Maths
Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
8 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
8 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
8 months agoContributor-Level 10
The problem asks to evaluate S = ∑ (k=0 to 10) (k² + 3k) ¹? C? (Assuming typo in OCR is k²).
S = ∑k² ¹? C? + 3∑k ¹? C?
Using the identities ∑k? C? = n 2? ¹ and ∑k²? C? = n (n+1)2? ².
For n=10:
3∑k ¹? C? = 3 * 10 * 2? = 30 * 2?
∑k² ¹? C? = 10 (11)2? = 110 * 2?
S = 110 * 2? + 30 * 2? = 110 * 2? + 60 * 2? = 170 * 2? = 85 * 2?
The OCR seems to follow a different path with typos, but arrives at 19 * 2¹?
Let's follow the OCR's result: 19 * 2¹? = α * 3¹? + β * 2¹?
Comparing coefficients, we get α = 0 and β = 19.
α + β = 0 + 19 = 19.
New answer posted
8 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.
New answer posted
8 months agoContributor-Level 10
We are given bounds for a function f (t) on two intervals and need to find the range of g (3) = ∫? ³ f (t) dt.
We split the integral: g (3) = ∫? ¹ f (t)dt + ∫? ³ f (t)dt.
For the first interval t ∈ [0, 1], we have 1/3 ≤ f (t) ≤ 1. Integrating from 0 to 1 gives:
∫? ¹ (1/3) dt ≤ ∫? ¹ f (t)dt ≤ ∫? ¹ 1 dt ⇒ 1/3 ≤ ∫? ¹ f (t)dt ≤ 1.
For the second interval t ∈ (1, 3], we have 0 ≤ f (t) ≤ 1/2. Integrating from 1 to 3 gives:
∫? ³ 0 dt ≤ ∫? ³ f (t)dt ≤ ∫? ³ (1/2) dt ⇒ 0 ≤ ∫? ³ f (t)dt ≤ (1/2) (3-1) = 1.
Adding the inequalities for the two parts of the integral:
1/3 + 0 ≤ g (3) ≤ 1
New answer posted
8 months agoContributor-Level 10
For a system of linear homogeneous equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.
Δ = | 4 λ 2 |
| 2 -1 | = 0
| μ 2 3 |
To simplify, perform the row operation R? → R? - 2R? :
Δ = | 0 λ+2 0 |
| 2 -1 | = 0
| μ 2 3 |
Expand the determinant along the first row:
- (λ+2) * det (| 2 1 |, | μ 3 |) = 0.
- (λ+2) (2*3 - 1*μ) = 0.
(λ+2) (μ-6) = 0.
This implies that either λ+2 = 0 or μ-6 = 0.
So, the conditions are λ = -2 (for any μ) or μ = 6 (for any λ).
New answer posted
8 months agoContributor-Level 10
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)
New answer posted
8 months agoContributor-Level 10
The equation of the tangent to the ellipse x²/27 + y² = 1 at the point (3√3 cosθ, sinθ) is:
x (3√3 cosθ)/27 + y (sinθ)/1 = 1 ⇒ x/ (3√3) cosθ + y sinθ = 1.
To find the intercepts on the axes:
x-intercept (set y=0): x = 3√3 / cosθ = 3√3 secθ.
y-intercept (set x=0): y = 1 / sinθ = cosecθ.
The sum of the intercepts is z (θ) = 3√3 secθ + cosecθ.
To find the minimum value of z, we differentiate with respect to θ and set it to zero:
dz/dθ = 3√3 secθ tanθ - cosecθ cotθ = 0.
3√3 (sinθ/cos²θ) = cosθ/sin²θ.
3√3 sin³θ = cos³θ ⇒ tan³θ = 1/ (3√3).
⇒ tanθ = 1/√3.
Since θ ∈ (0, π/2), the solution
New answer posted
8 months agoContributor-Level 10
To find the values of a and b, we evaluate the left-hand limit (LHL) and right-hand limit (RHL) at x=0 and equate them.
LHL: lim (x→0) sin (a+3)/2 * x / x = (a+3)/2. The full limit evaluates to (a+3)/2. So, (a+3)/2 = b.
This gives the relation a - 2b + 3 = 0 — (I)
RHL: lim (x→0) [√ (x+bx³) - √x] / (bx²).
Rationalize the numerator:
lim (x→0) [ (x+bx³) - x] / [bx² (√ (x+bx³) + √x)]
= lim (x→0) bx³ / [bx² (√x (√ (1+bx²) + √x)]
= lim (x→0) x / [√x (√ (1+bx²) + 1)] = lim (x→0) √x / (√ (1+bx²) + 1) = 0/2 = 0.
So, b = 0.
Substituting b=0 into equation (I): a - 2 (0) + 3 = 0 ⇒ a = -3.
Taking 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
- 66k Colleges
- 1.2k Exams
- 690k Reviews
- 1850k Answers