117442
Let \(f: N \rightarrow N\) be a function such that \(f(m+n)=\) \(f(m)+f(n)\) for every \(m, n \in N\). If \(f(6)=18\), then \(f(2) . f(3)\) is equal to
117444
The number of integers satisfying the inequality \(\left|\mathrm{n}^2-100\right|\lt \mathbf{5 0}\) is
1 5
2 6
3 12
4 8
5 10
Explanation:
\(\begin{array}{l} \left|\mathrm{x}^2-100\right|\lt 50 \\ -50\lt \mathrm{x}^2-100\lt 50 \\ \Rightarrow-50+100\lt \mathrm{n}^2\lt 50+100 \\ \Rightarrow 50\lt \mathrm{n}^2\lt 150 \end{array}\) Possible values of \(\mathrm{n}^2 \rightarrow 64,81,100,121,144\) Number of integer values satisfying given inequality \(=\) 5
Kerala CEE-2020
Sets, Relation and Function
117445
The solution set of the rational inequality \(\frac{x+9}{x-6} \leq 0\) is
1 \((-\infty, 9) \cup(6, \infty)\)
2 \((-\infty, 9] \cup(6, \infty)\)
3 \((-\infty, 9] \cup[6, \infty)\)
4 \([-9,6)\)
5 \((-9,6]\)
Explanation:
D Given, \(\frac{x+9}{x-6}=0\) Apply wavy curve method- Let, \(f(x)=x+9\) and \(g(x)=x-6\) So, at their roots equations are - \(x=-9, x=6\) So, for \(\mathrm{x}\lt 9\) and \(\mathrm{x}>6\) both functions has the same sign so they give the positive value of over all function but for \(-9 \leq x\lt 6, f(x)=+\) and \(g(x)=-\) which gives a negative sign of overall function. So, \(x \in[-9,6)\)
Kerala CEE-2020
Sets, Relation and Function
117446
The domain of the function \(f\) given by \(f(x)=\sqrt{x-1}\) is
1 \((-\infty, \infty)\)
2 \((1, \infty)\)
3 \([1, \infty)\)
4 \([0, \infty)\)
5 \((0, \infty)\)
Explanation:
C Given, \(\begin{aligned} & \text { & f: R \rightarrow R \quad f(x)=\sqrt{x-1} \\ & x-1 \geq 0 \\ & x \geq 1 \Rightarrow f(x) \in R \\ & \therefore \text { Domain of } f(x)=[1, \infty) \end{aligned}\)
Kerala CEE-2020
Sets, Relation and Function
117447
The domain of definition of the function \(f(x)=\) \(\frac{\log _3(x+7)}{x^2-5 x+6}\) is
117442
Let \(f: N \rightarrow N\) be a function such that \(f(m+n)=\) \(f(m)+f(n)\) for every \(m, n \in N\). If \(f(6)=18\), then \(f(2) . f(3)\) is equal to
117444
The number of integers satisfying the inequality \(\left|\mathrm{n}^2-100\right|\lt \mathbf{5 0}\) is
1 5
2 6
3 12
4 8
5 10
Explanation:
\(\begin{array}{l} \left|\mathrm{x}^2-100\right|\lt 50 \\ -50\lt \mathrm{x}^2-100\lt 50 \\ \Rightarrow-50+100\lt \mathrm{n}^2\lt 50+100 \\ \Rightarrow 50\lt \mathrm{n}^2\lt 150 \end{array}\) Possible values of \(\mathrm{n}^2 \rightarrow 64,81,100,121,144\) Number of integer values satisfying given inequality \(=\) 5
Kerala CEE-2020
Sets, Relation and Function
117445
The solution set of the rational inequality \(\frac{x+9}{x-6} \leq 0\) is
1 \((-\infty, 9) \cup(6, \infty)\)
2 \((-\infty, 9] \cup(6, \infty)\)
3 \((-\infty, 9] \cup[6, \infty)\)
4 \([-9,6)\)
5 \((-9,6]\)
Explanation:
D Given, \(\frac{x+9}{x-6}=0\) Apply wavy curve method- Let, \(f(x)=x+9\) and \(g(x)=x-6\) So, at their roots equations are - \(x=-9, x=6\) So, for \(\mathrm{x}\lt 9\) and \(\mathrm{x}>6\) both functions has the same sign so they give the positive value of over all function but for \(-9 \leq x\lt 6, f(x)=+\) and \(g(x)=-\) which gives a negative sign of overall function. So, \(x \in[-9,6)\)
