Choice (b) is correct. The equation of any line parallel to the x-axis is y = k If it passes through the point (2, -3), then its coordinates must satisfy its equation -3 = k Putting the value of k, the equation of the required line is y = -3 This question related to Chapter 9 maths Class 11th NCERT.Read more
Choice (b) is correct.
The equation of any line parallel to the x-axis is y = k
If it passes through the point (2, -3), then its coordinates must satisfy its equation -3 = k
Putting the value of k, the equation of the required line is y = -3
This question related to Chapter 9 maths Class 11th NCERT. From the Chapter 9: Straight Lines. Give answer according to your understanding.
Choice (b) is correct. We know that in case of distinct numbers A.M. > G.M x + y /2 > √xy, y + z/2 >√yz and z + x /2 >√zx Multiplying above three inequalities, we get x + y/2 . y + z /2 . z + x /2 >√xy. √yz. √zx ⇒ (x + y) (y + z) (z + x) >8xyz For more please visit here: https:/Read more
Choice (b) is correct.
We know that in case of distinct numbers A.M. > G.M
x + y /2 > √xy, y + z/2 >√yz and z + x /2 >√zx
Multiplying above three inequalities, we get
x + y/2 . y + z /2 . z + x /2 >√xy. √yz. √zx
⇒ (x + y) (y + z) (z + x) >8xyz
Choice (c) is correct. Since x, 2y, 3z are in A.P. 2y = x + 3z/ 2⇒ 4y = x + 3z Again, x, y, z are in G.P. Let r be its common ratio. y = xr and z = yr = (xr)r = xr² Putting values of y and z from (2) in (1), we get 4xr = x +3xr² ⇒ 3r² - 4r + 1 = 0 ⇒ 3r² - 3r - r + 1= 0 ⇒ 3r(r - 1) - (r - 1) = 0Read more
Choice (c) is correct.
Since x, 2y, 3z are in A.P.
2y = x + 3z/ 2⇒ 4y = x + 3z
Again, x, y, z are in G.P.
Let r be its common ratio.
y = xr and z = yr = (xr)r = xr²
Putting values of y and z from (2) in (1), we get
4xr = x +3xr² ⇒ 3r² – 4r + 1 = 0 ⇒ 3r² – 3r – r + 1= 0
⇒ 3r(r – 1) – (r – 1) = 0 ⇒ (r – 1)(3r – 1) = 0 ⇒ r = 1 or r = 1/3
r = 1/3
Choice (d) is correct. Let a and r be the first term and the common ratio of the G.P. Given, any term of the G.P. is equal to the sum of the next two terms. Tₙ = Tₙ ₊ ₁ + Tₙ ₊ ₂ ⇒ arⁿ ⁻ ¹ = arⁿ + arⁿ ⁺ ¹ ⇒ 1 = r + r² ⇒ r² + r - 1 = 0 ⇒ r = -1 ±√ 1² - 4 × 1 × (-1)/ 2 × 1 = -1 ± √5/2 ⇒ r =√5 - 1/2 ⇒ Read more
Choice (d) is correct.
Let a and r be the first term and the common ratio of the G.P.
Given, any term of the G.P. is equal to the sum of the next two terms.
Tₙ = Tₙ ₊ ₁ + Tₙ ₊ ₂ ⇒ arⁿ ⁻ ¹ = arⁿ + arⁿ ⁺ ¹ ⇒ 1 = r + r²
⇒ r² + r – 1 = 0
⇒ r = -1 ±√ 1² – 4 × 1 × (-1)/ 2 × 1 = -1 ± √5/2
⇒ r =√5 – 1/2
⇒ r = 2( √5 – 1/2) = 2 sin 18°
Choice (a) is correct. The series is a G.P. with a = 8, r = 1/2, l = 1/128 The nth term lₙ from end is given by lₙ = 1/rⁿ ⁻ ¹ The fourth term l₄ from end is given by l₄ = 1/28/(1/2)³ = 8/128 = 1/16 For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-8
Choice (a) is correct.
The series is a G.P. with a = 8, r = 1/2, l = 1/128
The nth term lₙ from end is given by lₙ = 1/rⁿ ⁻ ¹
The fourth term l₄ from end is given by l₄ = 1/28/(1/2)³ = 8/128 = 1/16
The equation of the straight line passing through the point (2, -3) and parallel to the x-axis is
Choice (b) is correct. The equation of any line parallel to the x-axis is y = k If it passes through the point (2, -3), then its coordinates must satisfy its equation -3 = k Putting the value of k, the equation of the required line is y = -3 This question related to Chapter 9 maths Class 11th NCERT.Read more
Choice (b) is correct.
