Ncert Solutions Maths class 12th
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New answer posted
a month agoContributor-Level 9
Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.
New answer posted
a month agoContributor-Level 9
f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-1)
New answer posted
a month agoContributor-Level 9
f (x) = sin²x (λ + sinx)
f' (x) = 2sinxcosx (λ + sinx) + sin²x (cosx) = sinxcosx (2λ + 3sinx)
For extrema, f' (x) = 0
sinx = 0, cosx = 0, or sinx = -2λ/3
For more than 2 points of extrema in the interval, sinx = -2λ/3 must have solutions other than where sinx=0 or cosx=0.
-1 < -2/3 < 1 and -2/3 0
This gives λ ∈ (-3/2, 3/2) - {0}
New answer posted
a month agoContributor-Level 9
Let y = (ex)?
ln y = x ln (ex) = x [1 + ln x]
(1/y) (dy/dx) = (1) (1 + ln x) + x (1/x) = 2 + ln x
⇒ dy = (ex)? (2 + ln x)dx
∫? ² (ex)? (2 + log? x)dx = [ (ex)? ]? ² = (2e)² - (1e)¹ = 4e² - e
New answer posted
a month agoContributor-Level 10
sum 6 → (1,5), (5,1), (3,3), (2,4), (4,2)
sum 7 → (1,6), (6,1), (5,2), (2,5), (3,4), (4,3)
= P (A) + P (? ) · P (B) · P (A) + P (? )P (B)P (? ) · P (B) · P (A) + …
This is infinite G.P. with common ratio P (? ) * P (B)
Probability of A wins = P (A) / (1 - P (? )P (B? )
= (5/36) / (1 - (31/36)* (30/36) = 30/61
New answer posted
a month agoContributor-Level 10
Equation PQ
(x-1)/2 = (y+2)/3 = (z-3)/ (-6) = λ
Let Q = (2λ + 1, 3λ − 2, −6λ + 3)
Q lies on x - y + z = 5
⇒ (2λ + 1) − (3λ − 2) + (−6λ + 3) = 5
⇒ λ = -1/7
Q = (5/7, -17/7, 15/7)
∴ PQ = √ (2/7)² + (3/7)² + (6/7)²)
New answer posted
a month agoContributor-Level 10
D = |1 -2 3; 2 1; 1 -7 a| = 0 ⇒ a = 8
also, D? = |9 -2 3; b 1; 24 -7 8| = 0 ⇒ b = 3
hence, a-b = 8-3=5
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
[x]² + 2 [x+2] - 7 = 0
⇒ [x]² + 2 [x] + 4 - 7 = 0
⇒ [x] = 1, -3
⇒ x ∈ [1,2) U [-3, -2)
New question posted
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