Class 12th
Get insights from 12k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
2 months agoContributor-Level 9
Let A = [a? ]? Sum of diagonal elements of A.A? is Tr (A.A? ) = ∑? ∑? a? ² = 9.
where each a? ∈ {0, 1, 2, 3}.
Case I: One of a? = 3 and rest are 0. (3²=9). There are? C? = 9 ways.
Case II: Two of a? are 2, one is 1, and rest are 0. (2² + 2² + 1² = 9). There are? C? *? C? = 36 * 7 = 252 ways.
Case III: One of a? = 2, five are 1, and rest are 0. (2² + 1²+1²+1²+1²+1² = 9). There are? C? *? C? = 9 * 56 = 504 ways.
Case IV: All nine a? = 1. (1² * 9 = 9). There is 1 way.
Total = 9 + 252 + 504 + 1 = 766.
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 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 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.
New answer posted
2 months agoContributor-Level 10
The function f (x) is non-differentiable at x=1, 3, 5.
Σ f (f (x) = f (f (1) + f (f (3) + f (f (5).
Assuming f (x) is defined such that f (1)=1, f (3)=1, f (5)=1 (based on context of absolute value functions).
Then Σ f (f (x) = f (1) + f (1) + f (1) = 1 + 1 + 1 = 3.
New answer posted
2 months agoContributor-Level 9
Two drawn cards are spades. There are 50 cards left.
The missing card could be a spade or not a spade.
P (missing card is spade) = 11/50 (since 11 spades remain out of 50 cards).
P (missing card is not spade) = 1 - 11/50 = 39/50
New answer posted
2 months agoContributor-Level 9
A (-3, -6,1), B (2,4, -3). Point P divides AB in ratio k:1.
P = [ (2k-3)/ (k+1), (4k-6)/ (k+1), (-3k+1)/ (k+1)]
P lies on the plane lx + my + nz = 0.
l (2k-3) + m (4k-6) + n (-3k+1) = 0
k (2l + 4m - 3n) = 3l + 6m - n
⇒ k = (3l + 6m - n) / (2l + 4m - 3n)
Plane contains the line (x-1)/-1 = (y+4)/2 = (z+2)/3.
The plane passes through (1, -4, -2) and its normal is perpendicular to the line's direction vector.
-l + 2m + 3n = 0
l (1) + m (-4) + n (-2) = 0 ⇒ l - 4m - 2n = 0
Solving these gives l/-8 = m/-1 = n/-2. Let l=8, m=1, n=2.
k = (3 (8) + 6 (1) - 2) / (2 (8) + 4 (1) - 3 (2) = (24+6-2)/ (16+4-6) = 28/14 = 2.
New answer posted
2 months agoContributor-Level 9
A = [i, -i], [-i, i]
A² = [-2, 2], [2, -2]
A? = [8, -8], [-8, 8]
A? = [-128, 128], [128, -128]
A? [x, y]? =?
-128x + 128y = 8 ⇒ -16x + 16y = 1 ⇒ x - y = -1/16 (I)
128x - 128y = 64 ⇒ 16x - 16y = 8 ⇒ x - y = 1/2 (II)
System is inconsistent hence No solution
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
- 65k Colleges
- 1.2k Exams
- 679k Reviews
- 1800k Answers