Ncert Solutions Maths class 12th

If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x³) + xg(x³) is divisible by x² + x + 1, then P(1) is equal to.

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P (x) = f (x³) + xg (x³) is divisible by x²+x+1. The roots of x²+x+1=0 are the complex cube roots of unity, ω and ω².
P (ω) = f (ω³) + ωg (ω³) = f (1) + ωg (1) = 0 — (I)
P (ω²) = f (ω²)³) + ω²g (ω²)³) = f (1) + ω²g (1) = 0 — (II)
Subtracting (II) from (I): (ω - ω²)g (1) = 0. Since ω ≠ ω², we must have g (1) = 0.
Substituting g (1)=0 into (I) gives f (1) = 0.
We need to find P (1) = f (1³) + 1*g (1³) = f (1) + g (1).
P (1) = 0 + 0 = 0.

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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.

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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, c) is found by the cross product of (4, -3, 7) and (3, 4, 2):
a = (-3) (2) - 7 (4) = -34.
b = 7 (3) - 4 (2) = 13.
c = 4 (4) - (-3) (3) = 25.
So the plane equation is -34 (x-1) + 13 (y+6) + 25 (z+5) = 0.
The point (1, -1, α) lies on this plane:
-34 (1-1) + 13 (-1+6) + 25 (α+5) = 0.
0 + 13 (5) + 25α + 125 = 0.
65 + 25α + 125 = 0 ⇒ 25α = -190

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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.

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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 the plane 2x - y + z - b = 0.
Substitute the coordinates of M: 2 (-1) - 4 + 3 - b = 0 ⇒ -2 - 4 + 3 = b ⇒ b = -3.

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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/p = 4 ⇒ q = 4p.
Using p+q=1, we have p+4p=1 ⇒ 5p=1 ⇒ p=1/5.
And q = 4/5.
We need to find P (X=3) =? C? p³q².
P (X=3) = 10 * (1/5)³ * (4/5)² = 10 * (1/125) * (16/25) = 160 / 3125 = 32 / 625.

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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 ______.

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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^ (n-3r)
T3 = T (2+1) = nC2 * a^2 * x^ (n-6)
T4 = T (3+1) = nC3 * a^3 * x^ (n-9)
T5 = T (4+1) = nC4 * a^4 * x^ (n-12)

The ratio of the coefficients is given:
(nC2 * a^2) / (nC3 * a^3) = 12/8 = 3/2
(n (n-1)/2) * a^2) / (n (n-1) (n-2)/6) * a^3) = 3/2
(3 / (n-2) * (1/a) = 3/2 => a (n-2) = 2 (i)

(nC3 * a^3) / (nC4 * a^4) = 8/3
(n (n-1) (n-2)/6) * a^3)

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Let f:[-3, 1] → R be given as f(x) = { min{(x+6), x²}, if -3 ≤ x ≤ 0; max{√x, x²}, if 0 ≤ x ≤ 1 }. If the area bounded by y=f(x) and x-axis is A, then the value of 6A is equal to ______.

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Kindly go through the solution

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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: We need to show that if ARB, then BRA.
ARB ⇒ PAP? ¹ = B.
To get the reverse, we need to express A in terms of B.
From PAP? ¹ = B, pre-multiply by P? ¹ and post-multiply by P:
P? ¹ (PAP? ¹)P = P? ¹BP ⇒ A = P? ¹BP. This shows BRA where the matrix is P? ¹. Thus, the relation is symmetric.

· Trans

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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:

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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:

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