117068
The function \(f: R \rightarrow R\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is......
1 Surjective but not injective
2 Bijective
3 Injective but not surjective
4 Neither injective nor surjective
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{\mathrm{x}_1}{\sqrt{1+\mathrm{x}_1^2}}=\frac{\mathrm{x}_2}{\sqrt{1+\mathrm{x}_2^2}}\) \(\mathrm{x}_1^2\left(1+\mathrm{x}_2^2\right)=\mathrm{x}_2^2\left(1+\mathrm{x}_1^2\right)\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is injective in nature Also, Let \(f(x)=y=\frac{x}{\sqrt{1+x^2}}\) Squaring both side, we get- \(y^2\left(1+x^2\right)=x^2\) \(y^2+y^2 x^2=x^2\) \(y^2=x^2-y^2 x^2\) \(y^2=x^2\left(1-y^2\right)\) \(\frac{y^2}{1-y^2}=x^2\) \(x=\frac{y}{\sqrt{1-y^2}}\) As \(\mathrm{y}^2 \leq 1 \Rightarrow-1 \leq|\mathrm{y}| \leq 1\) So, \(\mathrm{f}\) is non surjective.
Shift-II
Sets, Relation and Function
117069
Let \(A\) and \(B\) be finite sets and \(P_A\) and \(P_B\) respectively denote their power sets, If \(P_B\) has 112 elements more than those in \(P_A\), then the number of functions from \(A\) to \(B\) which are injective is
1 224
2 56
3 120
4 840
Explanation:
Exp: Let \(n(A)=m\) \(n(B)=n\) \(\text { According to question, }\) \(n(P(B))-n(P(A))=112\) \(2^{\mathrm{n}}-2^{\mathrm{m}}=112\) \(2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=16 \times 7\) \(\lvert\, 2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=2^4(2^3-1)\) \(\therefore \mathrm{m}=4, \text { and } \mathrm{n}-\mathrm{m}=3\) \(\mathrm{~m}=4, \mathrm{n}=7\) \(\therefore\) Number of injective function from \(A\) to \(B\). \({ }^{n(B)} P_{n(A)}={ }^7 P_4=\frac{7 !}{3 !}=840\)
Shift-II
Sets, Relation and Function
117070
If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\), then \(A\) and \(B\) are respectively
1 \((-\infty, \infty),(0, \infty)\)
2 \((-\infty, 0),[2, \infty)\)
3 \((0, \infty),(2, \infty)\)
4 \((-\infty, 0],(0, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) We know that, \(f(x)\) will be defined when \(|\mathrm{x}|-\mathrm{x}>0 \Rightarrow|\mathrm{x}|>\mathrm{x}\) \(\therefore \mathrm{x} \in(-\infty, 0)\) Now, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) \(=\sqrt{-2 \mathrm{x}}+\frac{1}{\sqrt{-2 \mathrm{x}}}\) \([\because \mathrm{x} \in(-\infty, 0)\) \(\therefore \mathrm{f}_{\min }=2 \sqrt{-2 \mathrm{x} \cdot \frac{1}{\sqrt{-2 \mathrm{x}}}}=2\) \([\because \mathrm{AM} \geq \mathrm{GM}]\) \(\therefore \mathrm{f}(\mathrm{x}) \in[2, \infty)\)
Shift-I
Sets, Relation and Function
117072
Given that \(f: S \rightarrow R\) is said to have a fixed point at \(c\) of \(S\) if \(f(c)=c\). Let \(f:[1, \infty] \rightarrow R\) be defined by \(f(x)=1+\sqrt{x}\). Then
1 f has no fixed point in \([1, \infty]\)
2 f has unique fixed point in \([1, \infty]\)
3 f has two fixed points in \([1, \infty]\)
4 f has infinitely many fixed points in \([1, \infty]\)
Explanation:
B Given that, \(\Rightarrow \quad \mathrm{x}=1+\sqrt{\mathrm{x}}\) \((\mathrm{x}-1)=\sqrt{\mathrm{x}}\) Squaring both side, we get - \(\Rightarrow(\mathrm{x}-1)^2=(\sqrt{\mathrm{x}})^2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=\mathrm{x}\) \(\lvert\, \Rightarrow x^2-2 x-x+1=0\) \(\Rightarrow 1 \cdot x^2-3 x+1=0\) \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1}\) \(=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\) There \(\mathrm{x}=\frac{3+\sqrt{5}}{2} \text { or } \mathrm{x}=\frac{3-\sqrt{5}}{2} \text {, }\) put \(\sqrt{5}=2.23\) \(\mathrm{x}=\frac{3-2.23}{2}=\frac{0.77}{2}=0.385\) \(\therefore 0.385\) does not belong to \([1, \infty]\) \(\mathrm{x}=\frac{3+\sqrt{5}}{2}=\frac{3+2.23}{2}=2.615\) 2.615 belong to \([1, \infty]\) Hence \(f\) has unique fixed point in \([1, \infty]\) Hence option (B) is correct.
