NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117300
Range of the function \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\), is
1 \(\left(0, \frac{\pi}{2}\right)\)
2 \(\left[0, \frac{\pi}{2}\right)\)
3 \(\left(0, \frac{\pi}{2}\right]\)
4 \(\left[0, \frac{\pi}{2}\right]\)
Explanation:
B \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) For \(y\) to be defined \(\left|\frac{x^2}{1+x^2}\right|\lt 1\) which is true for all \(x \in\) R. Now, \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) \(\Rightarrow \frac{x^2}{1+x^2}=\sin y \Rightarrow x=\sqrt{\frac{\sin y}{1-\sin y}}\) For the existence of \(x-\) \(\sin \mathrm{y} \geq 0\) and \(1-\sin \mathrm{y}>0\) \(\Rightarrow 0 \leq \sin \mathrm{y}\lt 1 \Rightarrow 0 \leq \mathrm{y}\lt \frac{\pi}{2}\) Thus, range of the given function is \(\left[0, \frac{\pi}{2}\right)\).
VITEEE-2010
Sets, Relation and Function
117301
If \(\left(1+\tan 1^0\right)\left(1+\tan 2^0\right) \ldots \ldots\left(1+\tan 45^0\right)=2^n\), then \(n\) is
1 22
2 24
3 23
4 12
Explanation:
C Given, \(\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots . .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\text { Consider one terms and solve, we get - }\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}+\left(-44^{\circ}\right)\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}-44^{\circ}\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\frac{\tan 45^{\circ}-\tan 44^{\circ}}{1+\tan 45^{\circ} \times \tan 44^{\circ}}\) \(1+\tan 1^{\circ}=1+\frac{1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{1+\tan 44^{\circ}+1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{2}{1+\tan 44^{\circ}}\) \(\text { So, }\left(1+\tan 1^{\circ}\right) \cdot\left(1+\tan 2^{\circ}\right) \ldots \ldots .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\frac{2}{\left(1+\tan 44^{\circ}\right)} \times \frac{2}{\left(1+\tan 43^{\circ}\right)} \times\left(1+\tan 43^{\circ}\right) \times(1+\tan\) \(\left.44^{\circ}\right) \times 2=2^{\mathrm{n}}\) \(\frac{2 \times 2 \times 2 \ldots . \times 2}{23 \operatorname{times}}=2^{\mathrm{n}}\) \(2^{23}=2^{\mathrm{n}}\) \(\text { So, } \operatorname{comparing} \text { both side } \mathrm{n}=23\) So, comparing both side \(\mathrm{n}=23\)
UPSEE-2018
Sets, Relation and Function
117303
The range of \(x\) for which the formula \(3 \sin ^{-1} x=\) \(\sin ^{-1}\left[x\left(3-4 x^2\right)\right]\) hold is
117304
If domain of the function \(f(x)=x^2-6 x+7\) is \((-\infty, \infty)\) then its range is
1 \([-2,3]\)
2 \((-\infty,-2]\)
3 \((-\infty, \infty)\)
4 \([-2, \infty)\)
Explanation:
D Given, function \(f(x)=x^2-6 x+7\) domain of the function is \((-\infty, \infty)\). Then for range, consider \(f(x)=y\) So, \(x^2-6 x+7=y\) \(\mathrm{x}^2-6 \mathrm{x}+7-\mathrm{y}=0\) Now, \(x=\frac{6 \pm \sqrt{36-4 \times 1(7-y)}}{2 \times 1}\) \(x=\frac{6 \pm \sqrt{36-28+4 y}}{2}=\frac{6 \pm \sqrt{8+4 y}}{2}\) \(x=\frac{6 \pm 2 \sqrt{y+2}}{2}\) \(x=3 \pm \sqrt{y+2}\) Then, \(f(x)\) is defined only when - \(y+2 \geq 0 \Rightarrow y \geq-2\)So, range of \(f(x)=[-2, \infty)\)
117300
Range of the function \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\), is
1 \(\left(0, \frac{\pi}{2}\right)\)
2 \(\left[0, \frac{\pi}{2}\right)\)
3 \(\left(0, \frac{\pi}{2}\right]\)
4 \(\left[0, \frac{\pi}{2}\right]\)
Explanation:
B \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) For \(y\) to be defined \(\left|\frac{x^2}{1+x^2}\right|\lt 1\) which is true for all \(x \in\) R. Now, \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) \(\Rightarrow \frac{x^2}{1+x^2}=\sin y \Rightarrow x=\sqrt{\frac{\sin y}{1-\sin y}}\) For the existence of \(x-\) \(\sin \mathrm{y} \geq 0\) and \(1-\sin \mathrm{y}>0\) \(\Rightarrow 0 \leq \sin \mathrm{y}\lt 1 \Rightarrow 0 \leq \mathrm{y}\lt \frac{\pi}{2}\) Thus, range of the given function is \(\left[0, \frac{\pi}{2}\right)\).
