bits questions

hmmm...got it...but in my post though a nasty 1...was it faulty???

What will this compound yeild on oxidative ozonolysis?

The products are CHO-CH₂-CHO and CHO-C(CH₃)₂-CHO.

ans pritish ur ans for the ozonolysis is wrong....the products will be
OHOC-CH₂-COOH and OHOC-C(CH₃)₂COOH....

it is oxidative workup....so ald will b converted to carboxylic acids....

Let (7+2√5)²ⁿ⁺¹ = 7²ⁿ⁺¹(1 + (2/7)√5)²ⁿ⁺¹ = I + f

now 7²ⁿ⁺¹(1 - (2/7)√5)²ⁿ⁺¹ = f' (0 < f '< 1)

now I + f + f ' = 7²ⁿ⁺¹ . 2 (something....) = even integer...

now 0 < f < 1

and 0 < f' < 1

so 0 < f + f' < 2

so f + f' = 1

so I + 1 = even integer

so I = odd integer....

K_p K_q qsn...

apply K_p=K_q(RT)^∂n_g

Oh yeah..oxidative workup..sorry lol.
If that's the answer to your Ph-CH...etcetc question then I guess the only way to find out is to make resonating structures of the radical.

and pritish....can u do nake the resonating structures plzzz...

which calculus prob?

for post #105...i think Cv value is wrong.......

@ ray

ya i got post num 107 and that gravitation one too....

thanks....[4]

and i was talking baout the post number 58......i got that derivative of teh given function is 0....but hw to find the value of the function itlsef???

i marked something don't know if it is always valid or not
but

in that question (7+2√5)²ⁿ⁺¹ that sandipan solved in post #100
2√5 = 4.472
and other is 7

we notice that 2√5 < 7
so we use f + f ' in the process as sandipan did

had it been ( 5√5 + 11 )²ⁿ⁺¹ we use f - f ' in the process to find if integral part is even or odd

NOTE. see i don't know if it is always valid but as far as i'm concerned jitne bhi question maine kiye hain unme ye valid hua hai

So this trick might help unless u know how to calculate square roots of numbers like 5 , 7 ,etc [1]

extemely sorry guys....... the ques was supposed to be
(4+2√5)²ⁿ⁺¹. i solved this sum months ago i got it even but when sandi asked if it is always odd then i opened the book and copied from its explanation without seeing that all 4s in the explanation was misprinted as 7.
SORRY GUYS. EXTREMELY SORRY.

Well, #58, the value is \frac{\pi }{4}

Let \phi (x)= k for all x

(here \phi (x) is the original function)

\Rightarrow \phi (\frac{\pi }{4})= k

\int_{0}^{\frac{1}{2}}{\sin^{-1} }\sqrt{x}dx+\int_{0}^{\frac{1}{2}}{\\cos^{-1} }\sqrt{x}dx=k

\int_{0}^{\frac{1}{2}}{\frac{\pi }{2}}dx=k

k=\frac{\pi }{4}

This gives,
\phi (x)=\frac{\pi }{4}

for all x.

How did you assume that it's a constant function, Ray?
Ah I get it...the Leibnitz differentiation. Nicely done :)

me still nt getting........[2]

pi by four kahan se aya...apne ap se suppose kara kya....so that we can combine the two integrals???

okay..thanks ray and pritish....

one more doubt....

using a scalar triple product...we can find the volume of a tetrahedron and a parallelopiped....are there any other geometric interpretations???