WB JEE-2021
Sets, Relation and Function
117073
Consider the function \(f_1(x)=x\), \(f_2(x)=2+\log _e x, x>0\). The graphs of the function intersect
1 Once in \((0,1)\) but never in \((1, \infty)\)
2 Once in \((0,1)\) and once in \(\left(\mathrm{e}^2, \infty\right)\)
3 Once in \((0,1)\) and once in (e, \(\left.\mathrm{e}^2\right)\)
4 More than twice in \((0, \infty)\)
Explanation:
C Given, \(f_1(x)=x \text { and } f_2(x)=2+\log _e x\) Consider, \(g(x)=f_2(x)-f_1(x)\) Now putting the value. \(g(x)=2+\log _e x-x\) Therefore, \(\mathrm{g}\left(0^{+}\right)\lt 0\) \(\mathrm{~g}(1)>0\) \(\mathrm{~g}(\mathrm{e})>0\) \(\mathrm{~g}\left(\mathrm{e}^2\right)\lt 0\) And, value of \(g(x)\) for all \(x \geq e^2\) is negative Hence, \(\mathrm{g}(\mathrm{x})=0\) has exactly two roots in \((0,1)\) and \(\left(\mathrm{e}, \mathrm{e}^2\right)\)
117068
The function \(f: R \rightarrow R\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is......
1 Surjective but not injective
2 Bijective
3 Injective but not surjective
4 Neither injective nor surjective
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{\mathrm{x}_1}{\sqrt{1+\mathrm{x}_1^2}}=\frac{\mathrm{x}_2}{\sqrt{1+\mathrm{x}_2^2}}\) \(\mathrm{x}_1^2\left(1+\mathrm{x}_2^2\right)=\mathrm{x}_2^2\left(1+\mathrm{x}_1^2\right)\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is injective in nature Also, Let \(f(x)=y=\frac{x}{\sqrt{1+x^2}}\) Squaring both side, we get- \(y^2\left(1+x^2\right)=x^2\) \(y^2+y^2 x^2=x^2\) \(y^2=x^2-y^2 x^2\) \(y^2=x^2\left(1-y^2\right)\) \(\frac{y^2}{1-y^2}=x^2\) \(x=\frac{y}{\sqrt{1-y^2}}\) As \(\mathrm{y}^2 \leq 1 \Rightarrow-1 \leq|\mathrm{y}| \leq 1\) So, \(\mathrm{f}\) is non surjective.
