117419
If \(f: R \rightarrow R\) be defined by \(f(x)=x+2|x+1|+2\) \(|\mathbf{x}-1|\), then the element in the co-domain, which has unique pre image in the domain is
1 3
2 1
3 2
4 5
Explanation:
A We have, \(\begin{aligned} & f(x)=x+2|x+1|+2|x-1| \\ & f(x)=\left\{\begin{array}{cc}x+2(-x-1)+2(-x+1), & x \leq-1 \\ x+2(x+1)+2(-x+1), & -1\lt x \leq 1 \\ x+2(x+1)+2(x-1), & x \geq 1\end{array}\right.\end{aligned}\) \(f(x)=\left\{\begin{array}{cc}-3 x & x \leq-1 \\ x+4, & -1\lt x\lt 1 \\ 5 x, & x \geq 1\end{array}\right.\) Graph of \(\mathrm{f}(\mathrm{x})\) is, Clearly, 3 has unique image in the domain.
TS EAMCET-14.09.2020
Sets, Relation and Function
117420
The range of the function \(f(x)=-\sqrt{-x^2-6 x-5}\) is
1 \([-2,0]\)
2 \([0,2]\)
3 \((\infty,-2]\)
4 \([-2,2]\)
Explanation:
A Given that, \(f(x)=-\sqrt{-x^2-6 x-5}\) Let, \(y=-\sqrt{-\left(x^2+6 x+5\right)}\) On squaring both the side, we get - \(y^2=-x^2+6 x+5\) \(x^2+6 x+5+y^2=0\) \(x=\frac{-6 \pm \sqrt{36-20-4 y^2}}{2}\) \(x=\frac{-6 \pm \sqrt{16-4 y^2}}{2}\) For real value, \(\mathrm{D} \geq 0\) According to question, \(y \in[-2,0]\)Rang of \(f(x) \in[-2,0]\)
TS EAMCET-07.05.2018
Sets, Relation and Function
117421
If \(x\) is real, then the interval in which no value of the expression \(\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) lies, is
1 \((2,5)\)
2 \((3,6)\)
3 \((3,4)\)
4 \((6,8)\)
Explanation:
D Let, \(y=\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) \(y(2 x-5)=2 x^2+4 x-22\) \(2 x y-5 y=2 x^2+4 x-22\) \(2 x^2-4 x-2 x y+5 y-22=0\) \(2 x^2-2 x(2-y)-22+5 y=0\) For the real value discriminate. \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\Rightarrow \quad {[2(2-y)]^2-4 \times 2[-(22-5 y)]=0}\) \(4(2-y)^2+8(22-5 y)=0\) \((2-y)^2+2(22-5 y)=0\) \(4+y^2-4 y+44-10 y=0\) \(y^2-14 y+48=0\) \((y-8)(y-6)=0\) \(y-8= 0 \text { and } y-6=0\) \(\text { So, } y=6 \text { and } y=8\)
TS EAMCET-04.05.2019
Sets, Relation and Function
117422
The domain of the function \(f(x)=\) \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) is
1 \([-1,1]\)
2 \([1,4]\)
3 \((0,16]\)
4 \([1,16]\)
Explanation:
D Given that, \(f(x)=\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) The function \(f(x)\) is defined, Case- I : \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]\) We know that range of \(\sin ^{-1} x \in[-1,1]\) \(-1 \leq \log _4\left(\frac{x}{4}\right) \leq 1\) \(4^{-1} \leq \frac{x}{4} \leq 4^1\) \(\frac{1}{4} \leq \frac{x}{4} \leq 4\) \(1 \leq \mathrm{x} \leq 16\) \(\mathrm{x} \in[1,16]\) Case-II : \(\sqrt{17 \mathrm{x}-\mathrm{x}^2-16}\) \(17 \mathrm{x}-\mathrm{x}^2-16 \geq 0\) \(\mathrm{x}^2-17 \mathrm{x}+16 \leq 0\) \(\mathrm{x}^2-16 \mathrm{x}-\mathrm{x}+16 \leq 0\) \((\mathrm{x}-16)(\mathrm{x}-1) \leq 0\) \(x \in[1,16]\) The domain of function \(\mathrm{f}(\mathrm{x})\) from equation (i) and (ii) \(\mathrm{x} \in[1,16]\).
