Ncert Solutions Maths class 12th
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The value of lim(n→∞) [ Σ(k=1 to n) (2k-1) + 8n ] / [ Σ(k=1 to n) (2k-1) + 4n ] is equal to:
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The value of lim(n→∞) [ Σ(k=1 to n) (2k-1) + 8n ] / [ Σ(k=1 to n) (2k-1) + 4n ] is equal to:
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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
=
Upvote 
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation √13.44, then the standard deviation of the second sample is:
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation √13.44, then the standard deviation of the second sample is:
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 =
Upvote Consider functions f : A → B and g : B → C (A,B,C ⊂ R) such that (gof)⁻¹ exists, then:
Consider functions f : A → B and g : B → C (A,B,C ⊂ R) such that (gof)⁻¹ exists, then:
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.
Upvote Let X be a random variable such that the probability function of a distribution is given by P(X=0) = 1/2, P(X = j) = (1/3)? (j = 1,2,3,.∞). Then the mean of the distribution and P(X is positive and even) respectively are:
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Let X be a random variable such that the probability function of a distribution is given by P(X=0) = 1/2, P(X = j) = (1/3)? (j = 1,2,3,.∞). Then the mean of the distribution and P(X is positive and even) respectively are:
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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
Upvote Let the plane passing through the point (-1, 0, -2) and perpendicular to each of the planes 2x + y - z = 2 and x - y - z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to:
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Let the plane passing through the point (-1, 0, -2) and perpendicular to each of the planes 2x + y - z = 2 and x - y - z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to:
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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
Upvote If |a| = 2, |b| = 5 and |a × b| = 8, then a . b is equal to:
If |a| = 2, |b| = 5 and |a × b| = 8, then a . b is equal to:
|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).
Upvote Let a = i + j + 2k and b = -i + 2j + 3k. Then the vector product (a + b) × ((a × ((a - b) × b)) × b) is equal to:
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Let a = i + j + 2k and b = -i + 2j + 3k. Then the vector product (a + b) × ((a × ((a - b) × b)) × b) is equal to:
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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) =
Upvote Let a, b and c be distinct positive numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk are co-planar, then c is equal to:
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Let a, b and c be distinct positive numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk are co-planar, then c is equal to:
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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
Upvote If [x] be the greatest integer less than or equal to x, then ∑₁₀₀ₙ₌₈ [(-1)ⁿ n]/2 is equal to:
If [x] be the greatest integer less than or equal to x, then ∑₁₀₀ₙ₌₈ [(-1)ⁿ n]/2 is equal to:
∑? [ (-1)? n / 2]
= [8/2] + [-9/2] + [10/2] + [-11/2] + . + [-99/2] + [100/2]
= 4 - 5 + 5 - 6 + . + (-50) + 50
= 4
Upvote 