117238
If \(f(x)=3 x+6, g(x)=4 x+k\) and fog \((x)=\) gof (x) then \(\mathrm{k}=\)
1 -9
2 18
3 \(\frac{1}{9}\)
4 9
Explanation:
D Given, \(f(x)=3 x+6, g(x)=4 x+k\) \(f \circ g(x)=f(g(x))=f(4 x+k)=3(4 x+k)+6=12 x+3 k+6\) \(g \circ f(x)=g(f(x))=g(3 x+6)=4(3 x+6)+k=12 x+24+k\) \(\because f o g(x)=\operatorname{gof}(x)\) \(12 x+3 k+6=12 x+24+k\) \(2 k=18\) \(k=9\)
MHT-CET 2020
Sets, Relation and Function
117239
If \(f: R \rightarrow R\) is defined by \(f(x)=2 x+3\), then \(f^{-1}\) (x)
1 does not exist because ' \(\mathrm{f}\) ' is not surjective
2 is given by \(\frac{x-3}{2}\)
3 is given by \(\frac{1}{2 x+3}\)
4 does not exist because ' \(\mathrm{f}\) ' is not injective
Explanation:
B Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) is defined by \(f(x)=2 x+3\) So, let \(f(x)=y\) \(y=2 x+3\) \(2 x=y-3\) \(x=\frac{y-3}{2}\) Hence, \(f^{-1}(x)=\frac{x-3}{2}\)
Karnataka CET 2012 BITSAT - 2006
Sets, Relation and Function
117240
\(f: R \rightarrow R, f(x)=3 x+2\) \(g: R \rightarrow R, g(x)=6 x+5\) for the given functions \(\left(g^{-1}\right)(10)=\)
117241
If \(f(x)=\frac{2 x-3}{3 x+4}\), then \(f^{-1}\left(\frac{-4}{3}\right)=\)
1 zero
2 \(\frac{3}{4}\)
3 \(-\frac{2}{3}\)
4 None of these
Explanation:
D Given, \(f(x)=\frac{2 x-3}{3 x+4}\) Let, \(\quad f(x)=y=\frac{2 x-3}{3 x+4}\) On cross multiplication, we get \(3 x y+4 y=2 x-3\) \(3 x y-2 x=-3-4 y\) \(x(3 y-2)=-3-4 y\) \(x=f^{-1}(y)=\frac{-3-4 y}{3 y-2}\) Put, \(y=-\frac{4}{3}\), we get - \(\mathrm{f}^{-1}\left(-\frac{4}{3}\right) =\frac{-3-4 \times\left(-\frac{4}{3}\right)}{3\left(-\frac{4}{3}\right)-2}\) \(=\frac{-3+\frac{16}{3}}{-4-2}=\frac{7}{3 \times(-6)}=-\frac{7}{18}\)
COMEDK 2014
Sets, Relation and Function
117242
\(\quad \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are two function such that \(f(x)=x^2\) and \(g(x)=\frac{1}{x^2}\), then \(x^4(f \circ g)(x)\) is equal to
1 0
2 1
3 \(\mathrm{x}^4\)
4 \(x^2\)
Explanation:
B Given that, \(f(x)=x^2\) \(g(x)=\frac{1}{x^2}\) Then, \((f \circ g) x=f(g(x))\) \(f(g(x))=\left(\frac{1}{x^2}\right)^2=\frac{1}{x^4}\) \(x^4(f \circ g)(x)=x^4 \times \frac{1}{x^4}=1\)
117238
If \(f(x)=3 x+6, g(x)=4 x+k\) and fog \((x)=\) gof (x) then \(\mathrm{k}=\)
1 -9
2 18
3 \(\frac{1}{9}\)
4 9
Explanation:
D Given, \(f(x)=3 x+6, g(x)=4 x+k\) \(f \circ g(x)=f(g(x))=f(4 x+k)=3(4 x+k)+6=12 x+3 k+6\) \(g \circ f(x)=g(f(x))=g(3 x+6)=4(3 x+6)+k=12 x+24+k\) \(\because f o g(x)=\operatorname{gof}(x)\) \(12 x+3 k+6=12 x+24+k\) \(2 k=18\) \(k=9\)
MHT-CET 2020
Sets, Relation and Function
117239
If \(f: R \rightarrow R\) is defined by \(f(x)=2 x+3\), then \(f^{-1}\) (x)
1 does not exist because ' \(\mathrm{f}\) ' is not surjective