Shift-II
Sets, Relation and Function
117069
Let \(A\) and \(B\) be finite sets and \(P_A\) and \(P_B\) respectively denote their power sets, If \(P_B\) has 112 elements more than those in \(P_A\), then the number of functions from \(A\) to \(B\) which are injective is
1 224
2 56
3 120
4 840
Explanation:
Exp: Let \(n(A)=m\) \(n(B)=n\) \(\text { According to question, }\) \(n(P(B))-n(P(A))=112\) \(2^{\mathrm{n}}-2^{\mathrm{m}}=112\) \(2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=16 \times 7\) \(\lvert\, 2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=2^4(2^3-1)\) \(\therefore \mathrm{m}=4, \text { and } \mathrm{n}-\mathrm{m}=3\) \(\mathrm{~m}=4, \mathrm{n}=7\) \(\therefore\) Number of injective function from \(A\) to \(B\). \({ }^{n(B)} P_{n(A)}={ }^7 P_4=\frac{7 !}{3 !}=840\)
Shift-II
Sets, Relation and Function
117070
If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\), then \(A\) and \(B\) are respectively
1 \((-\infty, \infty),(0, \infty)\)
2 \((-\infty, 0),[2, \infty)\)
3 \((0, \infty),(2, \infty)\)
4 \((-\infty, 0],(0, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) We know that, \(f(x)\) will be defined when \(|\mathrm{x}|-\mathrm{x}>0 \Rightarrow|\mathrm{x}|>\mathrm{x}\) \(\therefore \mathrm{x} \in(-\infty, 0)\) Now, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) \(=\sqrt{-2 \mathrm{x}}+\frac{1}{\sqrt{-2 \mathrm{x}}}\) \([\because \mathrm{x} \in(-\infty, 0)\) \(\therefore \mathrm{f}_{\min }=2 \sqrt{-2 \mathrm{x} \cdot \frac{1}{\sqrt{-2 \mathrm{x}}}}=2\) \([\because \mathrm{AM} \geq \mathrm{GM}]\) \(\therefore \mathrm{f}(\mathrm{x}) \in[2, \infty)\)
Shift-I
Sets, Relation and Function
117072
Given that \(f: S \rightarrow R\) is said to have a fixed point at \(c\) of \(S\) if \(f(c)=c\). Let \(f:[1, \infty] \rightarrow R\) be defined by \(f(x)=1+\sqrt{x}\). Then
1 f has no fixed point in \([1, \infty]\)
2 f has unique fixed point in \([1, \infty]\)
3 f has two fixed points in \([1, \infty]\)
4 f has infinitely many fixed points in \([1, \infty]\)
Explanation:
B Given that, \(\Rightarrow \quad \mathrm{x}=1+\sqrt{\mathrm{x}}\) \((\mathrm{x}-1)=\sqrt{\mathrm{x}}\) Squaring both side, we get - \(\Rightarrow(\mathrm{x}-1)^2=(\sqrt{\mathrm{x}})^2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=\mathrm{x}\) \(\lvert\, \Rightarrow x^2-2 x-x+1=0\) \(\Rightarrow 1 \cdot x^2-3 x+1=0\) \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1}\) \(=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\) There \(\mathrm{x}=\frac{3+\sqrt{5}}{2} \text { or } \mathrm{x}=\frac{3-\sqrt{5}}{2} \text {, }\) put \(\sqrt{5}=2.23\) \(\mathrm{x}=\frac{3-2.23}{2}=\frac{0.77}{2}=0.385\) \(\therefore 0.385\) does not belong to \([1, \infty]\) \(\mathrm{x}=\frac{3+\sqrt{5}}{2}=\frac{3+2.23}{2}=2.615\) 2.615 belong to \([1, \infty]\) Hence \(f\) has unique fixed point in \([1, \infty]\) Hence option (B) is correct.
