Class 12th
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New answer posted
2 months agoContributor-Level 10
Given r x a = b x r, which means r x a + r x b = 0, so r x (a+b) = 0.
This implies r is parallel to (a+b). So, r = λ (a+b).
a = I + 2j - 3k, b = 2i - 3j + 5k
a+b = 3i - j + 2k.
r = λ (3i - j + 2k).
Given r ⋅ (αi + 2j + k) = 3. The OCR is unclear, but the equation appears to be r ⋅ (αi + 2j + k) = 3. The solution works with r ⋅ (αi + 2j + k) = 3, but theOCR says r. (ai+2j+k). Let's assume it's α.
λ (3i - j + 2k) ⋅ (αi + 2j + k) = 3
λ (3α - 2 + 2) = 3 => λα = 1.
Given r ⋅ (2i + 5j - αk) = -1.
λ (3i - j + 2k) ⋅ (2i + 5j - αk) = -1
λ (6 - 5 - 2α) = -1 => λ (1 - 2α) = -1.
From λα = 1, α = 1/λ.
λ (1 - 2/λ) = -1 => λ
New answer posted
2 months agoContributor-Level 9
f (x) = (4a - 3) (x + ln5) + 2 (a - 7)cot (x/2)sin² (x/2)
= (4a - 3) (x + ln5) + (a - 7)sin (x)
f' (x) = (4a - 3) + (a - 7)cos (x)
For critical points f' (x) = 0
cos (x) = - (4a - 3) / (a - 7) = (3 - 4a) / (a - 7)
⇒ -1 ≤ (3 - 4a) / (a - 7) ≤ 1
Solving this inequality leads to:
⇒ a ∈ [4/3, 2]
New answer posted
2 months agoContributor-Level 10
P (x) = x² + bx + c.
Given ∫? ¹ P (x) dx = 1.
∫? ¹ (x² + bx + c) dx = [x³/3 + bx²/2 + cx] from 0 to 1 = 1/3 + b/2 + c = 1.
2 + 3b + 6c = 6 => 3b + 6c = 4 - (i)
When P (x) is divided by (x-2), the remainder is 5. So, P (2) = 5.
(2)² + b (2) + c = 5 => 4 + 2b + c = 5 => 2b + c = 1 - (ii)
From (ii), c = 1 - 2b. Substitute into (i):
3b + 6 (1 - 2b) = 4
3b + 6 - 12b = 4
-9b = -2 => b = 2/9.
c = 1 - 2 (2/9) = 1 - 4/9 = 5/9.
We need to find 9 (b+c).
9 (2/9 + 5/9) = 9 (7/9) = 7.
New answer posted
2 months agoContributor-Level 10
dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C => C = 0.
So, y * secx = ln (secx).
y = cosx * ln (secx).
At x = π/4:
y = cos (π/4) * ln (sec (π/4) = (1/√2) * ln (√2) = (1/√2) * (1/2)ln (2) = ln (2) / (2√2).
New answer posted
2 months agoContributor-Level 10
The equation of the circle is x²+y²+ax+2ay+c=0, with a<0.
x-intercept = 2√ (g² - c) = 2√ (a/2)² - c) = 2√2. So, a²/4 - c = 2 => a² = 8 + 4c - (i)
y-intercept = 2√ (f² - c) = 2√ (a² - c) = 2√5. So, a² - c = 5 => a² = 5 + c - (ii)
Equating (i) and (ii): 8 + 4c = 5 + c => 3c = -3 => c = -1.
Substituting c in (ii): a² = 5 - 1 = 4. Since a < 0, a = -2.
The equation of the circle is x² + y² - 2x - 4y - 1 = 0.
Completing the square: (x-1)² + (y-2)² = 1+4+1 = 6.
The center is (1,2) and radius is √6.
The tangent is perpendicular to the line x + 2y = 0 (slope -1/2).
So, the slope of the tangent is 2.
Equation of the tangent: (y-2) = 2 (x-1)
New answer posted
2 months agoContributor-Level 9
K.E = φ - φ?
φ? = 3 eV = 3 * 1.6 * 10? ¹? J = 4.8 * 10? ¹? J
φ = hc/λ = (6.63 * 10? ³? * 3 * 10? ) / (248 * 10? ) J = 8 * 10? ¹? J
K.E = 8 * 10? ¹? - 4.8 * 10? ¹? = 3.2 * 10? ¹? J
Now using, λ = h / √ (2 K.E m)
λ = (6.63 * 10? ³? ) / √ (2 * 3.2 * 10? ¹? * 9.1 * 10? ³¹) m
λ = (6.63 * 10? ³? ) / (7.63 * 10? ²? ) m = 0.87 * 10? m = 8.7 Å
So, the nearest integer is 9.
New answer posted
2 months agoContributor-Level 9
T? = 300 K; K? = 1 * 10? ³ s? ¹
T? = 200 K; K? =?
E_a = 11.488 kJ/mol
Using Arrhenius equation:
log (K? /K? ) = (E_a / 2.303R) * [ (T? - T? ) / (T? )]
log (K? / 10? ³) = (11.488 * 10³ / (2.303 * 8.314) * [ (-100) / (6 * 10? )]
log (K? / 10? ³) = -1
K? / 10? ³ = 10? ¹
K? = 10? s? ¹ or K? = 10 * 10? s? ¹
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