2 is given by \(\frac{x-3}{2}\)
3 is given by \(\frac{1}{2 x+3}\)
4 does not exist because ' \(\mathrm{f}\) ' is not injective
Explanation:
B Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) is defined by \(f(x)=2 x+3\) So, let \(f(x)=y\) \(y=2 x+3\) \(2 x=y-3\) \(x=\frac{y-3}{2}\) Hence, \(f^{-1}(x)=\frac{x-3}{2}\)
Karnataka CET 2012 BITSAT - 2006
Sets, Relation and Function
117240
\(f: R \rightarrow R, f(x)=3 x+2\) \(g: R \rightarrow R, g(x)=6 x+5\) for the given functions \(\left(g^{-1}\right)(10)=\)
117241
If \(f(x)=\frac{2 x-3}{3 x+4}\), then \(f^{-1}\left(\frac{-4}{3}\right)=\)
1 zero
2 \(\frac{3}{4}\)
3 \(-\frac{2}{3}\)
4 None of these
Explanation:
D Given, \(f(x)=\frac{2 x-3}{3 x+4}\) Let, \(\quad f(x)=y=\frac{2 x-3}{3 x+4}\) On cross multiplication, we get \(3 x y+4 y=2 x-3\) \(3 x y-2 x=-3-4 y\) \(x(3 y-2)=-3-4 y\) \(x=f^{-1}(y)=\frac{-3-4 y}{3 y-2}\) Put, \(y=-\frac{4}{3}\), we get - \(\mathrm{f}^{-1}\left(-\frac{4}{3}\right) =\frac{-3-4 \times\left(-\frac{4}{3}\right)}{3\left(-\frac{4}{3}\right)-2}\) \(=\frac{-3+\frac{16}{3}}{-4-2}=\frac{7}{3 \times(-6)}=-\frac{7}{18}\)
COMEDK 2014
Sets, Relation and Function
117242
\(\quad \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are two function such that \(f(x)=x^2\) and \(g(x)=\frac{1}{x^2}\), then \(x^4(f \circ g)(x)\) is equal to
1 0
2 1
3 \(\mathrm{x}^4\)
4 \(x^2\)
Explanation:
B Given that, \(f(x)=x^2\) \(g(x)=\frac{1}{x^2}\) Then, \((f \circ g) x=f(g(x))\) \(f(g(x))=\left(\frac{1}{x^2}\right)^2=\frac{1}{x^4}\) \(x^4(f \circ g)(x)=x^4 \times \frac{1}{x^4}=1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117238
If \(f(x)=3 x+6, g(x)=4 x+k\) and fog \((x)=\) gof (x) then \(\mathrm{k}=\)
1 -9
2 18
3 \(\frac{1}{9}\)
4 9
Explanation:
D Given, \(f(x)=3 x+6, g(x)=4 x+k\) \(f \circ g(x)=f(g(x))=f(4 x+k)=3(4 x+k)+6=12 x+3 k+6\) \(g \circ f(x)=g(f(x))=g(3 x+6)=4(3 x+6)+k=12 x+24+k\) \(\because f o g(x)=\operatorname{gof}(x)\) \(12 x+3 k+6=12 x+24+k\) \(2 k=18\) \(k=9\)
MHT-CET 2020
Sets, Relation and Function
117239
If \(f: R \rightarrow R\) is defined by \(f(x)=2 x+3\), then \(f^{-1}\) (x)
1 does not exist because ' \(\mathrm{f}\) ' is not surjective
2 is given by \(\frac{x-3}{2}\)
3 is given by \(\frac{1}{2 x+3}\)
4 does not exist because ' \(\mathrm{f}\) ' is not injective
Explanation:
B Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) is defined by \(f(x)=2 x+3\) So, let \(f(x)=y\) \(y=2 x+3\) \(2 x=y-3\) \(x=\frac{y-3}{2}\) Hence, \(f^{-1}(x)=\frac{x-3}{2}\)
Karnataka CET 2012 BITSAT - 2006
Sets, Relation and Function
117240
\(f: R \rightarrow R, f(x)=3 x+2\) \(g: R \rightarrow R, g(x)=6 x+5\) for the given functions \(\left(g^{-1}\right)(10)=\)