WB JEE-2021
Sets, Relation and Function
117073
Consider the function \(f_1(x)=x\), \(f_2(x)=2+\log _e x, x>0\). The graphs of the function intersect
1 Once in \((0,1)\) but never in \((1, \infty)\)
2 Once in \((0,1)\) and once in \(\left(\mathrm{e}^2, \infty\right)\)
3 Once in \((0,1)\) and once in (e, \(\left.\mathrm{e}^2\right)\)
4 More than twice in \((0, \infty)\)
Explanation:
C Given, \(f_1(x)=x \text { and } f_2(x)=2+\log _e x\) Consider, \(g(x)=f_2(x)-f_1(x)\) Now putting the value. \(g(x)=2+\log _e x-x\) Therefore, \(\mathrm{g}\left(0^{+}\right)\lt 0\) \(\mathrm{~g}(1)>0\) \(\mathrm{~g}(\mathrm{e})>0\) \(\mathrm{~g}\left(\mathrm{e}^2\right)\lt 0\) And, value of \(g(x)\) for all \(x \geq e^2\) is negative Hence, \(\mathrm{g}(\mathrm{x})=0\) has exactly two roots in \((0,1)\) and \(\left(\mathrm{e}, \mathrm{e}^2\right)\)
117068
The function \(f: R \rightarrow R\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is......
1 Surjective but not injective
2 Bijective
3 Injective but not surjective
4 Neither injective nor surjective
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{\mathrm{x}_1}{\sqrt{1+\mathrm{x}_1^2}}=\frac{\mathrm{x}_2}{\sqrt{1+\mathrm{x}_2^2}}\) \(\mathrm{x}_1^2\left(1+\mathrm{x}_2^2\right)=\mathrm{x}_2^2\left(1+\mathrm{x}_1^2\right)\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is injective in nature Also, Let \(f(x)=y=\frac{x}{\sqrt{1+x^2}}\) Squaring both side, we get- \(y^2\left(1+x^2\right)=x^2\) \(y^2+y^2 x^2=x^2\) \(y^2=x^2-y^2 x^2\) \(y^2=x^2\left(1-y^2\right)\) \(\frac{y^2}{1-y^2}=x^2\) \(x=\frac{y}{\sqrt{1-y^2}}\) As \(\mathrm{y}^2 \leq 1 \Rightarrow-1 \leq|\mathrm{y}| \leq 1\) So, \(\mathrm{f}\) is non surjective.
Shift-II
Sets, Relation and Function
117069
Let \(A\) and \(B\) be finite sets and \(P_A\) and \(P_B\) respectively denote their power sets, If \(P_B\) has 112 elements more than those in \(P_A\), then the number of functions from \(A\) to \(B\) which are injective is
1 224
2 56
3 120
4 840
Explanation:
Exp: Let \(n(A)=m\) \(n(B)=n\) \(\text { According to question, }\) \(n(P(B))-n(P(A))=112\) \(2^{\mathrm{n}}-2^{\mathrm{m}}=112\) \(2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=16 \times 7\) \(\lvert\, 2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=2^4(2^3-1)\) \(\therefore \mathrm{m}=4, \text { and } \mathrm{n}-\mathrm{m}=3\) \(\mathrm{~m}=4, \mathrm{n}=7\) \(\therefore\) Number of injective function from \(A\) to \(B\). \({ }^{n(B)} P_{n(A)}={ }^7 P_4=\frac{7 !}{3 !}=840\)
Shift-II
Sets, Relation and Function
117070
If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\), then \(A\) and \(B\) are respectively
1 \((-\infty, \infty),(0, \infty)\)
2 \((-\infty, 0),[2, \infty)\)
3 \((0, \infty),(2, \infty)\)
4 \((-\infty, 0],(0, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) We know that, \(f(x)\) will be defined when \(|\mathrm{x}|-\mathrm{x}>0 \Rightarrow|\mathrm{x}|>\mathrm{x}\) \(\therefore \mathrm{x} \in(-\infty, 0)\) Now, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) \(=\sqrt{-2 \mathrm{x}}+\frac{1}{\sqrt{-2 \mathrm{x}}}\) \([\because \mathrm{x} \in(-\infty, 0)\) \(\therefore \mathrm{f}_{\min }=2 \sqrt{-2 \mathrm{x} \cdot \frac{1}{\sqrt{-2 \mathrm{x}}}}=2\) \([\because \mathrm{AM} \geq \mathrm{GM}]\) \(\therefore \mathrm{f}(\mathrm{x}) \in[2, \infty)\)
