NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117224
Let \(f(x)=\frac{4 x+3}{x+2}\). Then the value of \(f^{-1}(-2)\) is equal to
1 \(\frac{7}{5}\)
2 \(\frac{-7}{6}\)
3 \(\frac{-7}{5}\)
4 \(\frac{7}{6}\)
5 \(\frac{5}{6}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+3}{\mathrm{x}+2}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{4 x+3}{x+2}\) \(\mathrm{xy}+2 \mathrm{y}=4 \mathrm{x}+3\) \(\mathrm{xy}-4 \mathrm{x}=3-2 \mathrm{y}\) \(\mathrm{x}(\mathrm{y}-4)=3-2 \mathrm{y}\) \(\mathrm{x}=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\mathrm{f}^{-1}(\mathrm{y})=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{3-2 \mathrm{x}}{\mathrm{x}-4}\) So, \(\quad f^{-1}(-2)=\frac{3-2(-2)}{-2-4}=\frac{7}{-6}=\frac{-7}{6}\)
Kerala CEE-2021
Sets, Relation and Function
117225
Let \(\odot\) be a binary operation on \(Q-\{0\}\) defined by \(\mathbf{a} \odot \mathbf{b}=\frac{\mathbf{a}}{\mathbf{b}}\). Then \(\mathbf{1} \odot(\mathbf{2} \odot \mathbf{( 3} \odot \mathbf{4})\) is equal to
117145
Let \(A=(u, v, w, z)\) and \(B=\{3,5\}\), then the number of relations from \(A\) to \(B\) is
1 256
2 1024
3 512
4 64
Explanation:
Exp: (A) : Given, \(\mathrm{A}=\{(\mathrm{u}, \mathrm{v}, \mathrm{w}, \mathrm{z})\}\) \(B=\{3,5\}\) Then, number of elements in \(A=4=m\) And, number of elements in \(\mathrm{B}=2=\mathrm{n}\) So, the number of relations from \(A\) to \(B\) is \(2^{\mathrm{m} \times \mathrm{n}}=2^{2 \times 4}=2^8=256\)
117224
Let \(f(x)=\frac{4 x+3}{x+2}\). Then the value of \(f^{-1}(-2)\) is equal to
1 \(\frac{7}{5}\)
2 \(\frac{-7}{6}\)
3 \(\frac{-7}{5}\)
4 \(\frac{7}{6}\)
5 \(\frac{5}{6}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+3}{\mathrm{x}+2}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{4 x+3}{x+2}\) \(\mathrm{xy}+2 \mathrm{y}=4 \mathrm{x}+3\) \(\mathrm{xy}-4 \mathrm{x}=3-2 \mathrm{y}\) \(\mathrm{x}(\mathrm{y}-4)=3-2 \mathrm{y}\) \(\mathrm{x}=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\mathrm{f}^{-1}(\mathrm{y})=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{3-2 \mathrm{x}}{\mathrm{x}-4}\) So, \(\quad f^{-1}(-2)=\frac{3-2(-2)}{-2-4}=\frac{7}{-6}=\frac{-7}{6}\)
Kerala CEE-2021
Sets, Relation and Function
117225
Let \(\odot\) be a binary operation on \(Q-\{0\}\) defined by \(\mathbf{a} \odot \mathbf{b}=\frac{\mathbf{a}}{\mathbf{b}}\). Then \(\mathbf{1} \odot(\mathbf{2} \odot \mathbf{( 3} \odot \mathbf{4})\) is equal to
117145
Let \(A=(u, v, w, z)\) and \(B=\{3,5\}\), then the number of relations from \(A\) to \(B\) is
1 256
2 1024
3 512
4 64
Explanation:
Exp: (A) : Given, \(\mathrm{A}=\{(\mathrm{u}, \mathrm{v}, \mathrm{w}, \mathrm{z})\}\) \(B=\{3,5\}\) Then, number of elements in \(A=4=m\) And, number of elements in \(\mathrm{B}=2=\mathrm{n}\) So, the number of relations from \(A\) to \(B\) is \(2^{\mathrm{m} \times \mathrm{n}}=2^{2 \times 4}=2^8=256\)
117224
