Educational

Find f(f(1/x)) Find \( f(f(1/x)) \) Question: If \[ f(x)=1-\frac1x, \] then write the value of \[ f\left(f\left(\frac1x\right)\right). \] Solution: First, \[ f\left(\frac1x\right) = 1-\frac{1}{1/x} \] \[ =1-x \] Now, \[ f(f(1/x)) = f(1-x) \] \[ =1-\frac1{1-x} \] \[ =\frac{1-x-1}{1-x} \] \[ =\frac{x}{x-1} \] Therefore, \[ \boxed{f(f(1/x))=\frac{x}{x-1}} \] Next Question / Full Exercise

Find the Value of a Find the Value of \( a \) Question: Let \[ f(x)=\frac{ax}{x+1}, \qquad x\ne-1 \] Then write the value of \(a\) satisfying \[ f(f(x))=x \] for all \(x\ne-1\). Solution: Given, \[ f(x)=\frac{ax}{x+1} \] Therefore, \[ f(f(x)) = \frac{a\left(\frac{ax}{x+1}\right)} {\frac{ax}{x+1}+1} \] \[ = \frac{a^2x}{(a+1)x+1} \] Since \[ f(f(x))=x \] we get \[

Range of e^(x-[x]) Find the Range of the Function Question: Write the range of the function \[ f(x)=e^{x-[x]}, \qquad x\in R \] Solution: Since \[ 0\le x-[x]

Range of cos[x] Find the Range of the Function Question: Write the range of the function \[ f(x)=\cos[x] \] where \[ -\frac{\pi}{4}

Find the Value of f(π) Find the Value of \( f(\pi) \) Question: If \[ f(x)=\cos[\pi^2]x+\cos[-\pi^2]x \] where \([x]\) denotes the greatest integer less than or equal to \(x\), then write the value of \[ f(\pi) \] Solution: Since \[ \pi^2\approx9.86 \] \[ [\pi^2]=9 \] and \[ [-\pi^2]=-10 \] Therefore, \[ f(x)=\cos9x+\cos(-10x) \] \[ =\cos9x+\cos10x

Range of sin[x] Find the Range of the Function Question: Write the range of the function \[ f(x)=\sin[x] \] where \[ -\frac{\pi}{4}\le x\le \frac{\pi}{4} \] Solution: Since \[ -\frac{\pi}{4}\le x\le \frac{\pi}{4} \] we have \[ -1\le [x]\le0 \] Therefore, \[ [x]=-1 \quad \text{or} \quad 0 \] Hence, \[ f(x)=\sin(-1),\ \sin0 \] \[ =-\sin1,\ 0 \]

Find the Expression for f(x) Find the Expression for \( f(x) \) Question: If \(f\) is a real function satisfying \[ f\left(x+\frac1x\right) = x^2+\frac1{x^2} \] for all \[ x\in R-\{0\}, \] then write the expression for \(f(x)\). Solution: Let \[ t=x+\frac1x \] Squaring, \[ t^2=x^2+\frac1{x^2}+2 \] \[ x^2+\frac1{x^2}=t^2-2 \] Therefore, \[ f(t)=t^2-2 \] Replacing \(t\)

Range of |x| Find the Range of the Function Question: Write the range of the real function \[ f(x)=|x| \] Solution: Since \[ |x|\ge0 \] Minimum value is \(0\) at \(x=0\). Therefore, range is \[ \boxed{[0,\infty)} \] Next Question / Full Exercise

Domain of 1/√(x−|x|) Find the Domain of the Function Question: The domain of the function \[ f(x)=\frac1{\sqrt{x-|x|}} \] is (a) \(R_0\) (b) \(R^+\) (c) \(R^-\) (d) none of these Solution: Since square root is in denominator, \[ x-|x|>0 \] Case I: \(x\ge0\) \[ |x|=x \] \[ x-|x|=x-x=0 \] Not allowed. Case II: \(x

Find f(x)+f(1/x) Find \( f(x)+f(1/x) \) Question: If \[ f(x)=x^3-\frac1{x^3} \] then \[ f(x)+f\left(\frac1x\right) \] is equal to (a) \(2x^3\) (b) \(\frac2{x^3}\) (c) \(0\) (d) \(1\) Solution: \[ f\left(\frac1x\right) = \left(\frac1x\right)^3 – \frac1{\left(\frac1x\right)^3} \] \[ = \frac1{x^3}-x^3 \] Therefore, \[ f(x)+f\left(\frac1x\right) = \left(x^3-\frac1{x^3}\right) + \left(\frac1{x^3}-x^3\right) \] \[ =0 \] Hence, \[ \boxed{\text{Correct Answer: (c)}} \]