Ncert Solutions Maths class 12th
Get insights from 2.5k questions on Ncert Solutions Maths class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Maths class 12th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
2 months 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
2 months 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
2 months 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
2 months 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
2 months 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
2 months 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
2 months 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
2 months agoNew answer posted
2 months 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
2 months agoTaking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 679k Reviews
- 1800k Answers