341You know, it would be easier if you posted separate qns in separate threads and in the appropriate sub-fora.
let f(x) , x ε [0,∞) be a non negative continuous function.
If f '(x) cosx ≤ f(x) sinx for all x≥0 then value of f(2π) is-
a)0
b)1
c)Î
d)none
-
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23 Answers
24ya ans should be c,d in my opinio..but it would be good to take expert opinion on this one[1]
62S[n]=3/4 +15/16 +63/64 +......
find S[n] ?
a) n- 1/3*4^n - 1/3
b) n +1/3*4^(-n) -1/3
c) n + 1/3*4^(n) -1/3
d) n- 1/3*4^(-n) +1/3
S[n] = (1-1/4)+(1-1/16)+(1-1/64)+....
= n - (1/4+1/16+1/64....)
=n-\frac{1}{4}\times\frac{1-(1/4^n)}{1-1/4} =n-\frac{1}{3}\times[1-(1/4^n)]
Hence option b [1]
341
1S[n]=3/4 +15/16 +63/64 +......
find S[n] ?
a) n- 1/3*4^n - 1/3
b) n +1/3*4^(-n) -1/3
c) n + 1/3*4^(n) -1/3
d) n- 1/3*4^(-n) +1/3
1post #10 (b) option is correct
see the case of f(x)=(1-x3)1/3
points of intersection of the line and the inverse does not lie on line y=x
11/2\int_{0}^{\Pi /2}{f(2x)+f"(2x)}sin2xdx
let 2x=z
then I=1/4\int_{0}^{\Pi }{f(z)+f"(z)}sinzdz
ADDING AND SUBTRACTING f'(z)cosz
I=1/4\int_{0}^{\Pi }[{f(z)sinz-f'(z)cosz}+{f'(z)cosz+f"(z)sinz}]dz
I=1/4\int_{0}^{\Pi }[d/dz{-f(z)cosz}+d/dz{f'(Z)sinz}]dz
or I=[-f(z)cosz+f'(z)sinz]0 to π
putting limits I=0
because f(0)=f(Ï€)=0
and sin0=sinπ=0
thus option d is correct
3411st one:
f'(x) \cos x \le f(x) \sin x \Rightarrow \int_{\frac{\pi}{2}}^{2 \pi} f'(x) \cos x \ dx \le \int_{\frac{\pi}{2}}^{2 \pi} f(x) \sin x \ dx
\Rightarrow \left | f(x) \cos x \right|_{\frac{\pi}{2}}^{2 \pi} + \int_{\frac{\pi}{2}}^{2 \pi} f(x) \sin \ dx \le \int_{\frac{\pi}{2}}^{2 \pi} f(x) \sin x \ dx
\Rightarrow f(2 \pi) \le 0
But since f(x) is given to be non-negative we have f(2 \pi) \ge 0
From the above two inequalities, it is evident that f(2 \pi) = 0
4Ans is given as A B D
Even I was doubtful by the given ans
So, can v go with c d
24a) is wrong...if a fn is defined such that f(x)=2 .so it lies on x axis...and its mirror image in y=x will be on y axis (y=2)...and they are intersecting orthogonally clearly...
b) is also wrong,,,,f(X) and f inverse must intersect on y=x only
c) also looks true..but not sure
d) is true,,,,it was discussed some time back here..
4MULTIPLE ANSWER QS
If f(x) has finite slope everywhere and is an invertible function, then
(A) y = f(x) & y = f-1 (x) cannot intersect orthogonally
(B) y = f(x) & y = f-1 (x) may have points of intersection not on the line y=x
(C) If f(x) ≠x then ∫[a-->b] f(x)dx ≠∫[a-->b] f-1(x) dx (a≠b)
(D) ∫[a-->b] f(x) dx + ∫[f(a) ---> f(b)] f-1(x) dx = bf(b) - af(a) . Given that f'(x)>0
for all x belongs to R
4If f(x) is a continuous function in [0,pi] such that f(0) = f(pi) = 0 then the values of
∫[0 to pi/2] [ f(2x) + f' ' (2x) ] sinx cosx dx is equal to
a) pi
b) 2 pi
c) 3 pi
d)None
3see f''(x) cannot be less than zero as it will cause the slope to become negative and hence f(x) which contradicts ""non negative"" !!!!!!!!! so f''(x)=0!! thus slope is constant
and f'(0)<=0 !!!!! so f'(x) = 0!!!!
also we have 0<=f(x)sinx ..... but!! sinx can fluctuate betw + and negative which is not desireable .!!!!! therfore f(x) =0!!!
1and pls tell how u got 0
i keep getting 1....
pls tell something apart from the fiitjee solution