Ncert Solutions Maths class 12th
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Let P be a plane containing the line (x-1)/3 = (y+6)/4 = (z+5)/2 and parallel to the line (x-3)/4 = (y-2)/-3 = (z+5)/7. If the point (1,-1,α) lies on the plane P, then the value of |5α| is equal to.
Let P be a plane containing the line (x-1)/3 = (y+6)/4 = (z+5)/2 and parallel to the line (x-3)/4 = (y-2)/-3 = (z+5)/7. If the point (1,-1,α) lies on the plane P, then the value of |5α| is equal to.
The equation of a plane is determined by a point it passes through and a normal vector. The plane passes through (1, -6, -5). Its normal vector (a, b, c) is perpendicular to two other vectors, derived from the given equations:
4a - 3b + 7c = 0
3a + 4b + 2c = 0
The direction of the normal vector (a, b,
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Let the mirror image of the point (1,3,a) with respect to the plane r.(2i - j + k) - b = 0 be (-3,5,2). Then, the value of |a+b| is equal to.
Let the mirror image of the point (1,3,a) with respect to the plane r.(2i - j + k) - b = 0 be (-3,5,2). Then, the value of |a+b| is equal to.
The vector PQ is given by Q - P = (-3-1, 5-3, 2-a) = (-4, 2, 2-a).
This vector is collinear with 2i - j + k = (2, -1, 1).
This means their components are proportional: -4/2 = 2/ (-1) = (2-a)/1.
From -2 = 2-a, we find a=4.
The midpoint M of PQ is (-3+1)/2, (5+3)/2, (2+4)/2) = (-1, 4, 3).
M lies on
Upvote Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :
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Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :
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This is a binomial probability problem with n=5. Let p be the probability of success and q be the probability of failure.
Given P (X=1) =? C? p¹q? = 5pq? = 0.4096 — (I)
Given P (X=2) =? C? p²q³ = 10p²q³ = 0.2048 — (II)
Divide (I) by (II): (5pq? ) / (10p²q³) = 0.4096 / 0.2048 = 2.
(1/2) * (q/p) = 2 ⇒ q/
Upvote Let the coefficients of third, fourth and fifth terms in the expansion of (x + a/x²)^n, x ≠ 0, be in the ratio 12:8:3. Then the term independent of x in the expansion is equal to ______.
Let the coefficients of third, fourth and fifth terms in the expansion of (x + a/x²)^n, x ≠ 0, be in the ratio 12:8:3. Then the term independent of x in the expansion is equal to ______.
The general term Tr+1 in the expansion of (a/x + x)^n (assuming from context, as it is not explicitly stated) is:
Tr+1 = nCr * (a/x)^r * x^ (n-r) = nCr * a^r * x^ (n-2r)
(The text shows (a/x^2) leading to x^ (n-3r)
Assuming the term is (x + a/x^2)^n:
Tr+1 = nCr * x^ (n-r) * (a/x^2)^r = nCr * a^r * x^ (
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Define a relation R over a class of n x n real matrices A and B as "ARB iff there exists a non-singular matrix P such that PAP-1 = B". Then which of the following is true?
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Define a relation R over a class of n x n real matrices A and B as "ARB iff there exists a non-singular matrix P such that PAP-1 = B". Then which of the following is true?
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A relation R is defined as ARB if PAP? ¹ = B for a non-singular matrix P.
· Reflexive: ARA requires PAP? ¹ = A. This holds if P is the identity matrix I, as IAI? ¹ = A. Assuming P can be I, the relation is reflexive.
· Symmetric:
Upvote Two tangents are drawn from a point P to the circle x² + y² - 2x - 4y + 4 = 0, such that the angle between these tangents is tan⁻¹(12/5), where tan⁻¹(12/5) ∈ (0,π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of ΔPAB and ΔCAB is:
Two tangents are drawn from a point P to the circle x² + y² - 2x - 4y + 4 = 0, such that the angle between these tangents is tan⁻¹(12/5), where tan⁻¹(12/5) ∈ (0,π). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of ΔPAB and ΔCAB is:
If the equation of plane passing through the mirror image of a point (2,3,1) with respect to line (x+1)/2 = (y-3)/1 = (z+2)/-1 and containing the line (x-2)/3 = (1-y)/2 = (z+1)/1 is, αx + βy + γz = 24 then α + β + γ is equal to:
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If the equation of plane passing through the mirror image of a point (2,3,1) with respect to line (x+1)/2 = (y-3)/1 = (z+2)/-1 and containing the line (x-2)/3 = (1-y)/2 = (z+1)/1 is, αx + βy + γz = 24 then α + β + γ is equal to:
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Let f:R→R be defined as f(x) = e⁻ˣ sin x. If F:→R is a differentiable function such that F(x) = ∫[0 to x] f(t)dt, then the value of ∫[0 to 1] (F'(x) + f(x))eˣdx lies in the interval:
Let f:R→R be defined as f(x) = e⁻ˣ sin x. If F:→R is a differentiable function such that F(x) = ∫[0 to x] f(t)dt, then the value of ∫[0 to 1] (F'(x) + f(x))eˣdx lies in the interval:
Given f(x) = e^x sin(x).
Let F(x) = ∫[0 to x] f(t) dt.
By the Fundamental Theorem of Calculus, F'(x) = f(x) = e^x sin(x).
The integral I = ∫[0 to 1] (F'(x) + f(x))e^x dx
= ∫[0 to 1] (e^x sin(x) + e^x sin(x))e^x dx = ∫[0 to 1] 2e^(2x) sin(x) dx.
The text computes I = ∫[0 to 1] 2 sin(x) dx = [-2cos(x)] fro
Upvote The number of solutions of the equation sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x², for x ∈ [-1,1] and [x] denotes the greatest integer less than or equal to x, is:
The number of solutions of the equation sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x², for x ∈ [-1,1] and [x] denotes the greatest integer less than or equal to x, is:
sin?¹(x² + 1/3) + cos?¹(x² - 2/3) = x²
The domains of sin?¹ and cos?¹ require:
-1 ≤ x² + 1/3 ≤ 1 ⇒ -4/3 ≤ x² ≤ 2/3. Since x² ≥ 0, we have 0 ≤ x² ≤ 2/3.
-1 ≤ x² - 2/3 ≤ 1 ⇒ -1/3 ≤ x² ≤ 5/3.
The intersection of these domains is 0 ≤ x² ≤ 2/3.
The range of sin?¹ is [-π/2, π/2] and cos?¹ is [0, π].
Let A = sin
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