117241
If \(f(x)=\frac{2 x-3}{3 x+4}\), then \(f^{-1}\left(\frac{-4}{3}\right)=\)
1 zero
2 \(\frac{3}{4}\)
3 \(-\frac{2}{3}\)
4 None of these
Explanation:
D Given, \(f(x)=\frac{2 x-3}{3 x+4}\) Let, \(\quad f(x)=y=\frac{2 x-3}{3 x+4}\) On cross multiplication, we get \(3 x y+4 y=2 x-3\) \(3 x y-2 x=-3-4 y\) \(x(3 y-2)=-3-4 y\) \(x=f^{-1}(y)=\frac{-3-4 y}{3 y-2}\) Put, \(y=-\frac{4}{3}\), we get - \(\mathrm{f}^{-1}\left(-\frac{4}{3}\right) =\frac{-3-4 \times\left(-\frac{4}{3}\right)}{3\left(-\frac{4}{3}\right)-2}\) \(=\frac{-3+\frac{16}{3}}{-4-2}=\frac{7}{3 \times(-6)}=-\frac{7}{18}\)
COMEDK 2014
Sets, Relation and Function
117242
\(\quad \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are two function such that \(f(x)=x^2\) and \(g(x)=\frac{1}{x^2}\), then \(x^4(f \circ g)(x)\) is equal to
1 0
2 1
3 \(\mathrm{x}^4\)
4 \(x^2\)
Explanation:
B Given that, \(f(x)=x^2\) \(g(x)=\frac{1}{x^2}\) Then, \((f \circ g) x=f(g(x))\) \(f(g(x))=\left(\frac{1}{x^2}\right)^2=\frac{1}{x^4}\) \(x^4(f \circ g)(x)=x^4 \times \frac{1}{x^4}=1\)
117238
If \(f(x)=3 x+6, g(x)=4 x+k\) and fog \((x)=\) gof (x) then \(\mathrm{k}=\)
1 -9
2 18
3 \(\frac{1}{9}\)
4 9
Explanation:
D Given, \(f(x)=3 x+6, g(x)=4 x+k\) \(f \circ g(x)=f(g(x))=f(4 x+k)=3(4 x+k)+6=12 x+3 k+6\) \(g \circ f(x)=g(f(x))=g(3 x+6)=4(3 x+6)+k=12 x+24+k\) \(\because f o g(x)=\operatorname{gof}(x)\) \(12 x+3 k+6=12 x+24+k\) \(2 k=18\) \(k=9\)
MHT-CET 2020
Sets, Relation and Function
117239
If \(f: R \rightarrow R\) is defined by \(f(x)=2 x+3\), then \(f^{-1}\) (x)
1 does not exist because ' \(\mathrm{f}\) ' is not surjective
2 is given by \(\frac{x-3}{2}\)
3 is given by \(\frac{1}{2 x+3}\)
4 does not exist because ' \(\mathrm{f}\) ' is not injective
Explanation:
B Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) is defined by \(f(x)=2 x+3\) So, let \(f(x)=y\) \(y=2 x+3\) \(2 x=y-3\) \(x=\frac{y-3}{2}\) Hence, \(f^{-1}(x)=\frac{x-3}{2}\)
Karnataka CET 2012 BITSAT - 2006
Sets, Relation and Function
117240
\(f: R \rightarrow R, f(x)=3 x+2\) \(g: R \rightarrow R, g(x)=6 x+5\) for the given functions \(\left(g^{-1}\right)(10)=\)
117241
If \(f(x)=\frac{2 x-3}{3 x+4}\), then \(f^{-1}\left(\frac{-4}{3}\right)=\)
1 zero
2 \(\frac{3}{4}\)
3 \(-\frac{2}{3}\)
4 None of these
Explanation:
D Given, \(f(x)=\frac{2 x-3}{3 x+4}\) Let, \(\quad f(x)=y=\frac{2 x-3}{3 x+4}\) On cross multiplication, we get \(3 x y+4 y=2 x-3\) \(3 x y-2 x=-3-4 y\) \(x(3 y-2)=-3-4 y\) \(x=f^{-1}(y)=\frac{-3-4 y}{3 y-2}\) Put, \(y=-\frac{4}{3}\), we get - \(\mathrm{f}^{-1}\left(-\frac{4}{3}\right) =\frac{-3-4 \times\left(-\frac{4}{3}\right)}{3\left(-\frac{4}{3}\right)-2}\) \(=\frac{-3+\frac{16}{3}}{-4-2}=\frac{7}{3 \times(-6)}=-\frac{7}{18}\)