117419
If \(f: R \rightarrow R\) be defined by \(f(x)=x+2|x+1|+2\) \(|\mathbf{x}-1|\), then the element in the co-domain, which has unique pre image in the domain is
1 3
2 1
3 2
4 5
Explanation:
A We have, \(\begin{aligned} & f(x)=x+2|x+1|+2|x-1| \\ & f(x)=\left\{\begin{array}{cc}x+2(-x-1)+2(-x+1), & x \leq-1 \\ x+2(x+1)+2(-x+1), & -1\lt x \leq 1 \\ x+2(x+1)+2(x-1), & x \geq 1\end{array}\right.\end{aligned}\) \(f(x)=\left\{\begin{array}{cc}-3 x & x \leq-1 \\ x+4, & -1\lt x\lt 1 \\ 5 x, & x \geq 1\end{array}\right.\) Graph of \(\mathrm{f}(\mathrm{x})\) is, Clearly, 3 has unique image in the domain.
TS EAMCET-14.09.2020
Sets, Relation and Function
117420
The range of the function \(f(x)=-\sqrt{-x^2-6 x-5}\) is
1 \([-2,0]\)
2 \([0,2]\)
3 \((\infty,-2]\)
4 \([-2,2]\)
Explanation:
A Given that, \(f(x)=-\sqrt{-x^2-6 x-5}\) Let, \(y=-\sqrt{-\left(x^2+6 x+5\right)}\) On squaring both the side, we get - \(y^2=-x^2+6 x+5\) \(x^2+6 x+5+y^2=0\) \(x=\frac{-6 \pm \sqrt{36-20-4 y^2}}{2}\) \(x=\frac{-6 \pm \sqrt{16-4 y^2}}{2}\) For real value, \(\mathrm{D} \geq 0\) According to question, \(y \in[-2,0]\)Rang of \(f(x) \in[-2,0]\)
TS EAMCET-07.05.2018
Sets, Relation and Function
117421
If \(x\) is real, then the interval in which no value of the expression \(\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) lies, is
1 \((2,5)\)
2 \((3,6)\)
3 \((3,4)\)
4 \((6,8)\)
Explanation:
D Let, \(y=\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) \(y(2 x-5)=2 x^2+4 x-22\) \(2 x y-5 y=2 x^2+4 x-22\) \(2 x^2-4 x-2 x y+5 y-22=0\) \(2 x^2-2 x(2-y)-22+5 y=0\) For the real value discriminate. \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\Rightarrow \quad {[2(2-y)]^2-4 \times 2[-(22-5 y)]=0}\) \(4(2-y)^2+8(22-5 y)=0\) \((2-y)^2+2(22-5 y)=0\) \(4+y^2-4 y+44-10 y=0\) \(y^2-14 y+48=0\) \((y-8)(y-6)=0\) \(y-8= 0 \text { and } y-6=0\) \(\text { So, } y=6 \text { and } y=8\)
TS EAMCET-04.05.2019
Sets, Relation and Function
117422
The domain of the function \(f(x)=\) \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) is
1 \([-1,1]\)
2 \([1,4]\)
3 \((0,16]\)
4 \([1,16]\)
Explanation:
D Given that, \(f(x)=\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) The function \(f(x)\) is defined, Case- I : \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]\) We know that range of \(\sin ^{-1} x \in[-1,1]\) \(-1 \leq \log _4\left(\frac{x}{4}\right) \leq 1\) \(4^{-1} \leq \frac{x}{4} \leq 4^1\) \(\frac{1}{4} \leq \frac{x}{4} \leq 4\) \(1 \leq \mathrm{x} \leq 16\) \(\mathrm{x} \in[1,16]\) Case-II : \(\sqrt{17 \mathrm{x}-\mathrm{x}^2-16}\) \(17 \mathrm{x}-\mathrm{x}^2-16 \geq 0\) \(\mathrm{x}^2-17 \mathrm{x}+16 \leq 0\) \(\mathrm{x}^2-16 \mathrm{x}-\mathrm{x}+16 \leq 0\) \((\mathrm{x}-16)(\mathrm{x}-1) \leq 0\) \(x \in[1,16]\) The domain of function \(\mathrm{f}(\mathrm{x})\) from equation (i) and (ii) \(\mathrm{x} \in[1,16]\).