VITEEE-2010
Sets, Relation and Function
117301
If \(\left(1+\tan 1^0\right)\left(1+\tan 2^0\right) \ldots \ldots\left(1+\tan 45^0\right)=2^n\), then \(n\) is
1 22
2 24
3 23
4 12
Explanation:
C Given, \(\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots . .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\text { Consider one terms and solve, we get - }\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}+\left(-44^{\circ}\right)\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}-44^{\circ}\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\frac{\tan 45^{\circ}-\tan 44^{\circ}}{1+\tan 45^{\circ} \times \tan 44^{\circ}}\) \(1+\tan 1^{\circ}=1+\frac{1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{1+\tan 44^{\circ}+1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{2}{1+\tan 44^{\circ}}\) \(\text { So, }\left(1+\tan 1^{\circ}\right) \cdot\left(1+\tan 2^{\circ}\right) \ldots \ldots .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\frac{2}{\left(1+\tan 44^{\circ}\right)} \times \frac{2}{\left(1+\tan 43^{\circ}\right)} \times\left(1+\tan 43^{\circ}\right) \times(1+\tan\) \(\left.44^{\circ}\right) \times 2=2^{\mathrm{n}}\) \(\frac{2 \times 2 \times 2 \ldots . \times 2}{23 \operatorname{times}}=2^{\mathrm{n}}\) \(2^{23}=2^{\mathrm{n}}\) \(\text { So, } \operatorname{comparing} \text { both side } \mathrm{n}=23\) So, comparing both side \(\mathrm{n}=23\)
UPSEE-2018
Sets, Relation and Function
117303
The range of \(x\) for which the formula \(3 \sin ^{-1} x=\) \(\sin ^{-1}\left[x\left(3-4 x^2\right)\right]\) hold is
117304
If domain of the function \(f(x)=x^2-6 x+7\) is \((-\infty, \infty)\) then its range is
1 \([-2,3]\)
2 \((-\infty,-2]\)
3 \((-\infty, \infty)\)
4 \([-2, \infty)\)
Explanation:
D Given, function \(f(x)=x^2-6 x+7\) domain of the function is \((-\infty, \infty)\). Then for range, consider \(f(x)=y\) So, \(x^2-6 x+7=y\) \(\mathrm{x}^2-6 \mathrm{x}+7-\mathrm{y}=0\) Now, \(x=\frac{6 \pm \sqrt{36-4 \times 1(7-y)}}{2 \times 1}\) \(x=\frac{6 \pm \sqrt{36-28+4 y}}{2}=\frac{6 \pm \sqrt{8+4 y}}{2}\) \(x=\frac{6 \pm 2 \sqrt{y+2}}{2}\) \(x=3 \pm \sqrt{y+2}\) Then, \(f(x)\) is defined only when - \(y+2 \geq 0 \Rightarrow y \geq-2\)So, range of \(f(x)=[-2, \infty)\)
117300
Range of the function \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\), is
1 \(\left(0, \frac{\pi}{2}\right)\)
2 \(\left[0, \frac{\pi}{2}\right)\)
3 \(\left(0, \frac{\pi}{2}\right]\)
4 \(\left[0, \frac{\pi}{2}\right]\)
Explanation:
B \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) For \(y\) to be defined \(\left|\frac{x^2}{1+x^2}\right|\lt 1\) which is true for all \(x \in\) R. Now, \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) \(\Rightarrow \frac{x^2}{1+x^2}=\sin y \Rightarrow x=\sqrt{\frac{\sin y}{1-\sin y}}\) For the existence of \(x-\) \(\sin \mathrm{y} \geq 0\) and \(1-\sin \mathrm{y}>0\) \(\Rightarrow 0 \leq \sin \mathrm{y}\lt 1 \Rightarrow 0 \leq \mathrm{y}\lt \frac{\pi}{2}\) Thus, range of the given function is \(\left[0, \frac{\pi}{2}\right)\).
