1Q1
do gp summation and finish it
1) Show that for a positive integer n , the co-efficient of xk (0 ≤ k ≤ n) in the expansion of :
1 + (1+x) + (1+x)2 +......+ (1+x)n is n+1Cn-k .
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2)If f is differentiable at x=a, find the value of
Lt x2 f(a) - a2 f(x)x - a
x--->a
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3)If m,n be integers then find the value of \int_{-\pi }^{\pi }{(cos(mx)-sin(nx))^{2} dx}
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4)\int_{0}^{1000}{(e^{x-[x]}) dx} is equal to :
a)e1000 - 1 e - 1 b)e1000 - 1 1000 c)e - 1 1000 d)1000(e - 1)
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5)The value of \int_{0}^{\propto }{\frac{dx}{(x^{2}+4)(x^{2}+9)}} is
a)\pi60 b)\pi20 c)\pi40 d)\pi80
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10 Answers
39The 5th question can be done quickly by partial fractions. Here powers of x are uniformly 2, so employ a fake substitution x2 = t, and separate by partial fractions. Now put back t = x2 and then integrate.
In Q2, apply L'Hospital Rule.
Limx→a 2f(a)x - a2f'(x)1 - 0
= 2af(a) - a2f'(a)
Anything beyond this?
In Q4, e{x} lies between e0 and e1. You could do this by drawing a graph to visualise...though I can't seem to remember how to draw FPF(fractional part function) graphs.
11ya got it PRITHISH.....except Q4....how integrate??
neways thnx for tryng........
ne1 wid Q1 and Q3??
1q-4 options seem like
an attempt to modify g.p summation formula
so answer by tukka is A
well i am not having pen and paper
we can proceed officialy by braking in intevals of integers
11@gordo.. thnx ..... and ya i also got a in 5 following PRITISH's method!!
106Q1. use summation of GP
Q4. \int_{0}^{1000}{e^\left\{x \right\}}dx = 1000\int_{0}^{1}{e^xdx}
11@ashis thnx for rplyin..
@pritish ya i think its coz of periodicity only.....[12]