Class 12th
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2 months agoNew answer posted
2 months agoContributor-Level 10
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.
New answer posted
2 months agoContributor-Level 10
In the ammonolysis reaction, HCl is produced as a byproduct. To neutralize this acidic impurity, the mixture is treated with NaOH.
New answer posted
2 months agoContributor-Level 10
P' (x) = a (x-1) (x+1) = a (x²-1).
P (x) = ∫ P' (x) dx = a (x³/3 - x) + b.
Given P (-3) = 0 ⇒ a (-9+3) + b = 0 ⇒ b = 6a.
Given ∫ P (x)dx = 18. Assuming the integration is over a symmetric interval like [-c, c] and using the fact that a (x³/3-x) is an odd function, ∫ (a (x³/3 - x)dx = 0. Then ∫ b dx = 18. If the interval is [-1, 1], this would be b (1 - (-1) = 2b = 18, so b=9.
With b=9, we find a = b/6 = 9/6 = 3/2.
So, P (x) = 3/2 (x³/3 - x) + 9 = x³/2 - 3x/2 + 9.
The sum of all coefficients is 1/2 - 3/2 + 9 = -1 + 9 = 8.
New answer posted
2 months agoContributor-Level 10
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
New answer posted
2 months agoContributor-Level 10
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.
New answer posted
2 months agoContributor-Level 10
The functional equation f (x+y) = f (x)f (y) implies f (x) = a? for some constant a.
Then f' (x) = a? ln (a).
Given f' (0) = 3, we have a? ln (a) = 3 ⇒ ln (a) = 3 ⇒ a = e³.
So, f (x) = e³?
We need to evaluate the limit: lim (x→0) (f (x)-1)/x = lim (x→0) (e³? -1)/x.
Using the standard limit lim (u→0) (e? -1)/u = 1, we can write:
lim (x→0) 3 * (e³? -1)/ (3x) = 3 * 1 = 3.
New answer posted
2 months agoContributor-Level 10
Given the matrix P = [2, -1], [5, -3].
The characteristic equation is det (P - λI) = 0, which is (2-λ) (-3-λ) - (-1) (5) = 0.
This simplifies to λ² + λ - 1 = 0.
By the Cayley-Hamilton theorem, the matrix P satisfies this equation: P² + P - I = 0, so P² = I - P.
To find P³: P³ = P * P² = P (I-P) = P - P² = P - (I-P) = 2P - I.
The problem asks for N=6, likely related to a higher power P? Continuing the pattern:
P? = 2P² - P = 2 (I-P) - P = 2I - 3P.
P? = 2P - 3P² = 2P - 3 (I-P) = 5P - 3I.
P? = 5P² - 3P = 5 (I-P) - 3P = 5I - 8P.
The solution N=6 must relate to a different question not fully transcribed, for example, if P^N = 5I - 8P.
New answer posted
2 months agoContributor-Level 10
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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