Ncert Solutions Maths class 11th
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New answer posted
4 weeks agoContributor-Level 9
(|x| - 3)|x + 4| = 6
(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)
Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.
Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.
New answer posted
4 weeks agoContributor-Level 9
log? /? [ (|z|+11)/ (|z|-1)²] < 2
(|z|+11)/ (|z|-1)² > (1/2)²
(|z|+11)/ (|z|-1)² > 1/4
⇒ 4|z| + 44 > |z|² - 2|z| + 1
⇒ |z|² - 6|z| - 43 < 0
⇒ |z| - 7 ≤ 0
∴ |z|max = 7
New answer posted
4 weeks agoContributor-Level 10
The inequality is experience ( (|z|+3) (|z|-1) / (|z|+1) * log?2 ) ≥ log√? 16.
The right side is log? (1/2) 16 = log? (2? ¹) 2? = (4/-1)log?2 = -4. This seems incorrect.
Let's assume the base of the log on the right is √2. log√? 16 = log? (1/2) 2? = 2 * log?2? = 8.
The inequality becomes: 2^ (|z|+3) (|z|-1) / (|z|+1) ≥ 8 = 2³.
So, (|z|+3) (|z|-1) / (|z|+1) ≥ 3.
Let |z| = t. (t+3) (t-1) / (t+1) ≥ 3
t² + 2t - 3 ≥ 3t + 3
t² - t - 6 ≥ 0
(t-3) (t+2) ≥ 0
Since t = |z| ≥ 0, we must have t-3 ≥ 0.
So, t ≥ 3, which means |z| ≥ 3.
The minimum value of |z| is 3.
New answer posted
a month agoContributor-Level 10
We want to evaluate S = ∑ (r=1 to 10) r! (r³ + 6r² + 2r + 5).
We can rewrite the polynomial r³ + 6r² + 2r + 5 as (r³+6r²+11r+6) - 9r - 1.
Note that (r+1) (r+2) (r+3) = r³+6r²+11r+6.
So the term is r! [ (r+1) (r+2) (r+3) - 9r - 1] = (r+3)! - (9r+1)r!
Rewrite 9r+1 as 9 (r+1) - 8.
The term is (r+3)! - [9 (r+1)-8]r! = (r+3)! - 9 (r+1)! + 8r!
Let T? = (r+3)! - 9 (r+1)! + 8r! This does not form a simple telescoping series.
Following the OCR's final calculation, the sum simplifies to 13! + 12! - 8 (11!).
= 11! (13*12 + 12 - 8) = 11! (156 + 4) = 160 (11!).
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