Let \(f(x)=\frac{4 x+3}{x+2}\). Then the value of \(f^{-1}(-2)\) is equal to
1 \(\frac{7}{5}\)
2 \(\frac{-7}{6}\)
3 \(\frac{-7}{5}\)
4 \(\frac{7}{6}\)
5 \(\frac{5}{6}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+3}{\mathrm{x}+2}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{4 x+3}{x+2}\) \(\mathrm{xy}+2 \mathrm{y}=4 \mathrm{x}+3\) \(\mathrm{xy}-4 \mathrm{x}=3-2 \mathrm{y}\) \(\mathrm{x}(\mathrm{y}-4)=3-2 \mathrm{y}\) \(\mathrm{x}=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\mathrm{f}^{-1}(\mathrm{y})=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{3-2 \mathrm{x}}{\mathrm{x}-4}\) So, \(\quad f^{-1}(-2)=\frac{3-2(-2)}{-2-4}=\frac{7}{-6}=\frac{-7}{6}\)
Kerala CEE-2021
Sets, Relation and Function
117225
Let \(\odot\) be a binary operation on \(Q-\{0\}\) defined by \(\mathbf{a} \odot \mathbf{b}=\frac{\mathbf{a}}{\mathbf{b}}\). Then \(\mathbf{1} \odot(\mathbf{2} \odot \mathbf{( 3} \odot \mathbf{4})\) is equal to
117145
Let \(A=(u, v, w, z)\) and \(B=\{3,5\}\), then the number of relations from \(A\) to \(B\) is
1 256
2 1024
3 512
4 64
Explanation:
Exp: (A) : Given, \(\mathrm{A}=\{(\mathrm{u}, \mathrm{v}, \mathrm{w}, \mathrm{z})\}\) \(B=\{3,5\}\) Then, number of elements in \(A=4=m\) And, number of elements in \(\mathrm{B}=2=\mathrm{n}\) So, the number of relations from \(A\) to \(B\) is \(2^{\mathrm{m} \times \mathrm{n}}=2^{2 \times 4}=2^8=256\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117224
Let \(f(x)=\frac{4 x+3}{x+2}\). Then the value of \(f^{-1}(-2)\) is equal to
1 \(\frac{7}{5}\)
2 \(\frac{-7}{6}\)
3 \(\frac{-7}{5}\)
4 \(\frac{7}{6}\)
5 \(\frac{5}{6}\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+3}{\mathrm{x}+2}\) Let, \(\quad \mathrm{f}(\mathrm{x})=\mathrm{y}\) Then, \(y=\frac{4 x+3}{x+2}\) \(\mathrm{xy}+2 \mathrm{y}=4 \mathrm{x}+3\) \(\mathrm{xy}-4 \mathrm{x}=3-2 \mathrm{y}\) \(\mathrm{x}(\mathrm{y}-4)=3-2 \mathrm{y}\) \(\mathrm{x}=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\mathrm{f}^{-1}(\mathrm{y})=\frac{3-2 \mathrm{y}}{\mathrm{y}-4}\) \(\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{3-2 \mathrm{x}}{\mathrm{x}-4}\) So, \(\quad f^{-1}(-2)=\frac{3-2(-2)}{-2-4}=\frac{7}{-6}=\frac{-7}{6}\)
Kerala CEE-2021
Sets, Relation and Function
117225
Let \(\odot\) be a binary operation on \(Q-\{0\}\) defined by \(\mathbf{a} \odot \mathbf{b}=\frac{\mathbf{a}}{\mathbf{b}}\). Then \(\mathbf{1} \odot(\mathbf{2} \odot \mathbf{( 3} \odot \mathbf{4})\) is equal to
117145
Let \(A=(u, v, w, z)\) and \(B=\{3,5\}\), then the number of relations from \(A\) to \(B\) is
1 256
2 1024
3 512
4 64
Explanation:
Exp: (A) : Given, \(\mathrm{A}=\{(\mathrm{u}, \mathrm{v}, \mathrm{w}, \mathrm{z})\}\) \(B=\{3,5\}\) Then, number of elements in \(A=4=m\) And, number of elements in \(\mathrm{B}=2=\mathrm{n}\) So, the number of relations from \(A\) to \(B\) is \(2^{\mathrm{m} \times \mathrm{n}}=2^{2 \times 4}=2^8=256\)