Ncert Solutions Maths class 12th
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New answer posted
a month agoContributor-Level 10
Circle: (x-1)² + (y-3)² = 4. C= (1,3), r=2.
Length of tangent from P (-1,1) is L = √ (-1)²+1²-2 (-1)-6 (1)+6) = √ (1+1+2-6+6) = √4 = 2.
Area of quadrilateral PACB = 2 * Area (PAC) = 2 * (1/2 * L * r) = 2*2=4.
AB is chord of contact. T=0 => -x+y- (x-1)-3 (y-1)+6=0 => -2x-2y+10=0 => x+y=5.
Distance of C from AB = |1+3-5|/√2 = 1/√2.
Length of AB = 2√ (r²-d²) = 2√ (4-1/2) = 2√ (7/2) = √14.
Area of ABD =?
New answer posted
a month agoContributor-Level 10
lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
= (1 + 2ln6) - (0 + 2ln4) = 1 + 2ln (6/4) = 1 + 2ln (3/2).
New answer posted
a month agoContributor-Level 10
1st sample: n? =100, x? =15, σ? =3. Σx? = 1500. Σx? ² = n? (σ? ²+x? ²) = 100 (9+225) = 23400.
Whole group: n=250, x? =15.6, σ²=13.44. Σx = 250*15.6 = 3900.
2nd sample: n? =150. Σy? = 3900 - 1500 = 2400. y? = 2400/150 = 16.
Σ (x+y)² = n (σ²+x? ²) = 250 (13.44+15.6²) = 250 (13.44+243.36) = 64200.
Σy? ² = 64200 - 23400 = 40800.
σ? ² = Σy? ²/n? - y? ² = 40800/150 - 16² = 272 - 256 = 16.
σ? = 4.
New answer posted
a month agoContributor-Level 10
If (gof)? ¹ exist then gof is a bijective function. For gof to be bijective, f must be one-one and g must be onto.
New answer posted
a month agoContributor-Level 10
Mean = Σx? p? = 0 (1/2) + Σ? ∞ j (1/3)? = (1/3)/ (1-1/3)² = (1/3)/ (4/9) = 3/4
P (X is positive and even) = P (X=2) + P (X=4) + .
= (1/3)² + (1/3)? + . = (1/9)/ (1-1/9) = (1/9)/ (8/9) = 1/8
New answer posted
a month agoContributor-Level 10
Normal to the required plane is perpendicular to the normals of the given planes.
n = n? * n? = (2i + j - k) * (i - j - k) = -2i + j - 3k.
Equation of the plane is -2 (x+1) + 1 (y-0) - 3 (z+2) = 0
-2x - 2 + y - 3z - 6 = 0
-2x + y - 3z - 8 = 0
2x - y + 3z + 8 = 0
Comparing with ax + by + cz + 8 = 0, we get a=2, b=-1, c=3.
a+b+c = 2-1+3 = 4.
New answer posted
a month agoContributor-Level 10
|a * b|² + |a . b|² = |a|²|b|²
8² + (a . b)² = 2² * 5²
64 + (a . b)² = 100
(a . b)² = 36
a . b = 6 (since angle seems acute from options, but could be -6).
New answer posted
a month agoContributor-Level 10
a = i + j + 2k
b = -i + 2j + 3k
a + b = 3j + 5k
a . b = -1 + 2 + 6 = 7
a * b = |i, j, k; 1, 2; -1, 2, 3| = -i - 5j + 3k
(a - b) * b) = (a * b) - (b * b) = a * b
(a * (a - b) * b) = a * (a * b) = (a . b)a - (a . a)b = 7a - 6b
. The expression becomes (a + b) * (7a - 6b) * b)
= (a + b) * (7 (a * b)
= 7 [ (a * (a * b) + (b * (a * b) ]
= 7 [ (
New answer posted
a month agoContributor-Level 10
ai+aj+ck, i+k and ci+cj+bk are co-planar,
|a c; 1 0 1; c b| = 0
a (0-c) - a (b-c) + c (c-0) = 0
-ac - ab + AC + c² = 0
c² = ab
c = √ab
New answer posted
a month agoContributor-Level 10
∑? [ (-1)? n / 2]
= [8/2] + [-9/2] + [10/2] + [-11/2] + . + [-99/2] + [100/2]
= 4 - 5 + 5 - 6 + . + (-50) + 50
= 4
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