Ncert Solutions Maths class 11th
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New answer posted
a month agoContributor-Level 10
x? + 2 (20)¹/? x³ + (20)¹/²x² + . No.
x² = - (20)¹/? x - (5)¹/².
α+β = - (20)¹/? , αβ=5¹/².
α²+β² = (α+β)²-2αβ = (20)¹/² - 2 (5)¹/² = 0.
α? +β? = (α²+β²)² - 2 (αβ)² = 0 - 2 (5) = -10.
α? +β? = (α? +β? )² - 2 (αβ)? = (-10)² - 2 (5)² = 100 - 50 = 50.
New answer posted
a month agoContributor-Level 10
C= (2,3), O= (0,0). r = OC = √13.
Slope of OC = 3/2. Slope of PQ = -2/3.
Let P= (x, y). Vector CP = (x-2, y-3). Vector OC = (2,3).
CP.OC=0 ⇒ 2 (x-2)+3 (y-3)=0 ⇒ 2x+3y=13.
Also (x-2)²+ (y-3)²=13.
From 2x=13-3y, x= (13-3y)/2.
(13-3y)/2-2)²+ (y-3)²=13 ⇒ (9-3y)/2)²+ (y-3)²=13
(9 (y-3)²/4) + (y-3)² = 13 ⇒ (13/4) (y-3)²=13 ⇒ (y-3)²=4 ⇒ y-3=±2.
y=5 or y=1.
If y=5, x=-1. P= (-1,5).
If y=1, x=5. Q= (5,1).
New answer posted
a month agoContributor-Level 10
? P? =? P? ⇒ n!/ (n-r)! = n!/ (n-r-1)! ⇒ n-r=1.
? C? =? C? ⇒ r + (r-1) = n ⇒ n = 2r-1.
Substitute n: (2r-1)-r=1 ⇒ r=2.
New answer posted
a month agoContributor-Level 10
Reflected point of (2,1) about y-axis is (-2,1).
Reflected ray passes through (-2,1) and (5,3).
Equation: (y-1)/ (x+2) = (3-1)/ (5+2) = 2/7 ⇒ 2x - 7y + 11 = 0.
This is one directrix. Let the other be 2x - 7y + α = 0.
Distance between directrices = 2a/e = |11-α|/√53.
Distance from focus to directrix = a/e - ae = 8/√53.
a/e (1-e²) = 8/√53.
e=1/3. a/e (8/9) = 8/√53 ⇒ a/e = 9/√53.
2a/e = 18/√53 = |11-α|/√53.
|11-α| = 18.
11-α = 18 ⇒ α = -7.
11-α = -18 ⇒ α = 29.
Other directrix: 2x-7y-7=0 or 2x-7y+29=0.
New answer posted
a month agoContributor-Level 10
cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2
New answer posted
a month agoContributor-Level 10
S? : x² + y² - x - y - 1/2 = 0, C? : (1/2, 1/2), r? = √ (1/4)+ (1/4)+ (1/2) = 1.
S? : x² + y² - 4y + 7/4 = 0, C? : (0, 2), r? = √ (4 - 7/4) = 3/2.
S? : (x-2)² + (y-1)² ≤ r², C? : (2, 1).
A ∪ B ⊂ C means both circles S? and S? must be inside S?
Distance C? = √ (2-1/2)² + (1-1/2)²) = √ (9/4 + 1/4) = √10/2.
Condition: r ≥ C? + r? ⇒ r ≥ √10/2 + 1.
Distance C? = √ (2-0)² + (1-2)²) = √5.
Condition: r ≥ C? + r? ⇒ r ≥ √5 + 3/2.
√10/2 + 1 ≈ 1.58 + 1 = 2.58.
√5 + 3/2 ≈ 2.23 + 1.5 = 3.73.
So minimum r = (√10+2)/2 and (2√5+3)/2. We need the maximum of these two.
Let's recheck the question logic f
New answer posted
a month agoContributor-Level 10
Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.
New answer posted
a month agoContributor-Level 10
sinθ + cosθ = 1/2
16 (sin (2θ) + cos (4θ) + sin (6θ)
= 16 [2sin (4θ)cos (2θ) + cos (4θ)]
= 16 [4sin (2θ)cos² (2θ) + 2cos² (2θ) - 1] . (i)
Now, sinθ + cosθ = 1/2, squaring on both sides, we get
1 + sin (2θ) = 1/4
sin (2θ) = -3/4
cos² (2θ) = 1 - sin² (2θ) = 1 - 9/16 = 7/16
From equation (i)
16 [4 (-3/4) (7/16) + 2 (7/16) - 1]
16 [-21/16 + 14/16 - 16/16] = 16 [-23/16] = -23
New answer posted
a month agoContributor-Level 10
S? : |z - 3 - 2i|² = 8
|z - (3 + 2i)| = 2√2
(x - 3)² + (y - 2)² = (2√2)²
S? : Re (z) ≥ 5
x ≥ 5
S? : |z - z? | ≥ 8
|2iy| ≥ 8
2|y| ≥ 8
|y| ≥ 4
y ≥ 4 or y ≤ -4
From the graph of the circle (S? ) and the regions (S? and S? ), we can see that there is one point of intersection at (5, 4).
∴ n (S? ∩ S? ∩ S? ) = 1
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