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New answer posted
2 months agoContributor-Level 9
A = lim (n→∞) (2/n) ∑ (r=1 to n) f (r/n + n/ (n²)
(The term n/n² seems intended to be part of the function argument, not simply added. The solution proceeds as if it's f (r/n)
A = lim (n→∞) (2/n) ∑ (r=1 to n) [ f (r/n) + f (1/n) + . + f (n-1)/n) ]
The expression in the image seems to be: A = lim (n→∞) (2/n) [ f (1/n) + f (2/n) + . + f (n-1)/n) ]
A = 2 ∫? ¹ f (x) dx = 2 ∫? ¹ log? (1 + tan (πx/4) dx
put πx/4 = t ⇒ dx = 4/π dt
A = 2 ∫? ^ (π/4) log? (1 + tan (t) * (4/π) dt = (8/π) ∫? ^ (π/4) log? (1 + tan (t) dt
Using the property ∫? f (x)dx = ∫? f (a-x)dx, the integral ∫? ^ (π/4) log (1 + tan (t)dt ev
New answer posted
2 months agoContributor-Level 9
lim (x→0) [a e? - b cos (x) + c e? ] / (x sin (x) = 2
Using Taylor expansions around x=0:
lim (x→0) [a (1+x+x²/2!+.) - b (1-x²/2!+.) + c (1-x+x²/2!+.)] / (x * x) = 2
lim (x→0) [ (a-b+c) + x (a-c) + x² (a/2+b/2+c/2) + O (x³)] / x² = 2
For the limit to exist, the coefficients of lower powers of x in the numerator must be zero.
a - b + c = 0
a - c = 0 ⇒ a = c
Substituting a=c into the first equation: 2a - b = 0 ⇒ b = 2a.
The limit becomes: lim (x→0) [x² (a/2 + b/2 + c/2)] / x² = (a+b+c)/2
(a + b + c) / 2 = 2 ⇒ a + b + c = 4.
New answer posted
2 months agoContributor-Level 9
|z+i|/|z-3i| = 1 ⇒ |z+i| = |z-3i|. This means z is on the perpendicular bisector of the segment from -i to 3i. The midpoint is i, so z = x+i.
w = z? - 2z + 2. Let z = x + iy.
w = (x² + y²) - 2 (x + iy) + 2 = (x² - 2x + 2 + y²) - 2iy.
Re (w) = x² - 2x + 2 + y² = (x - 1)² + 1 + y².
From the first condition, y=1. Re (w) = (x - 1)² + 1 + 1 = (x - 1)² + 2.
Re (w) is minimum for x = 1.
The common z is z = 1 + i.
w = (1+i) (1-i) - 2 (1+i) + 2 = 2 - 2 - 2i + 2 = 2 - 2i.
w² = (2 - 2i)² = 4 (1 - 2i - 1) = -8i.
w? = (-8i)² = -64 ∈ R.
∴ least n ∈ N for which w? ∈ R is n=4.
New answer posted
2 months agoContributor-Level 10
P will be the centroid of triangle ABC.
The centroid P is (x? +x? +x? )/3, (y? +y? +y? )/3).
The coordinates of P are given as (17/6, 8/3).
The coordinates of Q are not given, but a calculation is shown.
PQ = √ (24/6)² + (9/3)²) = √ (4² + 3²) = √ (16+9) = √25 = 5.
This implies the coordinates of Q are such that the difference in coordinates with P leads to this result. For example if P= (x? , y? ) and Q= (x? , y? ), then x? -x? =4 and y? -y? =3.
New answer posted
2 months agoContributor-Level 9
dy/dx = 2 (x + 1)
dy = 2 (x + 1)dx.
y = (x + 1)² - c (1)
For y = 0, (x+1)² = c ⇒ x = -1 ± √c.
Area A = ∫? √c? ¹? √c (0 - [ (x+1)² - c])dx = ∫? √c? ¹? √c (c - (x+1)²)dx
A = [cx - (x+1)³/3] from -1-√c to -1+√c
= (c (-1+√c) - (√c)³/3) - (c (-1-√c) - (-√c)³/3)
= -c+c√c - c√c/3 + c+c√c - c√c/3 = 2c√c - 2c√c/3 = (4/3)c√c.
Given A = 4√8 / 3 = 8√2 / 3.
(4/3)c^ (3/2) = 8√2 / 3 ⇒ c^ (3/2) = 2√2 = 2^ (3/2) ⇒ c = 2.
∴ y = (x + 1)² - 2.
∴ y (1) = (1 + 1)² - 2 = 2.
New answer posted
2 months agoContributor-Level 10
Let t = 3^ (x/2). As x→2, t→3^ (2/2) = 3.
The limit becomes lim (t→3) [ (t² + 27/t²) - 12 ] / [ (t - 3²/t) ].
lim (t→3) [ (t? - 12t² + 27)/t² ] / [ (t² - 9)/t ].
lim (t→3) [ (t²-9) (t²-3) / t² ] * [ t / (t²-9) ].
lim (t→3) [ (t²-3) / t ].
Substituting t=3: (3²-3)/3 = (9-3)/3 = 6.
(The provided solution arrives at 36, let's re-check the problem statement)
The denominator is t - 9/t, not t - 3²/t.
lim (t→3) [ (t²-9) (t²-3) / t² ] * [ t / (t-3) (t+3)/t ]
This leads to the same cancellation. Let's re-examine the image's steps.
lim (t-3) (t³ - 27)/ (t-3) . The algebra in the image is hard to follow but seems to manipul
New answer posted
2 months agoContributor-Level 10
Var (1, 2, ., n) = (Σn²/n) - (Σn/n)² = 10.
(n (n+1) (2n+1)/6n) - (n (n+1)/2n)² = 10.
(n+1) (2n+1)/6 - (n+1)/2)² = 10.
(n+1)/12 * [2 (2n+1) - 3 (n+1)] = 10.
(n+1)/12 * (4n+2 - 3n-3) = 10.
(n+1) (n-1)/12 = 10.
n² - 1 = 120 ⇒ n² = 121 ⇒ n = 11.
Var (2, 4, ., 2m) = Var (2* (1, 2, ., m) = 2² * Var (1, 2, ., m) = 16.
4 * Var (1, 2, ., m) = 16.
Var (1, 2, ., m) = 4.
Using the formula from above: (m²-1)/12 = 4.
m² - 1 = 48 ⇒ m² = 49 ⇒ m = 7.
m + n = 7 + 11 = 18.
New answer posted
2 months agoContributor-Level 9
f (x) + f (x + 1) = 2 (1)
replace x with x + 1: f (x + 1) + f (x + 2) = 2 (2)
(2) - (1) ⇒ f (x + 2) = f (x)
∴ f (x) is periodic with period 2.
I? = ∫? f (x)dx = 4 ∫? ² f (x)dx.
I? = ∫? ³ f (x)dx = ∫? f (u-1)du. Let u = x+1.
I? = ∫? f (x-1)dx = 2 ∫? ² f (x-1)dx.
From (1), f (x-1) + f (x) = 2.
I? + 2I? = 4∫? ² f (x)dx + 2 (2∫? ² f (x-1)dx) = 4∫? ² f (x)dx + 4∫? ² (2 - f (x)dx
= 4∫? ² (f (x) + 2 - f (x)dx = 4∫? ² 2 dx = 4 [2x] from 0 to 2 = 16.
New answer posted
2 months agoContributor-Level 9
y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406| = 406.
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