VITEEE-2010
Sets, Relation and Function
117301
If \(\left(1+\tan 1^0\right)\left(1+\tan 2^0\right) \ldots \ldots\left(1+\tan 45^0\right)=2^n\), then \(n\) is
1 22
2 24
3 23
4 12
Explanation:
C Given, \(\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots . .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\text { Consider one terms and solve, we get - }\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}+\left(-44^{\circ}\right)\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}-44^{\circ}\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\frac{\tan 45^{\circ}-\tan 44^{\circ}}{1+\tan 45^{\circ} \times \tan 44^{\circ}}\) \(1+\tan 1^{\circ}=1+\frac{1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{1+\tan 44^{\circ}+1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{2}{1+\tan 44^{\circ}}\) \(\text { So, }\left(1+\tan 1^{\circ}\right) \cdot\left(1+\tan 2^{\circ}\right) \ldots \ldots .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\frac{2}{\left(1+\tan 44^{\circ}\right)} \times \frac{2}{\left(1+\tan 43^{\circ}\right)} \times\left(1+\tan 43^{\circ}\right) \times(1+\tan\) \(\left.44^{\circ}\right) \times 2=2^{\mathrm{n}}\) \(\frac{2 \times 2 \times 2 \ldots . \times 2}{23 \operatorname{times}}=2^{\mathrm{n}}\) \(2^{23}=2^{\mathrm{n}}\) \(\text { So, } \operatorname{comparing} \text { both side } \mathrm{n}=23\) So, comparing both side \(\mathrm{n}=23\)
UPSEE-2018
Sets, Relation and Function
117303
The range of \(x\) for which the formula \(3 \sin ^{-1} x=\) \(\sin ^{-1}\left[x\left(3-4 x^2\right)\right]\) hold is
117304
If domain of the function \(f(x)=x^2-6 x+7\) is \((-\infty, \infty)\) then its range is
1 \([-2,3]\)
2 \((-\infty,-2]\)
3 \((-\infty, \infty)\)
4 \([-2, \infty)\)
Explanation:
D Given, function \(f(x)=x^2-6 x+7\) domain of the function is \((-\infty, \infty)\). Then for range, consider \(f(x)=y\) So, \(x^2-6 x+7=y\) \(\mathrm{x}^2-6 \mathrm{x}+7-\mathrm{y}=0\) Now, \(x=\frac{6 \pm \sqrt{36-4 \times 1(7-y)}}{2 \times 1}\) \(x=\frac{6 \pm \sqrt{36-28+4 y}}{2}=\frac{6 \pm \sqrt{8+4 y}}{2}\) \(x=\frac{6 \pm 2 \sqrt{y+2}}{2}\) \(x=3 \pm \sqrt{y+2}\) Then, \(f(x)\) is defined only when - \(y+2 \geq 0 \Rightarrow y \geq-2\)So, range of \(f(x)=[-2, \infty)\)
117300
Range of the function \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\), is
1 \(\left(0, \frac{\pi}{2}\right)\)
2 \(\left[0, \frac{\pi}{2}\right)\)
3 \(\left(0, \frac{\pi}{2}\right]\)
4 \(\left[0, \frac{\pi}{2}\right]\)
Explanation:
B \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) For \(y\) to be defined \(\left|\frac{x^2}{1+x^2}\right|\lt 1\) which is true for all \(x \in\) R. Now, \(y=\sin ^{-1}\left(\frac{x^2}{1+x^2}\right)\) \(\Rightarrow \frac{x^2}{1+x^2}=\sin y \Rightarrow x=\sqrt{\frac{\sin y}{1-\sin y}}\) For the existence of \(x-\) \(\sin \mathrm{y} \geq 0\) and \(1-\sin \mathrm{y}>0\) \(\Rightarrow 0 \leq \sin \mathrm{y}\lt 1 \Rightarrow 0 \leq \mathrm{y}\lt \frac{\pi}{2}\) Thus, range of the given function is \(\left[0, \frac{\pi}{2}\right)\).
