Class 12th
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New answer posted
10 months agoContributor-Level 9
f (x) = sin²x (λ + sinx)
f' (x) = 2sinxcosx (λ + sinx) + sin²x (cosx) = sinxcosx (2λ + 3sinx)
For extrema, f' (x) = 0
sinx = 0, cosx = 0, or sinx = -2λ/3
For more than 2 points of extrema in the interval, sinx = -2λ/3 must have solutions other than where sinx=0 or cosx=0.
-1 < -2/3 < 1 and -2/3 0
This gives λ ∈ (-3/2, 3/2) - {0}
New answer posted
10 months agoContributor-Level 10
Misch metal consists of Lanthanide metal (? 95%) and iron (? 5%) and traces of S, C, Ca and Al.
New answer posted
10 months agoContributor-Level 9
Given curves are y = x² - 1 and y = 1 - x² so intersection points are (±1,0). Bounded area =
4∫? ¹ (1 - x²)dx = 4 [x - x³/3]? ¹
= 4 (1 - 1/3) = 8/3 sq. units
New answer posted
10 months agoContributor-Level 9
Let y = (ex)?
ln y = x ln (ex) = x [1 + ln x]
(1/y) (dy/dx) = (1) (1 + ln x) + x (1/x) = 2 + ln x
⇒ dy = (ex)? (2 + ln x)dx
∫? ² (ex)? (2 + log? x)dx = [ (ex)? ]? ² = (2e)² - (1e)¹ = 4e² - e
New answer posted
10 months agoContributor-Level 10
For concentration cell E? cell = 0
Anode: Cu (s) → Cu²? (aq)?
Cathode: Cu²? (aq)? → Cu (s)
Overall: Cu²? (aq)? → Cu²? (aq)?
As ΔG = −nFEcell
If ΔG = -ve than Ecell is positive
Ecell = E? cell − (0.059/2) log (C? /C? )
Ecell = − (0.059/2) log (C? /C? )
Ecell > 0 ⇒ C? > C?
New answer posted
10 months agoContributor-Level 10
Relative lowering in vapour pressure depends on no. of mole of solute greater the no. of mole of solute greater in RLVP and smaller will be vapour pressure. So order of vapour pressure is B > C > A.
New answer posted
10 months agoContributor-Level 10
Clearly, ∫ [0 to n] {x}dx = n/2
∫ [0 to n] [x]dx = 1 + 2 + 3 . . . n − 1
= n (n-1)/2
∴ (n (n-1)/2)² = n/2 {10n (n-1)}
Solving, n = 21
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