NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117389
The domain of the function \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\)
1 \((-\infty,-\mathbf{1})\)
2 \([-3,-1]\)
3 \(\mathrm{R}-[-3,-1]\)
4 \((-5, \infty)-\{-3,-1\}\)
Explanation:
D Given, \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\) \(\left\{\begin{array}{l}x^2+4 x+3=x^2+3 x+x+3 \\ =x(x+3)+1(x+3) \\ =(x+3)(x+1)\end{array}\right\}\) \(\therefore f(x)=\frac{\log (x+5)}{(x+3)(x+1)}\) \(f(x)\) will be defined, if \(x+3 \neq 0\) and \(x>-5\) and \(x+5>0\) i.e. \(x \neq-3\) \(\mathrm{x} \neq-1\) and \(\mathrm{x}>-5\) So, domain of function is \((-5, \infty)-\{-3,-1\}\).
J and K CET-2017
Sets, Relation and Function
117390
The domain of the function \(f(x)=\sqrt{2 x^3-5 x^2-4 x+10}\) is
117392
Let \(A=\{1,2,3,4,5\}\). Find the domain in the relation from \(A\) to \(A\) by \(R=\{(x, y): y=2 x-1\}\).
1 \(\{1,2,3\}\)
2 \(\{1,2\}\)
3 \(\{1,3,5\}\)
4 \(\{2,4\}\)
Explanation:
A Given, \(\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=2 \mathrm{x}-1\}\) \(\therefore \mathrm{y}=2 \mathrm{x}-1 \Rightarrow \mathrm{x}=\frac{\mathrm{y}+1}{2}\) Since, relation is from \(\mathrm{A}\) to \(\mathrm{A}\) and \(\mathrm{A}=\{1,2,3,4,5\}\) \(\therefore\) For \(\mathrm{y}=1, \mathrm{x}=\frac{1+1}{2}=1 \in \mathrm{A}\) For \(\mathrm{y}=2, \mathrm{x}=\frac{3}{2} \notin \mathrm{A}\) For \(\mathrm{y}=3, \mathrm{x}=\frac{4}{2}=2 \in \mathrm{A}\) For \(\mathrm{y}=4, \mathrm{x}=\frac{5}{2} \notin \mathrm{A}\) For, \(y=5, x=3 \in A\) So, domain \(=\{1,2,3\}\).
117389
The domain of the function \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\)
1 \((-\infty,-\mathbf{1})\)
2 \([-3,-1]\)
3 \(\mathrm{R}-[-3,-1]\)
4 \((-5, \infty)-\{-3,-1\}\)
Explanation:
D Given, \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\) \(\left\{\begin{array}{l}x^2+4 x+3=x^2+3 x+x+3 \\ =x(x+3)+1(x+3) \\ =(x+3)(x+1)\end{array}\right\}\) \(\therefore f(x)=\frac{\log (x+5)}{(x+3)(x+1)}\) \(f(x)\) will be defined, if \(x+3 \neq 0\) and \(x>-5\) and \(x+5>0\) i.e. \(x \neq-3\) \(\mathrm{x} \neq-1\) and \(\mathrm{x}>-5\) So, domain of function is \((-5, \infty)-\{-3,-1\}\).
J and K CET-2017
Sets, Relation and Function
117390
The domain of the function \(f(x)=\sqrt{2 x^3-5 x^2-4 x+10}\) is
117392
Let \(A=\{1,2,3,4,5\}\). Find the domain in the relation from \(A\) to \(A\) by \(R=\{(x, y): y=2 x-1\}\).
1 \(\{1,2,3\}\)
2 \(\{1,2\}\)
3 \(\{1,3,5\}\)
4 \(\{2,4\}\)
Explanation:
A Given, \(\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=2 \mathrm{x}-1\}\) \(\therefore \mathrm{y}=2 \mathrm{x}-1 \Rightarrow \mathrm{x}=\frac{\mathrm{y}+1}{2}\) Since, relation is from \(\mathrm{A}\) to \(\mathrm{A}\) and \(\mathrm{A}=\{1,2,3,4,5\}\) \(\therefore\) For \(\mathrm{y}=1, \mathrm{x}=\frac{1+1}{2}=1 \in \mathrm{A}\) For \(\mathrm{y}=2, \mathrm{x}=\frac{3}{2} \notin \mathrm{A}\) For \(\mathrm{y}=3, \mathrm{x}=\frac{4}{2}=2 \in \mathrm{A}\) For \(\mathrm{y}=4, \mathrm{x}=\frac{5}{2} \notin \mathrm{A}\) For, \(y=5, x=3 \in A\) So, domain \(=\{1,2,3\}\).