Kerala CEE-2020
Sets, Relation and Function
117446
The domain of the function \(f\) given by \(f(x)=\sqrt{x-1}\) is
1 \((-\infty, \infty)\)
2 \((1, \infty)\)
3 \([1, \infty)\)
4 \([0, \infty)\)
5 \((0, \infty)\)
Explanation:
C Given, \(\begin{aligned} & \text { & f: R \rightarrow R \quad f(x)=\sqrt{x-1} \\ & x-1 \geq 0 \\ & x \geq 1 \Rightarrow f(x) \in R \\ & \therefore \text { Domain of } f(x)=[1, \infty) \end{aligned}\)
Kerala CEE-2020
Sets, Relation and Function
117447
The domain of definition of the function \(f(x)=\) \(\frac{\log _3(x+7)}{x^2-5 x+6}\) is
117442
Let \(f: N \rightarrow N\) be a function such that \(f(m+n)=\) \(f(m)+f(n)\) for every \(m, n \in N\). If \(f(6)=18\), then \(f(2) . f(3)\) is equal to
117444
The number of integers satisfying the inequality \(\left|\mathrm{n}^2-100\right|\lt \mathbf{5 0}\) is
1 5
2 6
3 12
4 8
5 10
Explanation:
\(\begin{array}{l} \left|\mathrm{x}^2-100\right|\lt 50 \\ -50\lt \mathrm{x}^2-100\lt 50 \\ \Rightarrow-50+100\lt \mathrm{n}^2\lt 50+100 \\ \Rightarrow 50\lt \mathrm{n}^2\lt 150 \end{array}\) Possible values of \(\mathrm{n}^2 \rightarrow 64,81,100,121,144\) Number of integer values satisfying given inequality \(=\) 5
Kerala CEE-2020
Sets, Relation and Function
117445
The solution set of the rational inequality \(\frac{x+9}{x-6} \leq 0\) is
1 \((-\infty, 9) \cup(6, \infty)\)
2 \((-\infty, 9] \cup(6, \infty)\)
3 \((-\infty, 9] \cup[6, \infty)\)
4 \([-9,6)\)
5 \((-9,6]\)
Explanation:
D Given, \(\frac{x+9}{x-6}=0\) Apply wavy curve method- Let, \(f(x)=x+9\) and \(g(x)=x-6\) So, at their roots equations are - \(x=-9, x=6\) So, for \(\mathrm{x}\lt 9\) and \(\mathrm{x}>6\) both functions has the same sign so they give the positive value of over all function but for \(-9 \leq x\lt 6, f(x)=+\) and \(g(x)=-\) which gives a negative sign of overall function. So, \(x \in[-9,6)\)
Kerala CEE-2020
Sets, Relation and Function
117446
The domain of the function \(f\) given by \(f(x)=\sqrt{x-1}\) is
1 \((-\infty, \infty)\)
2 \((1, \infty)\)
3 \([1, \infty)\)
4 \([0, \infty)\)
5 \((0, \infty)\)
Explanation:
C Given, \(\begin{aligned} & \text { & f: R \rightarrow R \quad f(x)=\sqrt{x-1} \\ & x-1 \geq 0 \\ & x \geq 1 \Rightarrow f(x) \in R \\ & \therefore \text { Domain of } f(x)=[1, \infty) \end{aligned}\)
Kerala CEE-2020
Sets, Relation and Function
117447
The domain of definition of the function \(f(x)=\) \(\frac{\log _3(x+7)}{x^2-5 x+6}\) is
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Sets, Relation and Function
117442
Let \(f: N \rightarrow N\) be a function such that \(f(m+n)=\) \(f(m)+f(n)\) for every \(m, n \in N\). If \(f(6)=18\), then \(f(2) . f(3)\) is equal to
117444
The number of integers satisfying the inequality \(\left|\mathrm{n}^2-100\right|\lt \mathbf{5 0}\) is
1 5
2 6
3 12
4 8
5 10
Explanation:
\(\begin{array}{l} \left|\mathrm{x}^2-100\right|\lt 50 \\ -50\lt \mathrm{x}^2-100\lt 50 \\ \Rightarrow-50+100\lt \mathrm{n}^2\lt 50+100 \\ \Rightarrow 50\lt \mathrm{n}^2\lt 150 \end{array}\) Possible values of \(\mathrm{n}^2 \rightarrow 64,81,100,121,144\) Number of integer values satisfying given inequality \(=\) 5
