Class 12th
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New answer posted
10 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
10 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
10 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
10 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
10 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.
New answer posted
10 months agoContributor-Level 10
(a+b+c)² = a²+b²+c²+2 (ab+bc+ca)
1² = a²+b²+c²+2 (2) ⇒ a²+b²+c² = -3.
a²b²+b²c²+c²a² = (ab+bc+ca)² - 2abc (a+b+c) = 2² - 2 (3) (1) = -2.
a? +b? +c? = (a²+b²+c²)² - 2 (a²b²+b²c²+c²a²) = (-3)² - 2 (-2) = 9+4=13.
New answer posted
10 months agoContributor-Level 10
Δ = |1,1, -1; 1,2, α 2, -1,1| = 1 (2+α)-1 (1-2α)-1 (-1-4) = 2+α+2α-1+5 = 3α+6=0 ⇒ α=-2.
Δ? = |2,1, -1; 1,2, α β, -1,1| = 2 (2+α)-1 (1-αβ)-1 (-1-2β) = 4+2α-1+αβ+1+2β = 4+2α+αβ+2β=0.
4-4-2β+2β=0. This holds.
Δ? = |1,2, -1; 1,1, α 2, β,1| = 1 (1-αβ)-2 (1-2α)-1 (β-2) = 1-αβ-2+4α-β+2 = 1+4α-αβ-β=0.
1-8+2β-β=0 ⇒ -7+β=0 ⇒ β=7.
α+β = -2+7 = 5.
New answer posted
10 months agoContributor-Level 10
sec y dy/dx = 2sinxcosy.
sec²y dy = 2sinx dx.
tan y = -2cosx + C.
y (0)=0 ⇒ 0=-2+C ⇒ C=2.
tan y = 2-2cosx.
y' = (-2sinx)/sec²y.
5y' (π/2) = 5 (2sin (π/2)/sec² (π/2)
sec²y dy/dx = 2sinx.
y' (π/2)? At x=π/2, tan y = 2. sec²y = 1+tan²y = 5.
5 (2sin (π/2) = 5 (2)=10.
New answer posted
10 months agoContributor-Level 10
P (x) = a (x-2)² + b (x-2) + c.
lim (x→2) P (x)/sin (x-2) = lim (x→2) P (x)/ (x-2) = P' (2) = 7.
P' (x) = 2a (x-2) + b. P' (2) = b = 7.
P' (x) = 2a.
P (3) = a (1)² + b (1) + c = a+b+c = 9.
Continuity at x=2 means lim f (x) = f (2).
lim (x→2) (a (x-2)²+b (x-2)+c)/ (x-2) = P' (2) = b=7. This is given.
The problem states f (2)=7.
P (x) = (x-2) (ax+b) form used in solution. Let's use this.
lim (x→2) (x-2) (ax+b)/sin (x-2) = lim (x→2) ax+b = 2a+b = 7.
P (3) = (3-2) (3a+b) = 3a+b=9.
Solving: a=2, b=3.
P (x) = (x-2) (2x+3).
P (5) = (5-2) (2*5+3) = 3 * 13 = 39.
New answer posted
10 months agoContributor-Level 10
Using L'Hopital Rule:
lim (x→2) (2xf (2) - 4f' (x)/1 = 2 (2)f (2) - 4f' (2) = 4 (4) - 4 (1) = 12.
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