = 5 units.
Hence, y = 5x
For the graph:
Putting x = 0, we have, y = 5 × 0 = 0
Putting x = 1, we have, y = 5 × 1 = 5
Putting x = 2, we have, y = 5 × 2 = 10
Hence, A(0, 0), B(1, 5) and C(2, 10) are the three solutions of the
given equation.
(i) If the distance travelled is 2 units, the work done is 10 units.
(ii) If the distance travelled is 0 units, the work done is 0 units.
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following cases: p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
(ii) p(x) = x³ + 3x² + 3x + 1 and g(x) = x + 2 Putting x + 2 = 0, we get, x = 2 Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by (g(x) = x + 2, remainder is given by p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 Since, remainder p(-2) ≠ 0, hence g(x) is not a factor oRead more
(ii) p(x) = x³ + 3x² + 3x + 1 and g(x) = x + 2
See lessPutting x + 2 = 0, we get, x = 2
Using remainder theorem, when p(x) = x³ + 3x² + 3x + 1 is divided by (g(x) = x + 2, remainder is given by p(-2)
= (-2)³ + 3(-2)² + 3(-2) + 1
= -8 + 12 – 6 + 1
= -1
Since, remainder p(-2) ≠ 0, hence g(x) is not a factor of p(x).
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following cases: x³– 4x² + x + 6, g(x) = x – 3
(iii) p(x) = x³ - 4x² + x + 6 and g(x) = x - 3 Putting x - 3 = 0, we get, x = 3 Using remainder theorm, when p(x) = x³ - 4x² + x + 6 is divided by g(x) = x -3, remainder is given by p(3) = (3)³ - 4(3)² + (3) + 6 = 27 - 36 + 3 + 6 = 0 Since, remainder p(3) = 0, hence g(x) is a factor of p(x).
(iii) p(x) = x³ – 4x² + x + 6 and g(x) = x – 3
See lessPutting x – 3 = 0, we get, x = 3
Using remainder theorm, when p(x) = x³ – 4x² + x + 6 is divided by g(x) = x -3,
remainder is given by p(3)
= (3)³ – 4(3)² + (3) + 6
= 27 – 36 + 3 + 6 = 0
Since, remainder p(3) = 0, hence g(x) is a factor of p(x).
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = x² + x + k
(i) p(x) = x² + x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = x² + x + k is divided by x - 1, remainder is given by p(1) = (1)² + (1) + k = 2 + k Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ 2 + k = 0 ⇒ k = -2
(i) p(x) = x² + x + k
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = x² + x + k is divided by x – 1, remainder is given by p(1)
= (1)² + (1) + k
= 2 + k
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ 2 + k = 0
⇒ k = -2
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = 2x² + kx + √2
(ii) p(x) = 2x² + kx + √2 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x - 1, remainder is given by p(1) = 2(1)² + k(1) + √2 = 2 + k + √2 Since x - 1 is a fctor of p(x), hence remainder p(1) = 0 ⇒ 2 + k + √2 = 0 ⇒ k = -2 - √2
(ii) p(x) = 2x² + kx + √2
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x – 1, remainder is given by p(1)
= 2(1)² + k(1) + √2
= 2 + k + √2
Since x – 1 is a fctor of p(x), hence remainder p(1) = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 – √2
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = kx² – √2 x + 1
(iii) p(x) = kx² – √2 x + 1 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x - 1, remainder is given by p(1) = k(1)² – √2(1) + 1 = k – √2 + 1 Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ k - √2 + 1 = 0 ⇒ k = √2 - 1
(iii) p(x) = kx² – √2 x + 1
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x – 1, remainder is given by p(1)
= k(1)² – √2(1) + 1
= k – √2 + 1
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ k – √2 + 1 = 0
⇒ k = √2 – 1
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = kx² – 3x + k
(iv) p(x) = kx² - 3x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² - 3x + k is divided by x - 1, remainder is given by p(1) = k(1)² - 3(1) + k = 2k - 3 Since x - 1 is a factor p(x), hence remainder p(1) = 0 ⇒ 2k - 3 = 0 ⇒ k = 3/2
(iv) p(x) = kx² – 3x + k
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – 3x + k is divided by x – 1, remainder is given by p(1)
= k(1)² – 3(1) + k
= 2k – 3
Since x – 1 is a factor p(x), hence remainder p(1) = 0
⇒ 2k – 3 = 0
⇒ k = 3/2