Domain of Square Root Rational Function Find the Domain of the Function Question: The domain of the function \[ f(x)=\sqrt{\frac{(x+1)(x-3)}{x-2}} \] is (a) \([ -1,2)\cup[3,\infty)\) (b) \(( -1,2)\cup[3,\infty)\) (c) \([ -1,2]\cup[3,\infty)\) (d) none of these Solution: For square root function, \[ \frac{(x+1)(x-3)}{x-2}\ge0 \] Critical points are \[ -1,\;2,\;3 \] Using sign analysis, \[ \frac{(x+1)(x-3)}{x-2}\ge0 \] […]

Domain of Rational Square Root Function Find the Domain of the Function Question: The domain of definition of \[ f(x)=\sqrt{\frac{x+3}{(2-x)(x-5)}} \] is (a) \((-\infty,-3]\cup(2,5)\) (b) \((-\infty,-3)\cup(2,5)\) (c) \((-\infty,-3]\cup[2,5]\) (d) none of these Solution: For square root function, \[ \frac{x+3}{(2-x)(x-5)}\ge0 \] Critical points are \[ x=-3,\;2,\;5 \] Using sign analysis, \[ \frac{x+3}{(2-x)(x-5)}\ge0 \] for \[ (-\infty,-3]\cup(2,5)

Domain of √(2−2x−x²) Find the Domain of \( f(x)=\sqrt{2-2x-x^2} \) Question: The domain of the function \[ f(x)=\sqrt{2-2x-x^2} \] is (a) \([-\sqrt3,\sqrt3]\) (b) \([-1-\sqrt3,-1+\sqrt3]\) (c) \([-2,2]\) (d) \([-2-\sqrt3,-2+\sqrt3]\) Solution: For square root function, \[ 2-2x-x^2\ge0 \] \[ x^2+2x-2\le0 \] Solving, \[ x=\frac{-2\pm\sqrt{4+8}}{2} \] \[ x=-1\pm\sqrt3 \] Since quadratic expression is \(\le0\), \[ -1-\sqrt3\le x\le -1+\sqrt3

Find the Value of the Given Function Find the Correct Option Question: If \[ f(x)=\sin[\pi^2]x+\sin[-\pi^2]x \] where \([x]\) denotes the greatest integer less than or equal to \(x\), then (a) \(f\left(\frac{\pi}{2}\right)=1\) (b) \(f(\pi)=2\) (c) \(f\left(\frac{\pi}{4}\right)=-1\) (d) none of these Solution: Since \[ \pi^2\approx9.86 \] \[ [\pi^2]=9 \] Also, \[ [-\pi^2]=-10 \] Therefore, \[ f(x)=\sin 9x+\sin(-10x)

Relation Between f(x) and f(1−x) Find the Relation Between \( f(x) \) and \( f(1-x) \) Question: If \[ f(x)=\frac{4^x}{4^x+2}, \qquad x\in R \] then (a) \(f(x)=f(1-x)\) (b) \(f(x)+f(1-x)=0\) (c) \(f(x)+f(1-x)=1\) (d) \(f(x)+f(x-1)=1\) Solution: \[ f(1-x) = \frac{4^{1-x}}{4^{1-x}+2} \] \[ = \frac{4/4^x}{4/4^x+2} \] \[ = \frac{2}{2+4^x} \] Now, \[ f(x)+f(1-x) = \frac{4^x}{4^x+2} + \frac{2}{4^x+2} \]

Find f(x) Find \( f(x) \) Question: If \[ 3f(x)+5f\left(\frac1x\right)=\frac1x-3 \] for all non-zero \(x\), then \(f(x)=\) (a) \(\frac1{14}\left(\frac3x+5x-6\right)\) (b) \(\frac1{14}\left(-\frac3x+5x-6\right)\) (c) \(\frac1{14}\left(-\frac3x+5x+6\right)\) (d) none of these Solution: Replace \(x\) by \(\frac1x\), \[ 3f\left(\frac1x\right)+5f(x)=x-3 \] Given, \[ 3f(x)+5f\left(\frac1x\right)=\frac1x-3 \] Solving the equations, \[ 14f(x)=5x-\frac3x-6 \] Therefore, \[ f(x)=\frac1{14}\left(-\frac3x+5x-6\right) \] \[ \boxed{\text{Correct Answer: (b)}} \] Next

Find f(α) and f(β) Find \( f(\alpha) \) and \( f(\beta) \) Question: If \[ f(x)=64x^3+\frac1{x^3} \] and \(\alpha,\beta\) are the roots of \[ 4x+\frac1x=3, \] then (a) \(f(\alpha)=f(\beta)=-9\) (b) \(f(\alpha)=f(\beta)=63\) (c) \(f(\alpha)\ne f(\beta)\) (d) none of these Solution: Given, \[ 4x+\frac1x=3 \] Cubing both sides, \[ \left(4x+\frac1x\right)^3=3^3 \] \[ 64x^3+\frac1{x^3} +12\left(4x\right)\left(\frac1x\right)\left(4x+\frac1x\right) =27 \] \[

Find f(α) and f(β) Find \( f(\alpha) \) and \( f(\beta) \) Question: If \[ f(x)=27x^3+\frac1{x^3} \] and \(\alpha,\beta\) are roots of \[ 3x+\frac1x=12, \] then (a) \(f(\alpha)\ne f(\beta)\) (b) \(f(\alpha)=10\) (c) \(f(\beta)=-10\) (d) none of these Solution: Given, \[ 3x+\frac1x=12 \] Cubing both sides, \[ \left(3x+\frac1x\right)^3=12^3 \] \[ 27x^3+\frac1{x^3} +9\left(3x+\frac1x\right) =1728 \] \[ f(x)+9(12)=1728

Find the Value of k Find the Value of \( k \) Question: If \[ e^{f(x)}=\frac{10+x}{10-x}, \qquad x\in(-10,10) \] and \[ f(x)=k\,f\left(\frac{200x}{100+x^2}\right), \] then \(k=\) (a) \(0.5\) (b) \(0.6\) (c) \(0.7\) (d) \(0.8\) Solution: Given, \[ e^{f(x)}=\frac{10+x}{10-x} \] Therefore, \[ f(x)=\log\left(\frac{10+x}{10-x}\right) \] Now, \[ f\left(\frac{200x}{100+x^2}\right) = \log\left( \frac{ 10+\frac{200x}{100+x^2} }{ 10-\frac{200x}{100+x^2} } \right) \] \[

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