Ncert Solutions Maths class 12th
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Let the mirror image of the point (a, b, c), with respect to the plane 3x – 4y + 12z + 19 = 0 be (a−6, β, γ). If a + b + c = 5, then 7β-9γ is equal to.
Let the mirror image of the point (a, b, c), with respect to the plane 3x – 4y + 12z + 19 = 0 be (a−6, β, γ). If a + b + c = 5, then 7β-9γ is equal to.
(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = -2 (3a-4b+12c+19)/ (3²+ (-4)²+12²)
(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = (-6a+8b-24c-38)/169
(x, y, z) = (a–6, β, γ)
(a-b)-a)/3 = (β-b)/ (-4) = (γ-c)/12 = (-6a+8b-24c-38)/169
(β-b)/ (-4) = -2
=> β = 8+b
=> 3a – 4b + 12c = 150 . (i)
a + b + c = 5
=> 3a + 3b + 3c = 15 . (i
Upvote 
Let f : R → R be a function defined by f(x) = 2e²ˣ / (e²ˣ + e). Then f(1/100) + f(2/100) + f(3/100) + . + f(99/100) is equal to.
Let f : R → R be a function defined by f(x) = 2e²ˣ / (e²ˣ + e). Then f(1/100) + f(2/100) + f(3/100) + . + f(99/100) is equal to.
f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99
Upvote Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is:
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Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is:
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Given P (X=3) = 5P (X=4) and n=7
=>? C? p³q? = 5? C? p? q³
=> q = 5p and also p + q = 1
=> p = 1/6 and q = 5/6
Mean = 7/6 and variance = 35/36
Mean + Variance
= 7/6 + 35/36 = 77/36
Upvote Five numbers x₁, x₂, x₃, x₄, x₅ are randomly selected from the numbers 1, 2, 3,., 18 and are arranged in the increasing order (x₁ < x₂ < x₃ < x₄ < x₅). The probability that x₂ = 7 and x₄ = 11 is:
Five numbers x₁, x₂, x₃, x₄, x₅ are randomly selected from the numbers 1, 2, 3,., 18 and are arranged in the increasing order (x₁ < x₂ < x₃ < x₄ < x₅). The probability that x₂ = 7 and x₄ = 11 is:
Total case = ¹? C?
Favourable cases
? C? (Select x? )
³C? (Select x? )
? C? (Select x? )
P = (6.3.7)/¹? C? = 1/68
Upvote If two straight lines whose direction cosines are given by the relations l+m-n=0, 3l2 +m2 +cnl=0 are parallel, then the positive value of c is:
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If two straight lines whose direction cosines are given by the relations l+m-n=0, 3l2 +m2 +cnl=0 are parallel, then the positive value of c is:
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l + m – n = 0 => n = l + m
3l² + m² + cnl = 0
3l² + m² + cl (l+m) = 0
= (3+c) (l/m)² + c (l/m) + 1 = 0
? Lines are parallel
D = 0
c = 4 (as c > 0)
Upvote Let a be an integer such that lim (x→a) [18 - [1-x]] / [x-3a] exists, where [t] is greatest integer ≤ t. Then a is equal to:
Let a be an integer such that lim (x→a) [18 - [1-x]] / [x-3a] exists, where [t] is greatest integer ≤ t. Then a is equal to:
lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
LHL = RHL
25/ (7-3a) = 8/ (2-a)
∴ a = -6
Upvote Let A(1,0), B(6,2) and C(3/2, 6) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point (-7/6, -1/3) is
Let A(1,0), B(6,2) and C(3/2, 6) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point (-7/6, -1/3) is
P will be the centroid of triangle ABC.
The centroid P is (x? +x? +x? )/3, (y? +y? +y? )/3).
The coordinates of P are given as (17/6, 8/3).
The coordinates of Q are not given, but a calculation is shown.
PQ = √ (24/6)² + (9/3)²) = √ (4² + 3²) = √ (16+9) = √25 = 5.
This implies the coordinates of Q are s
Upvote If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to.
If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to.
y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a
Upvote A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
Two drawn cards are spades. There are 50 cards left.
The missing card could be a spade or not a spade.
P (missing card is spade) = 11/50 (since 11 spades remain out of 50 cards).
P (missing card is not spade) = 1 - 11/50 = 39/50
Upvote Let P be a plane lx + my + nz = 0 containing the line, (1 - x)/1 = (y + 4)/2 = (z + 2)/3. If plane P divides the line segment AB joining points A(-3, -6, 1) and B(2, 4, -3) in ratio k : 1 then the value of k is equal to:
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Let P be a plane lx + my + nz = 0 containing the line, (1 - x)/1 = (y + 4)/2 = (z + 2)/3. If plane P divides the line segment AB joining points A(-3, -6, 1) and B(2, 4, -3) in ratio k : 1 then the value of k is equal to:
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A (-3, -6,1), B (2,4, -3). Point P divides AB in ratio k:1.
P = [ (2k-3)/ (k+1), (4k-6)/ (k+1), (-3k+1)/ (k+1)]
P lies on the plane lx + my + nz = 0.
l (2k-3) + m (4k-6) + n (-3k+1) = 0
k (2l + 4m - 3n) = 3l + 6m - n
⇒ k = (3l + 6m - n) / (2l + 4m - 3n)
Plane contains the line (x-1)/-1 = (y+4)/2 = (z+2)/3
Upvote