117462
The domain of the function \(\log _{10}\left(x^2-5 x+6\right)\) is
1 \((-\infty,-1) \cup(-3, \infty)\)
2 \((-\infty, 0) \cup(3, \infty)\)
3 \((-\infty, 2) \cup(3, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
C Given, the function \(\log _{10}\left(x^2-5 x+6\right)\). This function is defined for \(x^2-5 x+6>0\) or \begin{align*} \begin{aligned} & x^2-3 x-2 x+6>0 \\ & x(x-3)-2(x-3)>0 \\ & (x-2)(x-3)>0 \\ & x>3 \end{aligned} \end{align*} \begin{align*} \therefore \quad \mathrm{x} \in(3, \infty) \end{align*} Other possibility is \(\mathrm{x}\lt 2\) \begin{align*} \begin{array}{ll} \therefore & \mathrm{x} \in(-\infty 2) \\ \therefore & \mathrm{x} \in(-\infty, 2) \cup(3, \infty) \end{array} \end{align*}
MHT CET-2021
Sets, Relation and Function
117463
The domain of the function \(f(x)=\) \(\sqrt{\mathrm{x}-1}+\sqrt{6-\mathrm{x}}\) is
1 \([1,6]\)
2 \((-\infty, 6)\)
3 \([1, \infty]\)
4 \((-\infty, 6]\)
Explanation:
A Given, the function \begin{align*} f(x)=\sqrt{x-1}+\sqrt{6-x} \end{align*} The function \(f(x)\) is defined as, \begin{align*} x-1 \geq 0 \end{align*} And, \(\quad 6-x \geq 0\) So, we get \begin{align*} \begin{aligned} & x \geq 1 \text { and } x \leq 6 \\ & \therefore \quad 1 \leq \mathrm{x} \leq 6 \\ & \text { So, } \quad x \in[1,6] \\ & \end{aligned} \end{align*}
MHT CET-2021
Sets, Relation and Function
117464
The solution set of the inequality \(5(4 x+6)\lt \) \(25 \mathrm{x}+10\) is
117465
Let \(f:[-4,2] \rightarrow R\) be given by \(f(x)=\sqrt{16-x^2}\). Then the range of the function \(f\) is
1 \([0,2]\)
2 \([0,2 \sqrt{3}]\)
3 \([0,4]\)
4 \([-2,2]\)
Explanation:
C Let, \(\mathrm{f}:[-4,2] \rightarrow \mathrm{R}\) Given, \begin{align*} f(x)=\sqrt{16-x^2} \end{align*} For domain under root should not be negative quantity, \(16-x^2 \geq 0\) \begin{align*} 16 \geq x^2 \end{align*} Therefore, \(\mathrm{x} \leq 4\) or \(\mathrm{x} \geq-4\) Then, Domain \([-4,4]\) Range : \(f(x)\) is maximum at \(x=0, f(x)=4\) And, \(f(x)\) is minimum at \(x=4, f(x)=0\) So, range is \([0,4]\).
