Compare f(xy) and f(x)f(y) Compare \( f(xy) \) and \( f(x)f(y) \) Question: Let \[ f(x)=\sqrt{x^2+1} \] Then which of the following is correct? (a) \(f(xy)=f(x)f(y)\) (b) \(f(xy)\ge f(x)f(y)\) (c) \(f(xy)\le f(x)f(y)\) (d) none of these Solution: \[ f(xy)=\sqrt{x^2y^2+1} \] and \[ f(x)f(y) = \sqrt{x^2+1}\sqrt{y^2+1} \] Squaring both sides, \[ [f(x)f(y)]^2 = (x^2+1)(y^2+1) \] \[ […]

Range of |x−1| Find the Range of the Function Question: The range of the function \[ f(x)=|x-1| \] is (a) \((-\infty,0)\) (b) \([0,\infty)\) (c) \((0,\infty)\) (d) \(R\) Solution: Since modulus is always non-negative, \[ |x-1|\ge0 \] Minimum value occurs at \[ x=1 \] \[ |1-1|=0 \] There is no maximum value. Therefore, range is \[

Range of (x+2)/|x+2| Find the Range of the Function Question: The range of the function \[ f(x)=\frac{x+2}{|x+2|}, \qquad x\ne-2 \] is (a) \(\{-1,1\}\) (b) \(\{-1,0,1\}\) (c) \(\{1\}\) (d) \((0,\infty)\) Solution: If \(x+2>0\), \[ \frac{x+2}{|x+2|}=1 \] If \(x+2

Range of x/|x| Find the Range of the Function Question: The range of the function \[ f(x)=\frac{x}{|x|} \] is (a) \(R-\{0\}\) (b) \(R-\{-1,1\}\) (c) \(\{-1,1\}\) (d) none of these Solution: If \(x>0\), \[ \frac{x}{|x|}=1 \] If \(x

Domain of √(5|x|−x²−6) Find the Domain of the Function Question: The domain of the function \[ f(x)=\sqrt{5|x|-x^2-6} \] is (a) \(( -3,-2)\cup(2,3)\) (b) \([ -3,-2]\cup[2,3)\) (c) \([ -3,-2]\cup[2,3]\) (d) none of these Solution: For square root function, \[ 5|x|-x^2-6\ge0 \] Case I: \(x\ge0\) \[ 5x-x^2-6\ge0 \] \[ x^2-5x+6\le0 \] \[ (x-2)(x-3)\le0 \] \[ 2\le x\le3

Domain of Nested Radical Function Find the Domain of the Function Question: The domain of definition of \[ f(x)= \sqrt{x-3-2\sqrt{x-4}} – \sqrt{x-3+2\sqrt{x-4}} \] is (a) \([4,\infty)\) (b) \((-\infty,4]\) (c) \((4,\infty)\) (d) \((-\infty,4)\) Solution: For the inner square root, \[ x-4\ge0 \] \[ x\ge4 \] Also, \[ x-3-2\sqrt{x-4} = (\sqrt{x-4}-1)^2\ge0 \] and \[ x-3+2\sqrt{x-4} = (\sqrt{x-4}+1)^2\ge0

Domain of √(4x−x²) Find the Domain of the Function Question: The domain of definition of \[ f(x)=\sqrt{4x-x^2} \] is (a) \(R-[0,4]\) (b) \(R-(0,4)\) (c) \((0,4)\) (d) \([0,4]\) Solution: For square root function, \[ 4x-x^2\ge0 \] \[ x(4-x)\ge0 \] Therefore, \[ 0\le x\le4 \] Hence, the domain is \[ \boxed{[0,4]} \] \[ \boxed{\text{Correct Answer: (d)}} \]

Domain of log|x| Find the Domain of \( \log|x| \) Question: The domain of definition of the function \[ f(x)=\log|x| \] is (a) \(R\) (b) \((-\infty,0)\) (c) \((0,\infty)\) (d) \(R-\{0\}\) Solution: For logarithmic function, \[ |x|>0 \] This is true for all real values except \[ x=0 \] Therefore, domain is \[ R-\{0\} \] \[

Domain of Radical Function Find the Domain of the Function Question: The domain of definition of the function \[ f(x)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}} \] is (a) \((-\infty,-2]\cup[2,\infty)\) (b) \([-1,1]\) (c) \(\phi\) (d) none of these Solution: For the first square root, \[ \frac{x-2}{x+2}\ge0 \] \[ x\in(-\infty,-2)\cup[2,\infty) \] For the second square root, \[ \frac{1-x}{1+x}\ge0 \] \[ x\in(-1,1] \]

Domain of √(x−1)+√(3−x) Find the Domain of the Function Question: The domain of definition of the function \[ f(x)=\sqrt{x-1}+\sqrt{3-x} \] is (a) \([1,\infty)\) (b) \((-\infty,3)\) (c) \((1,3)\) (d) \([1,3]\) Solution: For square roots to exist, \[ x-1\ge0 \] \[ x\ge1 \] Also, \[ 3-x\ge0 \] \[ x\le3 \] Therefore, \[ 1\le x\le3 \] Hence, the