Educational

Domain of Rational Function Find the Domain of the Function Question: The domain of the function \[ f(x)=\frac{x^2+2x+1}{x^2-x-6} \] is (a) \(R-\{-2,3\}\) (b) \(R-\{-3,2\}\) (c) \(R-\{-2,3]\) (d) \(R-(-2,3)\) Solution: For rational function, denominator \(\ne0\) \[ x^2-x-6\ne0 \] Factorizing, \[ (x-3)(x+2)\ne0 \] Therefore, \[ x\ne3,\,-2 \] Hence, domain is \[ \boxed{R-\{-2,3\}} \] \[ \boxed{\text{Correct Answer: (a)}} […]

Domain and Range of √(x−1) Find the Domain and Range of the Function Question: The domain and range of the real function \[ f(x)=\sqrt{x-1} \] are given by (a) Domain \(=(1,\infty)\), Range \(=(0,\infty)\) (b) Domain \(=[1,\infty)\), Range \(=(0,\infty)\) (c) Domain \(=[1,\infty)\), Range \(=[0,\infty)\) (d) Domain \(=[1,\infty)\), Range \(=[0,\infty)\) Solution: For square root function, \[ x-1\ge0

Domain and Range of Rational Function Find the Domain and Range of the Function Question: The domain and range of the real function \[ f(x)=\frac{4-x}{x-4} \] are given by (a) Domain \(=R\), Range \(=\{-1,1\}\) (b) Domain \(=R-\{1\}\), Range \(=R\) (c) Domain \(=R-\{4\}\), Range \(=\{-1\}\) (d) Domain \(=R-\{-4\}\), Range \(=\{-1,1\}\) Solution: \[ f(x)=\frac{4-x}{x-4} \] \[ =\frac{-(x-4)}{x-4}

Find a and b in Linear Function Find \( a \) and \( b \) Question: If \[ f(x)=ax+b \] where \(a\) and \(b\) are integers, \[ f(-1)=-5 \] and \[ f(3)=3, \] then \(a\) and \(b\) are equal to (a) \(a=-3,\; b=-1\) (b) \(a=2,\; b=-3\) (c) \(a=0,\; b=2\) (d) \(a=2,\; b=3\) Solution: Given, \[

Domain of √(a²−x²) Find the Domain of the Function Question: Domain of \[ f(x)=\sqrt{a^2-x^2}, \qquad a>0 \] is (a) \(( -a,a)\) (b) \([ -a,a]\) (c) \([0,a]\) (d) \(( -a,0]\) Solution: For square root function, \[ a^2-x^2\ge0 \] \[ x^2\le a^2 \] \[ -a\le x\le a \] Therefore, domain is \[ \boxed{[-a,a]} \] \[ \boxed{\text{Correct Answer:

Domain of Radical Rational Function Find the Domain of the Function Question: The domain of the function \[ f(x)=\sqrt{4-x}+\frac1{\sqrt{x^2-1}} \] is equal to (a) \((-\infty,-1)\cup(1,4)\) (b) \((-\infty,-1]\cup(1,4]\) (c) \((-\infty,-1)\cup[1,4]\) (d) \((-\infty,-1)\cup[1,4)\) Solution: For \[ \sqrt{4-x} \] we need \[ 4-x\ge0 \] \[ x\le4 \] For \[ \frac1{\sqrt{x^2-1}} \] denominator must be positive: \[ x^2-1>0 \]

Range of 1/(1−2cosx) Find the Range of the Function Question: The range of \[ f(x)=\frac{1}{1-2\cos x} \] is (a) \(\left[\frac13,1\right]\) (b) \(\left[-1,\frac13\right]\) (c) \((-\infty,-1]\cup\left[\frac13,\infty\right)\) (d) \(\left[-\frac13,1\right]\) Solution: Since \[ -1\le\cos x\le1 \] therefore \[ -2\le2\cos x\le2 \] \[ -1\le1-2\cos x\le3 \] Also, \[ 1-2\cos x\ne0 \] Let \[ t=1-2\cos x \] Then \[ t\in[-1,0)\cup(0,3] \]

Greatest Integer Function Equation Solve the Greatest Integer Function Equation Question: If \[ [x]^2-5[x]+6=0 \] where \([\,]\) denotes the greatest integer function, then (a) \(x\in[3,4]\) (b) \(x\in(2,3]\) (c) \(x\in[2,3]\) (d) \(x\in[2,4)\) Solution: Let \[ [x]=t \] Then, \[ t^2-5t+6=0 \] \[ (t-2)(t-3)=0 \] \[ t=2 \quad \text{or} \quad t=3 \] So, \[ [x]=2 \Rightarrow x\in[2,3)

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 \[