Kerala CEE-2020
Sets, Relation and Function
117445
The solution set of the rational inequality \(\frac{x+9}{x-6} \leq 0\) is
1 \((-\infty, 9) \cup(6, \infty)\)
2 \((-\infty, 9] \cup(6, \infty)\)
3 \((-\infty, 9] \cup[6, \infty)\)
4 \([-9,6)\)
5 \((-9,6]\)
Explanation:
D Given, \(\frac{x+9}{x-6}=0\) Apply wavy curve method- Let, \(f(x)=x+9\) and \(g(x)=x-6\) So, at their roots equations are - \(x=-9, x=6\) So, for \(\mathrm{x}\lt 9\) and \(\mathrm{x}>6\) both functions has the same sign so they give the positive value of over all function but for \(-9 \leq x\lt 6, f(x)=+\) and \(g(x)=-\) which gives a negative sign of overall function. So, \(x \in[-9,6)\)
Kerala CEE-2020
Sets, Relation and Function
117446
The domain of the function \(f\) given by \(f(x)=\sqrt{x-1}\) is
1 \((-\infty, \infty)\)
2 \((1, \infty)\)
3 \([1, \infty)\)
4 \([0, \infty)\)
5 \((0, \infty)\)
Explanation:
C Given, \(\begin{aligned} & \text { & f: R \rightarrow R \quad f(x)=\sqrt{x-1} \\ & x-1 \geq 0 \\ & x \geq 1 \Rightarrow f(x) \in R \\ & \therefore \text { Domain of } f(x)=[1, \infty) \end{aligned}\)
Kerala CEE-2020
Sets, Relation and Function
117447
The domain of definition of the function \(f(x)=\) \(\frac{\log _3(x+7)}{x^2-5 x+6}\) is
117442
Let \(f: N \rightarrow N\) be a function such that \(f(m+n)=\) \(f(m)+f(n)\) for every \(m, n \in N\). If \(f(6)=18\), then \(f(2) . f(3)\) is equal to
117444
The number of integers satisfying the inequality \(\left|\mathrm{n}^2-100\right|\lt \mathbf{5 0}\) is
1 5
2 6
3 12
4 8
5 10
Explanation:
\(\begin{array}{l} \left|\mathrm{x}^2-100\right|\lt 50 \\ -50\lt \mathrm{x}^2-100\lt 50 \\ \Rightarrow-50+100\lt \mathrm{n}^2\lt 50+100 \\ \Rightarrow 50\lt \mathrm{n}^2\lt 150 \end{array}\) Possible values of \(\mathrm{n}^2 \rightarrow 64,81,100,121,144\) Number of integer values satisfying given inequality \(=\) 5
Kerala CEE-2020
Sets, Relation and Function
117445
The solution set of the rational inequality \(\frac{x+9}{x-6} \leq 0\) is
1 \((-\infty, 9) \cup(6, \infty)\)
2 \((-\infty, 9] \cup(6, \infty)\)
3 \((-\infty, 9] \cup[6, \infty)\)
4 \([-9,6)\)
5 \((-9,6]\)
Explanation:
D Given, \(\frac{x+9}{x-6}=0\) Apply wavy curve method- Let, \(f(x)=x+9\) and \(g(x)=x-6\) So, at their roots equations are - \(x=-9, x=6\) So, for \(\mathrm{x}\lt 9\) and \(\mathrm{x}>6\) both functions has the same sign so they give the positive value of over all function but for \(-9 \leq x\lt 6, f(x)=+\) and \(g(x)=-\) which gives a negative sign of overall function. So, \(x \in[-9,6)\)
Kerala CEE-2020
Sets, Relation and Function
117446
The domain of the function \(f\) given by \(f(x)=\sqrt{x-1}\) is
1 \((-\infty, \infty)\)
2 \((1, \infty)\)
3 \([1, \infty)\)
4 \([0, \infty)\)
5 \((0, \infty)\)
Explanation:
C Given, \(\begin{aligned} & \text { & f: R \rightarrow R \quad f(x)=\sqrt{x-1} \\ & x-1 \geq 0 \\ & x \geq 1 \Rightarrow f(x) \in R \\ & \therefore \text { Domain of } f(x)=[1, \infty) \end{aligned}\)
Kerala CEE-2020
Sets, Relation and Function
117447
The domain of definition of the function \(f(x)=\) \(\frac{\log _3(x+7)}{x^2-5 x+6}\) is