The equation of any line parallel to the x-axis is y = k
If it passes through the point (2, -3), then its coordinates must satisfy its equation -3 = k
Putting the value of k, the equation of the required line is y = -3
This question related to Chapter 9 maths Class 11th NCERT. From the Chapter 9: Straight Lines. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-9
If x, y and z are distinct positive integers, then value of expression (x + y) (y + z) (z + x) is
Choice (b) is correct. We know that in case of distinct numbers A.M. > G.M x + y /2 > √xy, y + z/2 >√yz and z + x /2 >√zx Multiplying above three inequalities, we get x + y/2 . y + z /2 . z + x /2 >√xy. √yz. √zx ⇒ (x + y) (y + z) (z + x) >8xyz For more please visit here: https:/Read more
Choice (b) is correct.
We know that in case of distinct numbers A.M. > G.M
x + y /2 > √xy, y + z/2 >√yz and z + x /2 >√zx
Multiplying above three inequalities, we get
x + y/2 . y + z /2 . z + x /2 >√xy. √yz. √zx
⇒ (x + y) (y + z) (z + x) >8xyz
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-8
If x, 2y, 3z are in A.P. where the distinct numbers x, y, z are in G.P., then the common ratio of the G.P. is
Choice (c) is correct. Since x, 2y, 3z are in A.P. 2y = x + 3z/ 2⇒ 4y = x + 3z Again, x, y, z are in G.P. Let r be its common ratio. y = xr and z = yr = (xr)r = xr² Putting values of y and z from (2) in (1), we get 4xr = x +3xr² ⇒ 3r² - 4r + 1 = 0 ⇒ 3r² - 3r - r + 1= 0 ⇒ 3r(r - 1) - (r - 1) = 0Read more
Choice (c) is correct.
Since x, 2y, 3z are in A.P.
2y = x + 3z/ 2⇒ 4y = x + 3z
Again, x, y, z are in G.P.
Let r be its common ratio.
y = xr and z = yr = (xr)r = xr²
Putting values of y and z from (2) in (1), we get
4xr = x +3xr² ⇒ 3r² – 4r + 1 = 0 ⇒ 3r² – 3r – r + 1= 0
⇒ 3r(r – 1) – (r – 1) = 0 ⇒ (r – 1)(3r – 1) = 0 ⇒ r = 1 or r = 1/3
r = 1/3
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In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. is
Choice (d) is correct. Let a and r be the first term and the common ratio of the G.P. Given, any term of the G.P. is equal to the sum of the next two terms. Tₙ = Tₙ ₊ ₁ + Tₙ ₊ ₂ ⇒ arⁿ ⁻ ¹ = arⁿ + arⁿ ⁺ ¹ ⇒ 1 = r + r² ⇒ r² + r - 1 = 0 ⇒ r = -1 ±√ 1² - 4 × 1 × (-1)/ 2 × 1 = -1 ± √5/2 ⇒ r =√5 - 1/2 ⇒ Read more
Choice (d) is correct.
Let a and r be the first term and the common ratio of the G.P.
Given, any term of the G.P. is equal to the sum of the next two terms.
Tₙ = Tₙ ₊ ₁ + Tₙ ₊ ₂ ⇒ arⁿ ⁻ ¹ = arⁿ + arⁿ ⁺ ¹ ⇒ 1 = r + r²
⇒ r² + r – 1 = 0
⇒ r = -1 ±√ 1² – 4 × 1 × (-1)/ 2 × 1 = -1 ± √5/2
⇒ r =√5 – 1/2
⇒ r = 2( √5 – 1/2) = 2 sin 18°
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The fourth term from the end of the G.P. 8, 4, 2, ….., 1/128 is
Choice (a) is correct. The series is a G.P. with a = 8, r = 1/2, l = 1/128 The nth term lₙ from end is given by lₙ = 1/rⁿ ⁻ ¹ The fourth term l₄ from end is given by l₄ = 1/28/(1/2)³ = 8/128 = 1/16 For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-8
Choice (a) is correct.
The series is a G.P. with a = 8, r = 1/2, l = 1/128
The nth term lₙ from end is given by lₙ = 1/rⁿ ⁻ ¹
The fourth term l₄ from end is given by l₄ = 1/28/(1/2)³ = 8/128 = 1/16
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-8