Find the Required Set Find the Required Set Question: If \( f:[-2,2]\to R \) is defined by \[ f(x)= \begin{cases} -1, & -2\le x\le0 \\ x-1, & 0\le x\le2 \end{cases} \] then \[ \{x\in[-2,2]:x\le0 \text{ and } f(|x|)=x\} \] is equal to (a) \(\{-1\}\) (b) \(\{0\}\) (c) \(\left\{-\frac12\right\}\) (d) \(\phi\) Solution: Since \(x\le0\), \[ |x|=-x\ge0

Find x Such That g(f(x))=8 Find the Values of \( x \) Such That \( g(f(x))=8 \) Question: If \[ f(x)=2x+3 \] and \[ g(x)=x^2+7 \] then the values of \(x\) such that \[ g(f(x))=8 \] are (a) \(1,2\) (b) \(-1,2\) (c) \(-1,-2\) (d) \(1,-2\) Solution: \[ g(f(x))=(2x+3)^2+7 \] Given, \[ (2x+3)^2+7=8 \] \[ (2x+3)^2=1

Range of f(x)=|x|/x Find the Range of \( f(x)=\frac{|x|}{x} \) Question: Let \[ A=\{x\in R:x\ne0,\,-4\le x\le4\} \] and \[ f:A\to R,\qquad f(x)=\frac{|x|}{x} \] for \(x\in A\). Then the range of \(f\) is (a) \(\{1,-1\}\) (b) \(\{x:0\le x\le4\}\) (c) \(\{1\}\) (d) \(\{x:-4\le x\le0\}\) Solution: If \(x>0\), \[ \frac{|x|}{x}=1 \] If \(x

Range of f(x)=cos²x+sin⁴x Find the Range of \( f(x)=\cos^2x+\sin^4x \) Question: The function \[ f(x)=\cos^2x+\sin^4x \] is defined from \(R\to R\). Then, \[ f(R)=? \] (a) \(\left[\frac34,1\right)\) (b) \(\left(\frac34,1\right]\) (c) \(\left[\frac34,1\right]\) (d) \(\left(\frac34,1\right)\) Solution: Using \[ \cos^2x=1-\sin^2x \] \[ f(x)=1-\sin^2x+\sin^4x \] Let \[ \sin^2x=t \] where \[ 0\le t\le1 \] Then, \[ f(x)=t^2-t+1 \] \[

Find f(2002) Find \( f(2002) \) Question: If \[ f(x)=\frac{\sin^4x+\cos^2x}{\sin^2x+\cos^4x}, \qquad x\in R \] then \( f(2002) \) is equal to (a) \(1\) (b) \(2\) (c) \(3\) (d) \(4\) Solution: Let \[ \sin^2x=a,\qquad \cos^2x=b \] Then, \[ a+b=1 \] Now, \[ f(x)=\frac{a^2+b}{a+b^2} \] Since \[ b=1-a \] numerator: \[ a^2+b = a^2+1-a \] denominator: \[

Find h(x) Find \( h(x) \) Question: Let \[ f(x)=x,\qquad g(x)=\frac1x \] and \[ h(x)=f(x)g(x) \] Then, \[ h(x)=1 \] for (a) \(x\in R\) (b) \(x\in Q\) (c) \(x\in R-Q\) (d) \(x\in R,\; x\ne0\) Solution: \[ h(x)=f(x)g(x) \] \[ =x\cdot\frac1x \] \[ =1 \] This is defined only when \[ x\ne0 \] Therefore, \[ h(x)=1

Find the Value of the Given Expression Find the Value of the Given Expression Question: If \[ f(x)=\cos(\log_e x) \] then \[ f\left(\frac1x\right) f\left(\frac1y\right) – \frac12 \left\{ f(xy)+f\left(\frac{x}{y}\right) \right\} \] is equal to (a) \(\cos(x-y)\) (b) \(\log(\cos(x-y))\) (c) \(1\) (d) \(\cos(x+y)\) Solution: \[ f\left(\frac1x\right) = \cos\left(\log\frac1x\right) = \cos(-\log x) = \cos(\log x) \] Similarly, \[

Find f(f(f(2))) Find \( f(f(f(2))) \) Question: If \[ f(x)=\frac{x+1}{x-1}, \qquad x\ne1 \] then \[ f(f(f(2))) \] is equal to (a) \(1\) (b) \(2\) (c) \(3\) (d) \(4\) Solution: \[ f(2)=\frac{2+1}{2-1}=3 \] \[ f(f(2))=f(3)=\frac{3+1}{3-1}=2 \] \[ f(f(f(2)))=f(2)=3 \] \[ \boxed{\text{Correct Answer: (c)}} \] Next Question / Full Exercise

Range of Rational Function Find the Range of the Function Question: The range of the function \[ f(x)=\frac{x^2-x}{x^2+2x} \] is (a) \(R\) (b) \(R-\{1\}\) (c) \(R-\left\{-\frac12,1\right\}\) (d) none of these Solution: Let \[ y=\frac{x^2-x}{x^2+2x} \] Then, \[ yx^2+2yx=x^2-x \] \[ (y-1)x^2+(2y+1)x=0 \] \[ x\left[(y-1)x+(2y+1)\right]=0 \] Since \(x=0\) is not in domain, \[ (y-1)x+(2y+1)=0 \] \[

Find f(2x)+f(−x)−f(x) Find \( f(2x)+f(-x)-f(x) \) Question: Let \( f:R\to R \) be defined by \[ f(x)=2x+|x| \] Then \[ f(2x)+f(-x)-f(x) \] is equal to (a) \(2x\) (b) \(2|x|\) (c) \(-2x\) (d) \(-2|x|\) Solution: \[ f(2x)=4x+|2x| =4x+2|x| \] \[ f(-x)=-2x+|-x| =-2x+|x| \] \[ f(x)=2x+|x| \] Therefore, \[ f(2x)+f(-x)-f(x) \] \[ =(4x+2|x|)+(-2x+|x|)-(2x+|x|) \] \[ =2|x| \]