Shift-I
Sets, Relation and Function
117072
Given that \(f: S \rightarrow R\) is said to have a fixed point at \(c\) of \(S\) if \(f(c)=c\). Let \(f:[1, \infty] \rightarrow R\) be defined by \(f(x)=1+\sqrt{x}\). Then
1 f has no fixed point in \([1, \infty]\)
2 f has unique fixed point in \([1, \infty]\)
3 f has two fixed points in \([1, \infty]\)
4 f has infinitely many fixed points in \([1, \infty]\)
Explanation:
B Given that, \(\Rightarrow \quad \mathrm{x}=1+\sqrt{\mathrm{x}}\) \((\mathrm{x}-1)=\sqrt{\mathrm{x}}\) Squaring both side, we get - \(\Rightarrow(\mathrm{x}-1)^2=(\sqrt{\mathrm{x}})^2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=\mathrm{x}\) \(\lvert\, \Rightarrow x^2-2 x-x+1=0\) \(\Rightarrow 1 \cdot x^2-3 x+1=0\) \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1}\) \(=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\) There \(\mathrm{x}=\frac{3+\sqrt{5}}{2} \text { or } \mathrm{x}=\frac{3-\sqrt{5}}{2} \text {, }\) put \(\sqrt{5}=2.23\) \(\mathrm{x}=\frac{3-2.23}{2}=\frac{0.77}{2}=0.385\) \(\therefore 0.385\) does not belong to \([1, \infty]\) \(\mathrm{x}=\frac{3+\sqrt{5}}{2}=\frac{3+2.23}{2}=2.615\) 2.615 belong to \([1, \infty]\) Hence \(f\) has unique fixed point in \([1, \infty]\) Hence option (B) is correct.
WB JEE-2021
Sets, Relation and Function
117073
Consider the function \(f_1(x)=x\), \(f_2(x)=2+\log _e x, x>0\). The graphs of the function intersect
1 Once in \((0,1)\) but never in \((1, \infty)\)
2 Once in \((0,1)\) and once in \(\left(\mathrm{e}^2, \infty\right)\)
3 Once in \((0,1)\) and once in (e, \(\left.\mathrm{e}^2\right)\)
4 More than twice in \((0, \infty)\)
Explanation:
C Given, \(f_1(x)=x \text { and } f_2(x)=2+\log _e x\) Consider, \(g(x)=f_2(x)-f_1(x)\) Now putting the value. \(g(x)=2+\log _e x-x\) Therefore, \(\mathrm{g}\left(0^{+}\right)\lt 0\) \(\mathrm{~g}(1)>0\) \(\mathrm{~g}(\mathrm{e})>0\) \(\mathrm{~g}\left(\mathrm{e}^2\right)\lt 0\) And, value of \(g(x)\) for all \(x \geq e^2\) is negative Hence, \(\mathrm{g}(\mathrm{x})=0\) has exactly two roots in \((0,1)\) and \(\left(\mathrm{e}, \mathrm{e}^2\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117068
The function \(f: R \rightarrow R\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is......
1 Surjective but not injective
2 Bijective
3 Injective but not surjective
4 Neither injective nor surjective
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{\mathrm{x}_1}{\sqrt{1+\mathrm{x}_1^2}}=\frac{\mathrm{x}_2}{\sqrt{1+\mathrm{x}_2^2}}\) \(\mathrm{x}_1^2\left(1+\mathrm{x}_2^2\right)=\mathrm{x}_2^2\left(1+\mathrm{x}_1^2\right)\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is injective in nature Also, Let \(f(x)=y=\frac{x}{\sqrt{1+x^2}}\) Squaring both side, we get- \(y^2\left(1+x^2\right)=x^2\) \(y^2+y^2 x^2=x^2\) \(y^2=x^2-y^2 x^2\) \(y^2=x^2\left(1-y^2\right)\) \(\frac{y^2}{1-y^2}=x^2\) \(x=\frac{y}{\sqrt{1-y^2}}\) As \(\mathrm{y}^2 \leq 1 \Rightarrow-1 \leq|\mathrm{y}| \leq 1\) So, \(\mathrm{f}\) is non surjective.
