106arshad try taking x = pi/6.. we get π2/36 - π/6 which is negative so, it is less than the min value obtained
Q1 let f:R→R be defined as f(x)=x2+ax+1x2+x+1.the set of all exhasutive values of a for which f(X) is onto
ans given is φ..it is wrong naa ??
Q2 whats min value of (sin-1(sin(x)))2-sin-1(sin(x))
Q3 if f(x)=x(2-x), 0≤x≤2.If the definition of f is extended over set R~[0,2] by f(x+1)=f(x)..then period is ??
is it periodic ??
Q4 If k=0Σn f(x+ka)=0 where a>0,,then find period of f(x)
is it periodic ???
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17 Answers
106
1yup ans of question 2 is right
min value at x=(-pi/2)
so min value will be
(π2/4)+( π/2)
24At first look soln4 looked okay..but now [2]
sir ,can u explain 2nd and 3rd step please..... [7][7][12]
106Q2. Let sin-1(sinx) = t, t ε [-π/2,π/2]
then we require min value of t2-t
This is quadratic exp. whose min value is at -(-1/2) = 1/2 which is within the range of t
So the min valu = 1/4 - 1/2 = -1/4
624th one
k=0Σn f(x+ka)=0
k=0Σn f(x+a+ka)=0
k=1Σn+1 f(x+ka)=0
substract the 1st and third...
f(x+(n+1)a) = f(x)
hence the period is (n+1)a
24the same thing i thought...but strange thing is that fiitjee guys have given the answer[3]
623rd one is not periodic eureka.. bcos the right side and the left side of the y axis will be mirror images of each other...
1it is all right
and i think the answer is 1
just solve it using the minima and maxima concept
24@asish,,,u r wrong..
@prophet sir....thx for soln...
so whats the final ans ??
3411. basically you have to prove that the range of f(x) does not span R.
this is easy to see if you write the expression in terms of t = x + \frac{1}{x}
Then we have f(t) =1 + \frac{a-1}{t+1}
Now t \not \in (-2,2) \Rightarrow \frac{1}{t+1} \in \left[-1, \frac{1}{3} \right] so that f(t) is bounded by 2-a and \frac{a+2}{3} which doesnt cover the real line for any real a.