117389
The domain of the function \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\)
1 \((-\infty,-\mathbf{1})\)
2 \([-3,-1]\)
3 \(\mathrm{R}-[-3,-1]\)
4 \((-5, \infty)-\{-3,-1\}\)
Explanation:
D Given, \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\) \(\left\{\begin{array}{l}x^2+4 x+3=x^2+3 x+x+3 \\ =x(x+3)+1(x+3) \\ =(x+3)(x+1)\end{array}\right\}\) \(\therefore f(x)=\frac{\log (x+5)}{(x+3)(x+1)}\) \(f(x)\) will be defined, if \(x+3 \neq 0\) and \(x>-5\) and \(x+5>0\) i.e. \(x \neq-3\) \(\mathrm{x} \neq-1\) and \(\mathrm{x}>-5\) So, domain of function is \((-5, \infty)-\{-3,-1\}\).
J and K CET-2017
Sets, Relation and Function
117390
The domain of the function \(f(x)=\sqrt{2 x^3-5 x^2-4 x+10}\) is
117392
Let \(A=\{1,2,3,4,5\}\). Find the domain in the relation from \(A\) to \(A\) by \(R=\{(x, y): y=2 x-1\}\).
1 \(\{1,2,3\}\)
2 \(\{1,2\}\)
3 \(\{1,3,5\}\)
4 \(\{2,4\}\)
Explanation:
A Given, \(\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=2 \mathrm{x}-1\}\) \(\therefore \mathrm{y}=2 \mathrm{x}-1 \Rightarrow \mathrm{x}=\frac{\mathrm{y}+1}{2}\) Since, relation is from \(\mathrm{A}\) to \(\mathrm{A}\) and \(\mathrm{A}=\{1,2,3,4,5\}\) \(\therefore\) For \(\mathrm{y}=1, \mathrm{x}=\frac{1+1}{2}=1 \in \mathrm{A}\) For \(\mathrm{y}=2, \mathrm{x}=\frac{3}{2} \notin \mathrm{A}\) For \(\mathrm{y}=3, \mathrm{x}=\frac{4}{2}=2 \in \mathrm{A}\) For \(\mathrm{y}=4, \mathrm{x}=\frac{5}{2} \notin \mathrm{A}\) For, \(y=5, x=3 \in A\) So, domain \(=\{1,2,3\}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117389
The domain of the function \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\)
1 \((-\infty,-\mathbf{1})\)
2 \([-3,-1]\)
3 \(\mathrm{R}-[-3,-1]\)
4 \((-5, \infty)-\{-3,-1\}\)
Explanation:
D Given, \(f(x)=\frac{\log (x+5)}{x^2+4 x+3}\) \(\left\{\begin{array}{l}x^2+4 x+3=x^2+3 x+x+3 \\ =x(x+3)+1(x+3) \\ =(x+3)(x+1)\end{array}\right\}\) \(\therefore f(x)=\frac{\log (x+5)}{(x+3)(x+1)}\) \(f(x)\) will be defined, if \(x+3 \neq 0\) and \(x>-5\) and \(x+5>0\) i.e. \(x \neq-3\) \(\mathrm{x} \neq-1\) and \(\mathrm{x}>-5\) So, domain of function is \((-5, \infty)-\{-3,-1\}\).
J and K CET-2017
Sets, Relation and Function
117390
The domain of the function \(f(x)=\sqrt{2 x^3-5 x^2-4 x+10}\) is
117392
Let \(A=\{1,2,3,4,5\}\). Find the domain in the relation from \(A\) to \(A\) by \(R=\{(x, y): y=2 x-1\}\).
1 \(\{1,2,3\}\)
2 \(\{1,2\}\)
3 \(\{1,3,5\}\)
4 \(\{2,4\}\)
Explanation:
A Given, \(\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=2 \mathrm{x}-1\}\) \(\therefore \mathrm{y}=2 \mathrm{x}-1 \Rightarrow \mathrm{x}=\frac{\mathrm{y}+1}{2}\) Since, relation is from \(\mathrm{A}\) to \(\mathrm{A}\) and \(\mathrm{A}=\{1,2,3,4,5\}\) \(\therefore\) For \(\mathrm{y}=1, \mathrm{x}=\frac{1+1}{2}=1 \in \mathrm{A}\) For \(\mathrm{y}=2, \mathrm{x}=\frac{3}{2} \notin \mathrm{A}\) For \(\mathrm{y}=3, \mathrm{x}=\frac{4}{2}=2 \in \mathrm{A}\) For \(\mathrm{y}=4, \mathrm{x}=\frac{5}{2} \notin \mathrm{A}\) For, \(y=5, x=3 \in A\) So, domain \(=\{1,2,3\}\).