Shift-II
Sets, Relation and Function
117069
Let \(A\) and \(B\) be finite sets and \(P_A\) and \(P_B\) respectively denote their power sets, If \(P_B\) has 112 elements more than those in \(P_A\), then the number of functions from \(A\) to \(B\) which are injective is
1 224
2 56
3 120
4 840
Explanation:
Exp: Let \(n(A)=m\) \(n(B)=n\) \(\text { According to question, }\) \(n(P(B))-n(P(A))=112\) \(2^{\mathrm{n}}-2^{\mathrm{m}}=112\) \(2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=16 \times 7\) \(\lvert\, 2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=2^4(2^3-1)\) \(\therefore \mathrm{m}=4, \text { and } \mathrm{n}-\mathrm{m}=3\) \(\mathrm{~m}=4, \mathrm{n}=7\) \(\therefore\) Number of injective function from \(A\) to \(B\). \({ }^{n(B)} P_{n(A)}={ }^7 P_4=\frac{7 !}{3 !}=840\)
Shift-II
Sets, Relation and Function
117070
If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\), then \(A\) and \(B\) are respectively
1 \((-\infty, \infty),(0, \infty)\)
2 \((-\infty, 0),[2, \infty)\)
3 \((0, \infty),(2, \infty)\)
4 \((-\infty, 0],(0, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) We know that, \(f(x)\) will be defined when \(|\mathrm{x}|-\mathrm{x}>0 \Rightarrow|\mathrm{x}|>\mathrm{x}\) \(\therefore \mathrm{x} \in(-\infty, 0)\) Now, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) \(=\sqrt{-2 \mathrm{x}}+\frac{1}{\sqrt{-2 \mathrm{x}}}\) \([\because \mathrm{x} \in(-\infty, 0)\) \(\therefore \mathrm{f}_{\min }=2 \sqrt{-2 \mathrm{x} \cdot \frac{1}{\sqrt{-2 \mathrm{x}}}}=2\) \([\because \mathrm{AM} \geq \mathrm{GM}]\) \(\therefore \mathrm{f}(\mathrm{x}) \in[2, \infty)\)
Shift-I
Sets, Relation and Function
117072
Given that \(f: S \rightarrow R\) is said to have a fixed point at \(c\) of \(S\) if \(f(c)=c\). Let \(f:[1, \infty] \rightarrow R\) be defined by \(f(x)=1+\sqrt{x}\). Then
1 f has no fixed point in \([1, \infty]\)
2 f has unique fixed point in \([1, \infty]\)
3 f has two fixed points in \([1, \infty]\)
4 f has infinitely many fixed points in \([1, \infty]\)
Explanation:
B Given that, \(\Rightarrow \quad \mathrm{x}=1+\sqrt{\mathrm{x}}\) \((\mathrm{x}-1)=\sqrt{\mathrm{x}}\) Squaring both side, we get - \(\Rightarrow(\mathrm{x}-1)^2=(\sqrt{\mathrm{x}})^2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=\mathrm{x}\) \(\lvert\, \Rightarrow x^2-2 x-x+1=0\) \(\Rightarrow 1 \cdot x^2-3 x+1=0\) \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1}\) \(=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\) There \(\mathrm{x}=\frac{3+\sqrt{5}}{2} \text { or } \mathrm{x}=\frac{3-\sqrt{5}}{2} \text {, }\) put \(\sqrt{5}=2.23\) \(\mathrm{x}=\frac{3-2.23}{2}=\frac{0.77}{2}=0.385\) \(\therefore 0.385\) does not belong to \([1, \infty]\) \(\mathrm{x}=\frac{3+\sqrt{5}}{2}=\frac{3+2.23}{2}=2.615\) 2.615 belong to \([1, \infty]\) Hence \(f\) has unique fixed point in \([1, \infty]\) Hence option (B) is correct.
