Class 12th
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New answer posted
4 months agoContributor-Level 10
(a+3b). (7a-5b) = 7|a|² - 5ab + 21ab - 15|b|² = 7|a|²+16ab-15|b|²=0.
(a-4b). (7a-2b) = 7|a|² - 2ab - 28ab + 8|b|² = 7|a|²-30ab+8|b|²=0.
Subtracting: 46ab - 23|b|² = 0 ⇒ 2ab = |b|².
Substituting: 7|a|² + 8|b|² - 15|b|² = 0 ⇒ 7|a|² = 7|b|² ⇒ |a|=|b|.
cosθ = ab/ (|a|b|) = ab/|b|² = (1/2)|b|²/|b|² = 1/2.
θ = 60°.
New answer posted
4 months ago
Contributor-Level 10
Let rectangle have dimensions l, w.
Triangle above rectangle is similar to large triangle.
(H-w)/H = l/a.
l = a (H-w)/H = a (1-w/H).
Area A = lw = aw (1-w/H).
A' (w) = a (1-2w/H) = 0 ⇒ w = H/2 = √6/2.
l = a/2 = √2.
Max Area A = √2 * √6/2 = √3.
Square of area = 3.
New answer posted
4 months agoContributor-Level 10
log? (18x-x²-77)>0 ⇒ 18x-x²-77>1 ⇒ x²-18x+78<0. Roots are 93.
log? (.)>0 ⇒ log? (.)>1 ⇒ 18x-x²-77>3 ⇒ x²-18x+80<0 (x-8) (x-10)<0.
8
I = ∫? ¹? sin³x/ (sin³x+sin³ (18-x)dx. Using King's property.
I = ∫? ¹? sin³ (18-x)/ (sin³ (18-x)+sin³x)dx.
2I = ∫? ¹? dx = 2. I=1.
New answer posted
4 months agoContributor-Level 10
f (1)=1.
f (4)=f (2)²=1 or 4.
f (6)=f (2)f (3).
Possible functions determined by values at primes: f (2), f (3), f (5), f (7).
f (2) can be 1 or 2. f (3) can be 1 or 3. f (5)=1,5. f (7)=1,7.
If f (m)=m, f (mn)=mn. One function. f (x)=1 is another.
What if f (2)=1, f (3)=3? f (6)=3.
New answer posted
4 months agoContributor-Level 10
f (x) is discontinuous at integers x=1,2,3. P= {1,2,3}.
f (x) is not differentiable at integers and where x- [x]=1+ [x]-x ⇒ 2 (x- [x])=1 ⇒ {x}=1/2.
So at x=0.5, 1, 1.5, 2, 2.5.
Q= {0.5, 1, 1.5, 2, 2.5}. Sum of elements is not asked.
Number of elements in P=3, in Q=5. Sum = 8.
Let's check the solution. Q= {1/2, 1, 3/2, 5/2}.
The sum of number of elements: 3+5=8.
New answer posted
4 months agoContributor-Level 10
Determinant of vectors must be zero. Vector between points on lines: (-1-k, -2-2, -3-3). Vector directions: (1,2,3) and (3,2,1).
| -1-k, -4, -6; 1, 2, 3; 3, 2, 1 | = 0.
(-1-k) (2-6) - (-4) (1-9) + (-6) (2-6) = 0.
4 (1+k) - 32 + 24 = 0.
4+4k - 8 = 0. 4k=4 ⇒ k=1.
New answer posted
4 months agoContributor-Level 10
a*b=c ⇒ a.c=0, b.c=0.
|c|² = |a|²|b|² - (a.b)² = (3)|b|² - 1. |c|=√2. So |b|²=1, |b|=1.
Projection of b on a*c.
a*c = a* (a*b) = (a.b)a - (a.a)b = a - 3b.
|a-3b|² = |a|²+9|b|²-6 (a.b) = 3+9-6 = 6.
l = |b. (a-3b)|/|a-3b| = | (a.b)-3|b|²|/√6 = |1-3|/√6 = 2/√6.
3l² = 3 (4/6) = 2.
New answer posted
4 months agoContributor-Level 10
f (x) = |sin²x, -2+cos²x, cos2x; 2+sin²x, cos²x, cos2x; sin²x, cos²x, 1+cos2x|.
R? →R? -R? , R? →R? -R?
f (x) = |sin²x, -2+cos²x, cos2x; 2, 2-2cos²x, 0; 0, 2-2cos²x, 1|.
f (x) = sin²x (2-2cos²x) - (-2+cos²x) (2) + cos2x (2 (2-2cos²x).
This seems tedious. From the solution, f (x)=4+2cos2x.
Max value when cos2x=1, f (x)=6.
New answer posted
4 months agoContributor-Level 10
e? F (x) = ∫ (3t²+2t+4F' (t)dt.
e? F (x)+e? F' (x) = 3x²+2x+4F' (x).
(e? -4)F' (x) = 3x²+2x-e? F (x).
F' (4) = (48+8-e? F (4)/ (e? -4).
Also F (3)=0, F (x)= (x³+x²-36)/ (e? -4) from solution. F (4)= (64+16-36)/ (e? -4) = 44/ (e? -4).
F' (4) = (56-e? (44/ (e? -4)/ (e? -4) = (56 (e? -4)-44e? )/ (e? -4)² = (12e? -224)/ (e? -4)².
α=12, β=4. α+β=16.
New answer posted
4 months agoContributor-Level 10
Vector on plane: (3-2, 7-3, -7- (-2) = (1,4, -5).
Line direction vector (-3,2,1).
Normal to plane n = (1,4, -5)* (-3,2,1) = (14,14,14) or (1,1,1).
Plane: 1 (x-3)+1 (y-7)+1 (z+7)=0 ⇒ x+y+z-3=0.
d = |-3|/√3 = √3. d²=3.
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