117462
The domain of the function \(\log _{10}\left(x^2-5 x+6\right)\) is
1 \((-\infty,-1) \cup(-3, \infty)\)
2 \((-\infty, 0) \cup(3, \infty)\)
3 \((-\infty, 2) \cup(3, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
C Given, the function \(\log _{10}\left(x^2-5 x+6\right)\). This function is defined for \(x^2-5 x+6>0\) or \begin{align*} \begin{aligned} & x^2-3 x-2 x+6>0 \\ & x(x-3)-2(x-3)>0 \\ & (x-2)(x-3)>0 \\ & x>3 \end{aligned} \end{align*} \begin{align*} \therefore \quad \mathrm{x} \in(3, \infty) \end{align*} Other possibility is \(\mathrm{x}\lt 2\) \begin{align*} \begin{array}{ll} \therefore & \mathrm{x} \in(-\infty 2) \\ \therefore & \mathrm{x} \in(-\infty, 2) \cup(3, \infty) \end{array} \end{align*}
MHT CET-2021
Sets, Relation and Function
117463
The domain of the function \(f(x)=\) \(\sqrt{\mathrm{x}-1}+\sqrt{6-\mathrm{x}}\) is
1 \([1,6]\)
2 \((-\infty, 6)\)
3 \([1, \infty]\)
4 \((-\infty, 6]\)
Explanation:
A Given, the function \begin{align*} f(x)=\sqrt{x-1}+\sqrt{6-x} \end{align*} The function \(f(x)\) is defined as, \begin{align*} x-1 \geq 0 \end{align*} And, \(\quad 6-x \geq 0\) So, we get \begin{align*} \begin{aligned} & x \geq 1 \text { and } x \leq 6 \\ & \therefore \quad 1 \leq \mathrm{x} \leq 6 \\ & \text { So, } \quad x \in[1,6] \\ & \end{aligned} \end{align*}
MHT CET-2021
Sets, Relation and Function
117464
The solution set of the inequality \(5(4 x+6)\lt \) \(25 \mathrm{x}+10\) is
117465
Let \(f:[-4,2] \rightarrow R\) be given by \(f(x)=\sqrt{16-x^2}\). Then the range of the function \(f\) is
1 \([0,2]\)
2 \([0,2 \sqrt{3}]\)
3 \([0,4]\)
4 \([-2,2]\)
Explanation:
C Let, \(\mathrm{f}:[-4,2] \rightarrow \mathrm{R}\) Given, \begin{align*} f(x)=\sqrt{16-x^2} \end{align*} For domain under root should not be negative quantity, \(16-x^2 \geq 0\) \begin{align*} 16 \geq x^2 \end{align*} Therefore, \(\mathrm{x} \leq 4\) or \(\mathrm{x} \geq-4\) Then, Domain \([-4,4]\) Range : \(f(x)\) is maximum at \(x=0, f(x)=4\) And, \(f(x)\) is minimum at \(x=4, f(x)=0\) So, range is \([0,4]\).
117462
The domain of the function \(\log _{10}\left(x^2-5 x+6\right)\) is
1 \((-\infty,-1) \cup(-3, \infty)\)
2 \((-\infty, 0) \cup(3, \infty)\)
3 \((-\infty, 2) \cup(3, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
C Given, the function \(\log _{10}\left(x^2-5 x+6\right)\). This function is defined for \(x^2-5 x+6>0\) or \begin{align*} \begin{aligned} & x^2-3 x-2 x+6>0 \\ & x(x-3)-2(x-3)>0 \\ & (x-2)(x-3)>0 \\ & x>3 \end{aligned} \end{align*} \begin{align*} \therefore \quad \mathrm{x} \in(3, \infty) \end{align*} Other possibility is \(\mathrm{x}\lt 2\) \begin{align*} \begin{array}{ll} \therefore & \mathrm{x} \in(-\infty 2) \\ \therefore & \mathrm{x} \in(-\infty, 2) \cup(3, \infty) \end{array} \end{align*}
MHT CET-2021
Sets, Relation and Function
117463
The domain of the function \(f(x)=\) \(\sqrt{\mathrm{x}-1}+\sqrt{6-\mathrm{x}}\) is
1 \([1,6]\)
2 \((-\infty, 6)\)
3 \([1, \infty]\)
4 \((-\infty, 6]\)
Explanation:
A Given, the function \begin{align*} f(x)=\sqrt{x-1}+\sqrt{6-x} \end{align*} The function \(f(x)\) is defined as, \begin{align*} x-1 \geq 0 \end{align*} And, \(\quad 6-x \geq 0\) So, we get \begin{align*} \begin{aligned} & x \geq 1 \text { and } x \leq 6 \\ & \therefore \quad 1 \leq \mathrm{x} \leq 6 \\ & \text { So, } \quad x \in[1,6] \\ & \end{aligned} \end{align*}
MHT CET-2021
Sets, Relation and Function
117464
The solution set of the inequality \(5(4 x+6)\lt \) \(25 \mathrm{x}+10\) is
117465
Let \(f:[-4,2] \rightarrow R\) be given by \(f(x)=\sqrt{16-x^2}\). Then the range of the function \(f\) is
1 \([0,2]\)
2 \([0,2 \sqrt{3}]\)
3 \([0,4]\)
4 \([-2,2]\)
Explanation:
C Let, \(\mathrm{f}:[-4,2] \rightarrow \mathrm{R}\) Given, \begin{align*} f(x)=\sqrt{16-x^2} \end{align*} For domain under root should not be negative quantity, \(16-x^2 \geq 0\) \begin{align*} 16 \geq x^2 \end{align*} Therefore, \(\mathrm{x} \leq 4\) or \(\mathrm{x} \geq-4\) Then, Domain \([-4,4]\) Range : \(f(x)\) is maximum at \(x=0, f(x)=4\) And, \(f(x)\) is minimum at \(x=4, f(x)=0\) So, range is \([0,4]\).
