1is the answer for the second one (0,2) ?
its ans ellipse x2a2 + y24 = 1 ...
and the given point is the bottommost point on the curve....
* 1. Using the relation 2(1-\cos x)<x^2, x\neq 0 , or otherwise, prove that \sin(\tan x)\geq x, \text{ } \forall \text{ } x\in\left[0,\frac{\pi }{4} \right].
2. Find the point on the curve 4x2+a2y2=4a2, 4<a2<8 that is farthest from the point (0,-2) (need verification, so solution would be of great help)
* 3. Let A(p2, -p) , B(q2,q) and C(r2,-r) be the vertices of \triangle ABC. A parallelogram AFDE is drawn with vertices D,E & F on the line segment BC, CA & AB respectively. Show that maximum area of such a parallelogram is: \frac{1}{4}\left( p+q\right)\left(q+r \right)\left( p-r\right)
* 4. Using calculus prove that, \sin x\geq \frac{2x}{\pi} \text{ in } 0\leq x\leq \frac{\pi}{2}.
NOTE: Please provide only starting hints or important observations in questions marked with '*' .
Solutions maybe posted after sufficient hints have been posted :D Maybe after a day or two!
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11 Answers
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2622) just draw the ellipse.
given point is the lower end of minor axis .
max distance will be possible for ends of major axis or for other end of minor axis.
but for the given range of a, dist. is max for the other end of minor axis.
hence the point is (2,0)
11A more general inequality for the last sum is
|x|\le \pi \left|\sin \frac{x}{2} \right| for all x in [-\pi, \pi]
Also useful:
If \epsilon be so chosen such that 0<\frac{\epsilon}{2} <\pi-\frac{\epsilon}{2}<\pi
then, for all x in \left[\frac{\epsilon}{2}, \pi- \frac{\epsilon}{2} \right]
\sin x\ge \frac{2x}{\pi}
11For the last sum, take f(x)=Ï€sinx-2x.
Note that f(0)=f(Ï€2)=0
f'(x)=Ï€cosx-2.
f"(x)=-Ï€sinx<0 for all x in (0,Ï€2)
f'(0)>0 and f'(π/2)<0, so for some λ in that interval f'(x)=0. Thus f(x) is increasing (0,λ) and decreasing on (λ,π/2). From which it follows f(x)>0 for all all x in (0,π/2).
To extend the arguement to the generalised version, we use the fact that f(x)=-f(-x)....that does it.
21a better way (i feel)
take f(x) = sin x/x
f'(x)= (xcosx - sinx)/x2
the claim is sinx≥xcosx implies tanx≥x which is true when x is in [0,\pi/2]
thus f'(x)≤0 in the above interval
hence sinx/x ≥ sin(pi/2)/(pi/2) =2/pi :D
30Any hints for 3?
Aditya's method for 1 works [1] , but any hints to use the expression given to prove the inequality?
1for the 3rd one
let the sides ABC be some a , b and c respectively.
so the area of the given triangle (which is fixed) = sum of the areas of Δ CDE and Δ BDF and the parralellogram afde..
so the expression will be in terms of AE (= s) and AF (= t) and angle A...
so you'll get t = a(c-s)/c
put this in area of parralleogram
and differentiate wrt s (since its the only variable left)
so we get the answer to be half of area of the triangle [which shud be evident from the question]
111) I think the qsn setter meant something like this -
2(1-cosx)≤x2 implies cosx≥1-x22
Now put x2=a. Let 0≤x≤1
a(a-1)<0 since 0<a<1
So that gives
-a^2+2a-a+2>2\Rightarrow (2-a)(1+a)>2
Cross-multiplying, we have 1-\frac{a}{2}>\frac{1}{1+a}
So \cos x > \frac{1}{1+x^2 }
Put x=\tan x to have
\cos (\tan x ) \sec ^2 x-1>0
implying f'(x)>0
where f(x)=sin(tanx)-x.....
The inequality follows.
71Mujhe lagta hai mera dimag hi reverse ho jayega ye sab solution dekh kar..
aise ulta sochte kaise hai?
Nice!