COMEDK 2014
Sets, Relation and Function
117242
\(\quad \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are two function such that \(f(x)=x^2\) and \(g(x)=\frac{1}{x^2}\), then \(x^4(f \circ g)(x)\) is equal to
1 0
2 1
3 \(\mathrm{x}^4\)
4 \(x^2\)
Explanation:
B Given that, \(f(x)=x^2\) \(g(x)=\frac{1}{x^2}\) Then, \((f \circ g) x=f(g(x))\) \(f(g(x))=\left(\frac{1}{x^2}\right)^2=\frac{1}{x^4}\) \(x^4(f \circ g)(x)=x^4 \times \frac{1}{x^4}=1\)
117238
If \(f(x)=3 x+6, g(x)=4 x+k\) and fog \((x)=\) gof (x) then \(\mathrm{k}=\)
1 -9
2 18
3 \(\frac{1}{9}\)
4 9
Explanation:
D Given, \(f(x)=3 x+6, g(x)=4 x+k\) \(f \circ g(x)=f(g(x))=f(4 x+k)=3(4 x+k)+6=12 x+3 k+6\) \(g \circ f(x)=g(f(x))=g(3 x+6)=4(3 x+6)+k=12 x+24+k\) \(\because f o g(x)=\operatorname{gof}(x)\) \(12 x+3 k+6=12 x+24+k\) \(2 k=18\) \(k=9\)
MHT-CET 2020
Sets, Relation and Function
117239
If \(f: R \rightarrow R\) is defined by \(f(x)=2 x+3\), then \(f^{-1}\) (x)
1 does not exist because ' \(\mathrm{f}\) ' is not surjective
2 is given by \(\frac{x-3}{2}\)
3 is given by \(\frac{1}{2 x+3}\)
4 does not exist because ' \(\mathrm{f}\) ' is not injective
Explanation:
B Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) is defined by \(f(x)=2 x+3\) So, let \(f(x)=y\) \(y=2 x+3\) \(2 x=y-3\) \(x=\frac{y-3}{2}\) Hence, \(f^{-1}(x)=\frac{x-3}{2}\)
Karnataka CET 2012 BITSAT - 2006
Sets, Relation and Function
117240
\(f: R \rightarrow R, f(x)=3 x+2\) \(g: R \rightarrow R, g(x)=6 x+5\) for the given functions \(\left(g^{-1}\right)(10)=\)
117241
If \(f(x)=\frac{2 x-3}{3 x+4}\), then \(f^{-1}\left(\frac{-4}{3}\right)=\)
1 zero
2 \(\frac{3}{4}\)
3 \(-\frac{2}{3}\)
4 None of these
Explanation:
D Given, \(f(x)=\frac{2 x-3}{3 x+4}\) Let, \(\quad f(x)=y=\frac{2 x-3}{3 x+4}\) On cross multiplication, we get \(3 x y+4 y=2 x-3\) \(3 x y-2 x=-3-4 y\) \(x(3 y-2)=-3-4 y\) \(x=f^{-1}(y)=\frac{-3-4 y}{3 y-2}\) Put, \(y=-\frac{4}{3}\), we get - \(\mathrm{f}^{-1}\left(-\frac{4}{3}\right) =\frac{-3-4 \times\left(-\frac{4}{3}\right)}{3\left(-\frac{4}{3}\right)-2}\) \(=\frac{-3+\frac{16}{3}}{-4-2}=\frac{7}{3 \times(-6)}=-\frac{7}{18}\)
COMEDK 2014
Sets, Relation and Function
117242
\(\quad \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are two function such that \(f(x)=x^2\) and \(g(x)=\frac{1}{x^2}\), then \(x^4(f \circ g)(x)\) is equal to
1 0
2 1
3 \(\mathrm{x}^4\)
4 \(x^2\)
Explanation:
B Given that, \(f(x)=x^2\) \(g(x)=\frac{1}{x^2}\) Then, \((f \circ g) x=f(g(x))\) \(f(g(x))=\left(\frac{1}{x^2}\right)^2=\frac{1}{x^4}\) \(x^4(f \circ g)(x)=x^4 \times \frac{1}{x^4}=1\)