117462
The domain of the function \(\log _{10}\left(x^2-5 x+6\right)\) is
1 \((-\infty,-1) \cup(-3, \infty)\)
2 \((-\infty, 0) \cup(3, \infty)\)
3 \((-\infty, 2) \cup(3, \infty)\)
4 \((-\infty, \infty)\)
Explanation:
C Given, the function \(\log _{10}\left(x^2-5 x+6\right)\). This function is defined for \(x^2-5 x+6>0\) or \begin{align*} \begin{aligned} & x^2-3 x-2 x+6>0 \\ & x(x-3)-2(x-3)>0 \\ & (x-2)(x-3)>0 \\ & x>3 \end{aligned} \end{align*} \begin{align*} \therefore \quad \mathrm{x} \in(3, \infty) \end{align*} Other possibility is \(\mathrm{x}\lt 2\) \begin{align*} \begin{array}{ll} \therefore & \mathrm{x} \in(-\infty 2) \\ \therefore & \mathrm{x} \in(-\infty, 2) \cup(3, \infty) \end{array} \end{align*}
MHT CET-2021
Sets, Relation and Function
117463
The domain of the function \(f(x)=\) \(\sqrt{\mathrm{x}-1}+\sqrt{6-\mathrm{x}}\) is
1 \([1,6]\)
2 \((-\infty, 6)\)
3 \([1, \infty]\)
4 \((-\infty, 6]\)
Explanation:
A Given, the function \begin{align*} f(x)=\sqrt{x-1}+\sqrt{6-x} \end{align*} The function \(f(x)\) is defined as, \begin{align*} x-1 \geq 0 \end{align*} And, \(\quad 6-x \geq 0\) So, we get \begin{align*} \begin{aligned} & x \geq 1 \text { and } x \leq 6 \\ & \therefore \quad 1 \leq \mathrm{x} \leq 6 \\ & \text { So, } \quad x \in[1,6] \\ & \end{aligned} \end{align*}
MHT CET-2021
Sets, Relation and Function
117464
The solution set of the inequality \(5(4 x+6)\lt \) \(25 \mathrm{x}+10\) is
117465
Let \(f:[-4,2] \rightarrow R\) be given by \(f(x)=\sqrt{16-x^2}\). Then the range of the function \(f\) is
1 \([0,2]\)
2 \([0,2 \sqrt{3}]\)
3 \([0,4]\)
4 \([-2,2]\)
Explanation:
C Let, \(\mathrm{f}:[-4,2] \rightarrow \mathrm{R}\) Given, \begin{align*} f(x)=\sqrt{16-x^2} \end{align*} For domain under root should not be negative quantity, \(16-x^2 \geq 0\) \begin{align*} 16 \geq x^2 \end{align*} Therefore, \(\mathrm{x} \leq 4\) or \(\mathrm{x} \geq-4\) Then, Domain \([-4,4]\) Range : \(f(x)\) is maximum at \(x=0, f(x)=4\) And, \(f(x)\) is minimum at \(x=4, f(x)=0\) So, range is \([0,4]\).