117419
If \(f: R \rightarrow R\) be defined by \(f(x)=x+2|x+1|+2\) \(|\mathbf{x}-1|\), then the element in the co-domain, which has unique pre image in the domain is
1 3
2 1
3 2
4 5
Explanation:
A We have, \(\begin{aligned} & f(x)=x+2|x+1|+2|x-1| \\ & f(x)=\left\{\begin{array}{cc}x+2(-x-1)+2(-x+1), & x \leq-1 \\ x+2(x+1)+2(-x+1), & -1\lt x \leq 1 \\ x+2(x+1)+2(x-1), & x \geq 1\end{array}\right.\end{aligned}\) \(f(x)=\left\{\begin{array}{cc}-3 x & x \leq-1 \\ x+4, & -1\lt x\lt 1 \\ 5 x, & x \geq 1\end{array}\right.\) Graph of \(\mathrm{f}(\mathrm{x})\) is, Clearly, 3 has unique image in the domain.
TS EAMCET-14.09.2020
Sets, Relation and Function
117420
The range of the function \(f(x)=-\sqrt{-x^2-6 x-5}\) is
1 \([-2,0]\)
2 \([0,2]\)
3 \((\infty,-2]\)
4 \([-2,2]\)
Explanation:
A Given that, \(f(x)=-\sqrt{-x^2-6 x-5}\) Let, \(y=-\sqrt{-\left(x^2+6 x+5\right)}\) On squaring both the side, we get - \(y^2=-x^2+6 x+5\) \(x^2+6 x+5+y^2=0\) \(x=\frac{-6 \pm \sqrt{36-20-4 y^2}}{2}\) \(x=\frac{-6 \pm \sqrt{16-4 y^2}}{2}\) For real value, \(\mathrm{D} \geq 0\) According to question, \(y \in[-2,0]\)Rang of \(f(x) \in[-2,0]\)
TS EAMCET-07.05.2018
Sets, Relation and Function
117421
If \(x\) is real, then the interval in which no value of the expression \(\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) lies, is
1 \((2,5)\)
2 \((3,6)\)
3 \((3,4)\)
4 \((6,8)\)
Explanation:
D Let, \(y=\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) \(y(2 x-5)=2 x^2+4 x-22\) \(2 x y-5 y=2 x^2+4 x-22\) \(2 x^2-4 x-2 x y+5 y-22=0\) \(2 x^2-2 x(2-y)-22+5 y=0\) For the real value discriminate. \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\Rightarrow \quad {[2(2-y)]^2-4 \times 2[-(22-5 y)]=0}\) \(4(2-y)^2+8(22-5 y)=0\) \((2-y)^2+2(22-5 y)=0\) \(4+y^2-4 y+44-10 y=0\) \(y^2-14 y+48=0\) \((y-8)(y-6)=0\) \(y-8= 0 \text { and } y-6=0\) \(\text { So, } y=6 \text { and } y=8\)
TS EAMCET-04.05.2019
Sets, Relation and Function
117422
The domain of the function \(f(x)=\) \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) is
1 \([-1,1]\)
2 \([1,4]\)
3 \((0,16]\)
4 \([1,16]\)
Explanation:
D Given that, \(f(x)=\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) The function \(f(x)\) is defined, Case- I : \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]\) We know that range of \(\sin ^{-1} x \in[-1,1]\) \(-1 \leq \log _4\left(\frac{x}{4}\right) \leq 1\) \(4^{-1} \leq \frac{x}{4} \leq 4^1\) \(\frac{1}{4} \leq \frac{x}{4} \leq 4\) \(1 \leq \mathrm{x} \leq 16\) \(\mathrm{x} \in[1,16]\) Case-II : \(\sqrt{17 \mathrm{x}-\mathrm{x}^2-16}\) \(17 \mathrm{x}-\mathrm{x}^2-16 \geq 0\) \(\mathrm{x}^2-17 \mathrm{x}+16 \leq 0\) \(\mathrm{x}^2-16 \mathrm{x}-\mathrm{x}+16 \leq 0\) \((\mathrm{x}-16)(\mathrm{x}-1) \leq 0\) \(x \in[1,16]\) The domain of function \(\mathrm{f}(\mathrm{x})\) from equation (i) and (ii) \(\mathrm{x} \in[1,16]\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117419
If \(f: R \rightarrow R\) be defined by \(f(x)=x+2|x+1|+2\) \(|\mathbf{x}-1|\), then the element in the co-domain, which has unique pre image in the domain is
1 3
2 1
3 2
4 5
Explanation:
A We have, \(\begin{aligned} & f(x)=x+2|x+1|+2|x-1| \\ & f(x)=\left\{\begin{array}{cc}x+2(-x-1)+2(-x+1), & x \leq-1 \\ x+2(x+1)+2(-x+1), & -1\lt x \leq 1 \\ x+2(x+1)+2(x-1), & x \geq 1\end{array}\right.\end{aligned}\) \(f(x)=\left\{\begin{array}{cc}-3 x & x \leq-1 \\ x+4, & -1\lt x\lt 1 \\ 5 x, & x \geq 1\end{array}\right.\) Graph of \(\mathrm{f}(\mathrm{x})\) is, Clearly, 3 has unique image in the domain.