VITEEE-2010
Sets, Relation and Function
117301
If \(\left(1+\tan 1^0\right)\left(1+\tan 2^0\right) \ldots \ldots\left(1+\tan 45^0\right)=2^n\), then \(n\) is
1 22
2 24
3 23
4 12
Explanation:
C Given, \(\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots . .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\text { Consider one terms and solve, we get - }\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}+\left(-44^{\circ}\right)\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\tan \left(45^{\circ}-44^{\circ}\right)\) \(\left(1+\tan 1^{\circ}\right)=1+\frac{\tan 45^{\circ}-\tan 44^{\circ}}{1+\tan 45^{\circ} \times \tan 44^{\circ}}\) \(1+\tan 1^{\circ}=1+\frac{1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{1+\tan 44^{\circ}+1-\tan 44^{\circ}}{1+\tan 44^{\circ}}\) \(1+\tan 1^{\circ}=\frac{2}{1+\tan 44^{\circ}}\) \(\text { So, }\left(1+\tan 1^{\circ}\right) \cdot\left(1+\tan 2^{\circ}\right) \ldots \ldots .\left(1+\tan 45^{\circ}\right)=2^{\mathrm{n}}\) \(\frac{2}{\left(1+\tan 44^{\circ}\right)} \times \frac{2}{\left(1+\tan 43^{\circ}\right)} \times\left(1+\tan 43^{\circ}\right) \times(1+\tan\) \(\left.44^{\circ}\right) \times 2=2^{\mathrm{n}}\) \(\frac{2 \times 2 \times 2 \ldots . \times 2}{23 \operatorname{times}}=2^{\mathrm{n}}\) \(2^{23}=2^{\mathrm{n}}\) \(\text { So, } \operatorname{comparing} \text { both side } \mathrm{n}=23\) So, comparing both side \(\mathrm{n}=23\)
UPSEE-2018
Sets, Relation and Function
117303
The range of \(x\) for which the formula \(3 \sin ^{-1} x=\) \(\sin ^{-1}\left[x\left(3-4 x^2\right)\right]\) hold is
117304
If domain of the function \(f(x)=x^2-6 x+7\) is \((-\infty, \infty)\) then its range is
1 \([-2,3]\)
2 \((-\infty,-2]\)
3 \((-\infty, \infty)\)
4 \([-2, \infty)\)
Explanation:
D Given, function \(f(x)=x^2-6 x+7\) domain of the function is \((-\infty, \infty)\). Then for range, consider \(f(x)=y\) So, \(x^2-6 x+7=y\) \(\mathrm{x}^2-6 \mathrm{x}+7-\mathrm{y}=0\) Now, \(x=\frac{6 \pm \sqrt{36-4 \times 1(7-y)}}{2 \times 1}\) \(x=\frac{6 \pm \sqrt{36-28+4 y}}{2}=\frac{6 \pm \sqrt{8+4 y}}{2}\) \(x=\frac{6 \pm 2 \sqrt{y+2}}{2}\) \(x=3 \pm \sqrt{y+2}\) Then, \(f(x)\) is defined only when - \(y+2 \geq 0 \Rightarrow y \geq-2\)So, range of \(f(x)=[-2, \infty)\)