WB JEE-2021
Sets, Relation and Function
117073
Consider the function \(f_1(x)=x\), \(f_2(x)=2+\log _e x, x>0\). The graphs of the function intersect
1 Once in \((0,1)\) but never in \((1, \infty)\)
2 Once in \((0,1)\) and once in \(\left(\mathrm{e}^2, \infty\right)\)
3 Once in \((0,1)\) and once in (e, \(\left.\mathrm{e}^2\right)\)
4 More than twice in \((0, \infty)\)
Explanation:
C Given, \(f_1(x)=x \text { and } f_2(x)=2+\log _e x\) Consider, \(g(x)=f_2(x)-f_1(x)\) Now putting the value. \(g(x)=2+\log _e x-x\) Therefore, \(\mathrm{g}\left(0^{+}\right)\lt 0\) \(\mathrm{~g}(1)>0\) \(\mathrm{~g}(\mathrm{e})>0\) \(\mathrm{~g}\left(\mathrm{e}^2\right)\lt 0\) And, value of \(g(x)\) for all \(x \geq e^2\) is negative Hence, \(\mathrm{g}(\mathrm{x})=0\) has exactly two roots in \((0,1)\) and \(\left(\mathrm{e}, \mathrm{e}^2\right)\)
117068
The function \(f: R \rightarrow R\) defined by \(f(x)=\frac{x}{\sqrt{1+x^2}}\) is......
1 Surjective but not injective
2 Bijective
3 Injective but not surjective
4 Neither injective nor surjective
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{\mathrm{x}_1}{\sqrt{1+\mathrm{x}_1^2}}=\frac{\mathrm{x}_2}{\sqrt{1+\mathrm{x}_2^2}}\) \(\mathrm{x}_1^2\left(1+\mathrm{x}_2^2\right)=\mathrm{x}_2^2\left(1+\mathrm{x}_1^2\right)\) \(\mathrm{x}_1^2=\mathrm{x}_2^2\) \(\mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is injective in nature Also, Let \(f(x)=y=\frac{x}{\sqrt{1+x^2}}\) Squaring both side, we get- \(y^2\left(1+x^2\right)=x^2\) \(y^2+y^2 x^2=x^2\) \(y^2=x^2-y^2 x^2\) \(y^2=x^2\left(1-y^2\right)\) \(\frac{y^2}{1-y^2}=x^2\) \(x=\frac{y}{\sqrt{1-y^2}}\) As \(\mathrm{y}^2 \leq 1 \Rightarrow-1 \leq|\mathrm{y}| \leq 1\) So, \(\mathrm{f}\) is non surjective.
Shift-II
Sets, Relation and Function
117069
Let \(A\) and \(B\) be finite sets and \(P_A\) and \(P_B\) respectively denote their power sets, If \(P_B\) has 112 elements more than those in \(P_A\), then the number of functions from \(A\) to \(B\) which are injective is
1 224
2 56
3 120
4 840
Explanation:
Exp: Let \(n(A)=m\) \(n(B)=n\) \(\text { According to question, }\) \(n(P(B))-n(P(A))=112\) \(2^{\mathrm{n}}-2^{\mathrm{m}}=112\) \(2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=16 \times 7\) \(\lvert\, 2^{\mathrm{m}}\left(2^{\mathrm{n}-\mathrm{m}}-1\right)=2^4(2^3-1)\) \(\therefore \mathrm{m}=4, \text { and } \mathrm{n}-\mathrm{m}=3\) \(\mathrm{~m}=4, \mathrm{n}=7\) \(\therefore\) Number of injective function from \(A\) to \(B\). \({ }^{n(B)} P_{n(A)}={ }^7 P_4=\frac{7 !}{3 !}=840\)
Shift-II
Sets, Relation and Function
117070
If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\), then \(A\) and \(B\) are respectively
1 \((-\infty, \infty),(0, \infty)\)
2 \((-\infty, 0),[2, \infty)\)
