difficult@bansal

q no 1)
let n>=2 be a fixed integer.......
f(x) be a bounded function defined in f:(0,a)→R
satisfying f(x)=\frac{1}{n^2}\sum_{r=0}^{(n-1)a}{f(x+\frac{r}{n})}
then f(x)=
a)-f(x)
b)2f(x)
c)f(2x)
d)nf(x)

6 Answers

1
Arshad ~Died~ ·

Q no. 2)
Let f:[a,b]→R (0<a<b) be a continuous function in [a,b] and differentiable in (a,b) then there exista some c in (a,b) for which
a)f ' (c) > \frac{5}{a-c}

b)f ' (c)>\frac{3}{a-c}

c)f ' (c)>\frac{2}{a-c}

d)f ' (c)<\frac{2}{b-c}

1
Arshad ~Died~ ·

noteeeeeeeee----
all questions are multi correct ones.......

11
Tush Watts ·

Ans2) Let F(x) =(x-a)(x-b)e f(X); as function f(x) holds Rolle's theorm

F ' (c) = 0 implies, f'(c) = 1 / (a-c) + 1 / (b-c)

So, therefore, f'(c) > 2 / (a-c) and f ' (c) < 2 / (b-c)

So, ans is (c) , (d)

341
Hari Shankar ·

Q1 - let K be an upper bound

Then RHS ≤ Ka/n

Likewise if k is a lower bound then RHS≥ka/n

Thus ka/n ≤ f(x)≤Ka/n for all n

But since n can be made arbitrarily large, f(x) = 0.

1
Arshad ~Died~ ·

thank u tushar and thank u prophet sir...
:-)

1
Arshad ~Died~ ·

both of these are awesome questions for jee.........everyone should try to do them

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