662. Apply Rolle's theorem to the function
f(x)=ax^n+bx^{n-1}+cx
in the interval [0,1].
Here is a good enough test for all maths lovers , and I really liked the question patterns , so I am uploading some of them ---- A few are very confusing , so please help in solving those ---- 1 > Find the sum of the following series ---- cos x + 2 cos 2x + 3 cos 3x + 4 cos 4x ……….. + n cos nx
2 > If a + b + c = 0 , then prove that the equation naxn - 1 + ( n - 1 )bxn - 2 + c = 0 has at least one root in the interval ( 0 , 1 ).
3 > Prove that if p , q , k are all greater than 1 ; a1 , a2 , a3 , a4 …..are all numbers which are positive and whose sum is unity , then ( a1p + a1q )k + ( a2p + a2q )k ……. ( anp + anq ) ≥ ( np + nq )k( n )kp + kq - 1
4 > Find all the positive , integral , possible values of a and b for which the equality ab2 = ba holds.
5 > Find the range of values a natural number a can take so that for all real and positive x , x + a x -2 > 2 .
6 > a > Prove that , ( b - a ) sec 2 a < tan a - tan b < ( b -a ) sec 2 b Where 0 < a < b < Î 2
b > Prove that the polynomial 1 + x / 1! + x / 2! + x / 3! …. + x / n! = 0 cannot have a repeated root .
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5 Answers
661. Consider, instead, the following sum:
S=e^{ix}+2e^{i2x}+3e^{i3x}+\ldots + n e^{inx}
The required sum in the problem is, then, the real part of S. Multiply S with eix and subtract from S to obtain
(1-e^{ix})S=e^{ix}+e^{i2x}+e^{i3x}+\ldots + e^{inx} -ne^{i(n+1)x}
\Rightarrow\ (1-e^{ix})S=e^{ix}\dfrac{1-e^{inx}}{1-e^{ix}} -ne^{i(n+1)x}
As such we obtain,
S=e^{ix}\dfrac{1-e^{inx}}{(1-e^{ix})^2} -\dfrac{ne^{i(n+1)x}}{1-e^{ix}}
Taking the real part and simplifying, we get the required sum
\sum_{k=1}^nk\cos kx = \dfrac{n}{2}\ \dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{x}{2}}-\dfrac{1}{2}\ \dfrac{\sin^2\frac{nx}{2}}{\sin^2\frac{x}{2}}
66666. a) The correct inequality should be
(b-a)\sec^2a<\tan b-\tan a < (b-a)\sec^2b
To see this, we see that in (0,\pi/2), sec2x is an increasing function. As such, we get
\sec^2a< \dfrac{1}{(b-a)}\int_a^b \sec^2x\ \mathrm dx< \sec^2b
from where the result follows.
15 >
let f(x)=x+ax-2
so for maxima or minima , lets put f ' (x) = 0
we get 1 - 2ax-3 =0
or x =(2a)1/3
again f '' (x) = 6ax-4 > 0 for all positive x and all natural a .
so x = (2a)1/3 is the point of global minima .
thus (2a)1/3 + a . (2a)1/3 > 2
so we get a >32 / 27
or a >= 2
as a is a natural number
done
