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the rhs can be calculated for any x iff x is a natural no.
but while calculating the derivative of the rhs , f(x+h) = (x+h)+....(x+h) times is not defined since x+h is not a natural no.
hence derivative of the rhs does not exist ....and hence the contradiction....:))
nice to reply after such a long time.......
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v=u-gt = 20- 10*4 = -20 m /s in upward dir..
that means its 20 m /s in the downward direction..
now we know :
1/v + 1/u = 2/R
differentiating:
-1/v^2 dv/dt =1/u^2du/dt
now du/dt can b easily foud by taking v=u+gt ....and so u can easily get dv/dt ...:)
if there' any doubt do nudge me..:)
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if p >0 ; f(x) is increasing bcoz px is incres. ..so bijective
if p<0 , f(x) is decreasing bcoz px is decreasing....so bijective
but if p =0 f(x)=sinx which is many one...
hence p =R- {0}
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u cant solve this ques... some data is missing...
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see to make all possible arr. of ascending ordered nos ,
we can do it by taking any possible combination amd then arranging it in ascending order.
first of all we find all the poss ways to choose 3 nos.
next we arrange tyhem in ascending order
since there is only one way to arrange any particular set of 3 nos so the total wways possible=total ways to choose 3 diff no sets=6C3
so prob=6C3/6*6*6
:)
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3 nos can b chosen in 6C3 ways..
now they can b arranged in only one way side by side such that they are in ascending order.
so total ways such that each no is larger than the previous one =6C3..
now the total no of ways possible=6*6*6 ways
therefiore req prob = 6C3/6*6*6 = 20/ 6*6*6 = 5/ 54 .......
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the redq no of ways : n! { summation of (-1)^r * nCr}. where r varies from r=1 to r=n...
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the number of integral solns for a1
is equal to the coeff of x^n in the expression
(x^a1 + x^(a1+1)+...+ x^b1)( x^a2+ .....+x^b2).......till n terms..
the abv series form a gp series ... and so we get terms like (1-x^r)/(1-x)...
also the coefficient of x^r in (1-x)^-n is (n+r-1)Cr........
so from these u can get the coefficient of x^n...
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ya sorry for the mistake by me... it will b m and not n..
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a,ar,ar^2, (21-a)
now 2ar^2= 21-a + ar...
ar + ar^2 =18..
now solve both the eqn by sustituting the value of a in terms of r .
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(x-2)^(x-2)(x-4)>1
since (x-2) has to b >0 so we can take llog on both sides ..
taking log we get:
(x-2)(x-4)* log(x-2)>0...
now consider several cases:
i)2the abv lhs >0...so satisfies
ii) x=3
LHS=0 so does not satisfy..
iii)x>3
LHS<0 .. so doesnt satisfy.
iv) x=4
0.. so doesnt satisfy.
v) x>4
LHS>0 so satisfies...
so the solution is x =(2,3) U(4,infinity).....
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the first prize can b distributed in m ways..
simlarly the 2nd prize can also b distr. in m ways...
and so on ....
hence total no of ways og distributing all the n prizes = m*m*m*...= m^n ways..
but the abv calculation also includes the case when all the n prizes are given to a single person ..
total ways of dist. all the n prizes among m persons= n ways ..
so the reqd. no of ways so that each person entitled to receive at most (n-1) prizes, = m^n- n ways
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