3 \((0, \infty),(2, \infty)\)
4 \((-\infty, 0],(0, \infty)\)
Explanation:
B We have, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) We know that, \(f(x)\) will be defined when \(|\mathrm{x}|-\mathrm{x}>0 \Rightarrow|\mathrm{x}|>\mathrm{x}\) \(\therefore \mathrm{x} \in(-\infty, 0)\) Now, \(f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}\) \(=\sqrt{-2 \mathrm{x}}+\frac{1}{\sqrt{-2 \mathrm{x}}}\) \([\because \mathrm{x} \in(-\infty, 0)\) \(\therefore \mathrm{f}_{\min }=2 \sqrt{-2 \mathrm{x} \cdot \frac{1}{\sqrt{-2 \mathrm{x}}}}=2\) \([\because \mathrm{AM} \geq \mathrm{GM}]\) \(\therefore \mathrm{f}(\mathrm{x}) \in[2, \infty)\)
Shift-I
Sets, Relation and Function
117072
Given that \(f: S \rightarrow R\) is said to have a fixed point at \(c\) of \(S\) if \(f(c)=c\). Let \(f:[1, \infty] \rightarrow R\) be defined by \(f(x)=1+\sqrt{x}\). Then
1 f has no fixed point in \([1, \infty]\)
2 f has unique fixed point in \([1, \infty]\)
3 f has two fixed points in \([1, \infty]\)
4 f has infinitely many fixed points in \([1, \infty]\)
Explanation:
B Given that, \(\Rightarrow \quad \mathrm{x}=1+\sqrt{\mathrm{x}}\) \((\mathrm{x}-1)=\sqrt{\mathrm{x}}\) Squaring both side, we get - \(\Rightarrow(\mathrm{x}-1)^2=(\sqrt{\mathrm{x}})^2\) \(\Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=\mathrm{x}\) \(\lvert\, \Rightarrow x^2-2 x-x+1=0\) \(\Rightarrow 1 \cdot x^2-3 x+1=0\) \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1}\) \(=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}\) There \(\mathrm{x}=\frac{3+\sqrt{5}}{2} \text { or } \mathrm{x}=\frac{3-\sqrt{5}}{2} \text {, }\) put \(\sqrt{5}=2.23\) \(\mathrm{x}=\frac{3-2.23}{2}=\frac{0.77}{2}=0.385\) \(\therefore 0.385\) does not belong to \([1, \infty]\) \(\mathrm{x}=\frac{3+\sqrt{5}}{2}=\frac{3+2.23}{2}=2.615\) 2.615 belong to \([1, \infty]\) Hence \(f\) has unique fixed point in \([1, \infty]\) Hence option (B) is correct.
WB JEE-2021
Sets, Relation and Function
117073
Consider the function \(f_1(x)=x\), \(f_2(x)=2+\log _e x, x>0\). The graphs of the function intersect
1 Once in \((0,1)\) but never in \((1, \infty)\)
2 Once in \((0,1)\) and once in \(\left(\mathrm{e}^2, \infty\right)\)
3 Once in \((0,1)\) and once in (e, \(\left.\mathrm{e}^2\right)\)
4 More than twice in \((0, \infty)\)
Explanation:
C Given, \(f_1(x)=x \text { and } f_2(x)=2+\log _e x\) Consider, \(g(x)=f_2(x)-f_1(x)\) Now putting the value. \(g(x)=2+\log _e x-x\) Therefore, \(\mathrm{g}\left(0^{+}\right)\lt 0\) \(\mathrm{~g}(1)>0\) \(\mathrm{~g}(\mathrm{e})>0\) \(\mathrm{~g}\left(\mathrm{e}^2\right)\lt 0\) And, value of \(g(x)\) for all \(x \geq e^2\) is negative Hence, \(\mathrm{g}(\mathrm{x})=0\) has exactly two roots in \((0,1)\) and \(\left(\mathrm{e}, \mathrm{e}^2\right)\)