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, \[ \boxed{\text{Correct Answer: (c)}} \]
If f(x) = x³ − 1/x³ , then f(x) + f(1/x) is equal to(a) 2x³(b) 2/x³(c) 0(d) 1 Read More »
Domain and Range of 2−|x−5| Find the Domain and Range of the Function Question: The domain and range of the function \[ f(x)=2-|x-5| \] are (a) Domain \(=R^+\), Range \(=(-\infty,1]\) (b) Domain \(=R\), Range \(=(-\infty,2]\) (c) Domain \(=R\), Range \(=(-\infty,2)\) (d) Domain \(=R^+\), Range \(=(-\infty,2]\) Solution: Since modulus function is defined for all real \(x\),
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 f(x) = √(a² − x²), a > 0 is(a) (−a, a)(b) [−a, a](c) [0, a](d) (−a, 0] Read More »
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 \[ t\in[-1,0)\cup(0,3] \]