Class 12th
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New answer posted
10 months agoContributor-Level 10
(a + √2bcosx) (a - √2bcosy) = a² - b²
⇒ a² - √2abcosy + √2abcosx - 2b²cosxcosy = a² - b²
Differentiating both sides:
0 - √2ab (-siny dy/dx) + √2ab (-sinx) - 2b² [cosx (-siny dy/dx) + cosy (-sinx)] = 0
At (π/4, π/4):
ab dy/dx - ab - 2b² (-1/2 dy/dx + 1/2) = 0
⇒ dy/dx = (ab+b²)/ (ab-b²) = (a+b)/ (a-b); a, b > 0
New answer posted
10 months agoContributor-Level 10
f (x) = a? ⋅ (b? * c? ) = |x -2 3; -2 x -1; 7 -2 x|
= x³ - 27x + 26
f' (x) = 3x² - 27 = 0 ⇒ x = ±3 and f' (-3) < 0
⇒ local maxima at x = x? = -3
Thus, a? = -3i? - 2j? + 3k? , b? = 2i? - 3j? - k? , and c? = 7i? - 2j? - 3k?
⇒ a? ⋅ b? + b? ⋅ c? + c? ⋅ a? = 9 - 5 - 26 = -22
New answer posted
10 months agoContributor-Level 10
f (2)=8, f' (2)=5, f' (x) ≥ 1, f' (x) ≥ 4, ∀x ∈ (1,6)
Using LMVT
f' (x) = (f' (5) - f' (2)/ (5-2) ≥ 4 ⇒ f' (5) ≥ 17
f' (x) = (f (5) - f (2)/ (5-2) ≥ 1 ⇒ f (5) ≥ 11
Therefore f' (5) + f (5) ≥ 28
New answer posted
10 months agoContributor-Level 10
1/λmin = R [1/1² - 1/∞²] = R [n=∞ → n=1]
1/λmax = R [1/1² - 1/2²] = 3R/4 [n=2 → n=1]
⇒ Δλ ⇒ 4/3R - 1/R ⇒ 1/3R ⇒ 340
For Paschan ⇒ 1/λmin = R [1/9] [n=∞ → h=3]
⇒ 1/λmax = R [1/9 - 1/16] = 7R/144
⇒ Δλ = 81/7R
New answer posted
10 months agoContributor-Level 10
M = L/f? [1 + D/fe]
⇒ 100 = 20/Fe [1 + 25/Fe]
⇒ 1 + 25/Fe = 5
⇒ Fe = 25/4 = 6.25 cm
New answer posted
10 months agoContributor-Level 10
↑? (1) ε = 3
(2) ε – Ir = 2.5 V
⇒ Ir = 0.5
Now, IR = 2.5
⇒ R/r = 5.
⇒ PR/Pr = I²R/I²r = R/r = 5
⇒ Pr = 0.5/5 = 0.1
New answer posted
10 months agoContributor-Level 10
Uinitial = k(4q)(q)/(d/2) + k(q)(-q)/(d/2)
⇒ 6kq²/d
⇒ Ufinal = k(4q)(q)/((3d/2)) + k(q)(-q)/(d/2)
⇒ 2kq²/3d
⇒ ΔU = (2/3 - 6)kq²/d = -16kq²/3d
New answer posted
10 months agoContributor-Level 10
τ? = μ? * B?
⇒ 0.018 = μ(0.06)(sin 30°)
⇒ μ = 0.6
⇒ Work = Uf – Ui
⇒ 2μB
⇒ 7.2 * 10?² J.
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