TS EAMCET-14.09.2020
Sets, Relation and Function
117420
The range of the function \(f(x)=-\sqrt{-x^2-6 x-5}\) is
1 \([-2,0]\)
2 \([0,2]\)
3 \((\infty,-2]\)
4 \([-2,2]\)
Explanation:
A Given that, \(f(x)=-\sqrt{-x^2-6 x-5}\) Let, \(y=-\sqrt{-\left(x^2+6 x+5\right)}\) On squaring both the side, we get - \(y^2=-x^2+6 x+5\) \(x^2+6 x+5+y^2=0\) \(x=\frac{-6 \pm \sqrt{36-20-4 y^2}}{2}\) \(x=\frac{-6 \pm \sqrt{16-4 y^2}}{2}\) For real value, \(\mathrm{D} \geq 0\) According to question, \(y \in[-2,0]\)Rang of \(f(x) \in[-2,0]\)
TS EAMCET-07.05.2018
Sets, Relation and Function
117421
If \(x\) is real, then the interval in which no value of the expression \(\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) lies, is
1 \((2,5)\)
2 \((3,6)\)
3 \((3,4)\)
4 \((6,8)\)
Explanation:
D Let, \(y=\frac{2\left(x^2+2 x-11\right)}{2 x-5}\) \(y(2 x-5)=2 x^2+4 x-22\) \(2 x y-5 y=2 x^2+4 x-22\) \(2 x^2-4 x-2 x y+5 y-22=0\) \(2 x^2-2 x(2-y)-22+5 y=0\) For the real value discriminate. \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}\) \(\Rightarrow \quad {[2(2-y)]^2-4 \times 2[-(22-5 y)]=0}\) \(4(2-y)^2+8(22-5 y)=0\) \((2-y)^2+2(22-5 y)=0\) \(4+y^2-4 y+44-10 y=0\) \(y^2-14 y+48=0\) \((y-8)(y-6)=0\) \(y-8= 0 \text { and } y-6=0\) \(\text { So, } y=6 \text { and } y=8\)
TS EAMCET-04.05.2019
Sets, Relation and Function
117422
The domain of the function \(f(x)=\) \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) is
1 \([-1,1]\)
2 \([1,4]\)
3 \((0,16]\)
4 \([1,16]\)
Explanation:
D Given that, \(f(x)=\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}\) The function \(f(x)\) is defined, Case- I : \(\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]\) We know that range of \(\sin ^{-1} x \in[-1,1]\) \(-1 \leq \log _4\left(\frac{x}{4}\right) \leq 1\) \(4^{-1} \leq \frac{x}{4} \leq 4^1\) \(\frac{1}{4} \leq \frac{x}{4} \leq 4\) \(1 \leq \mathrm{x} \leq 16\) \(\mathrm{x} \in[1,16]\) Case-II : \(\sqrt{17 \mathrm{x}-\mathrm{x}^2-16}\) \(17 \mathrm{x}-\mathrm{x}^2-16 \geq 0\) \(\mathrm{x}^2-17 \mathrm{x}+16 \leq 0\) \(\mathrm{x}^2-16 \mathrm{x}-\mathrm{x}+16 \leq 0\) \((\mathrm{x}-16)(\mathrm{x}-1) \leq 0\) \(x \in[1,16]\) The domain of function \(\mathrm{f}(\mathrm{x})\) from equation (i) and (ii) \(\mathrm{x} \in[1,16]\).
