116909
\(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\), then \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) is equal to
1 1
2 2
3 3
4 \(\frac{1}{2}\)
Explanation:
A Given, \(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\) Then, \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}\) \(=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{x^2-x^2+1}\) \(==\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)\) We have, \(\mathrm{x}=\frac{1}{2} \times \frac{4}{\sqrt{3}}=\frac{2}{\sqrt{3}}\) \(\therefore \mathrm{x}^2-1=\frac{4}{3}-1=\frac{1}{3}\) \(\therefore \sqrt{\mathrm{x}^2-1}\left(\mathrm{x}+\sqrt{\mathrm{x}^2-1}\right)=\frac{1}{\sqrt{3}}\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)=\frac{1 \times 3}{\sqrt{3} \times \sqrt{3}}=1\)
AP EAMCET-2005
Sets, Relation and Function
116910
If \(F\) is function such that \(F(0)=2, F(1)=3\), \(F(x+2)=2 F(x)-F(x+1)\) for \(x \geq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 17
4 13
Explanation:
D Given, \(F(0)=2, F(1)=3\) \(F(x+2)=2 F(x)-F(x-1)\) Putting \(\mathrm{x}=0\), we get - \(F(2)=2 F(0)-F(1)\) \(F(2)=2(2)-3\) \(F(2)=4-3\) \(F(2)=1\) Putting \(\mathrm{x}=1\), in equation (i) we get - \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2(3)-1 \{\because \mathrm{F}(1)=3, \mathrm{~F}(2)=1\}\) \(\mathrm{F}(3)=5\) Putting \(\mathrm{x}=2\), in equation (i) we get - \(\mathrm{F}(4)=2 \mathrm{~F}(2)-\mathrm{F}(3)\) \(\mathrm{F}(4)=2(1)-5 \quad\{\because \mathrm{F}(2)=1, \mathrm{~F}(3)=5\}\) \(F(4)=-3\) Putting \(\mathrm{x}=3\), in equation (i) we get - \(\mathrm{F}(5)=2 \mathrm{~F}(3)-\mathrm{F}(4)\) \(\mathrm{F}(5)=2(5)+3\) \(\mathrm{~F}(5)=13\)
VITEEE-2010
Sets, Relation and Function
116912
Let \(f\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\), for \(x \geq\) 0 then \(f(x)\) for \(x\lt 0\) is
1 \(-3 \sin x+4 \cos x\)
2 \(-3 \sin x-4 \cos x\)
3 \(3 \sin x+4 \cos x\)
4 \(3 \sin x-4 \cos x\)
Explanation:
D Given, \(\mathrm{f}\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\) for \(x \geq 0\). Then, \(f(x)=3 \sin x+4 \cos x\) Since, \(f\) is odd function. Then, \(f(-x)=-f(x), \quad x \geq 0\) \(f(-x)=3 \sin (-x)+4 \cos (-x)\) \(f(-x)=-3 \sin x+4 \cos x\left[\begin{array}{r} \because \cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta \\ \end{array}\right]\) \(f(-x)=-(3 \sin x-4 \cos x)\) So, comparing \(f(-x)=-f(x)\) \(-f(x)=-(3 \sin x-4 \cos x)\) Hence, for odd function \(f(x)\) for \(x\lt 0\) is \(3 \sin x-4 \cos\) \(\mathrm{x}\).
UPSEE-2017
Sets, Relation and Function
116913
If the real valued function \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even, then \(n\) is equal to
1 2
2 \(\frac{2}{3}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
D Given function, \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even. We know that for even function. \(f(-x)=f(x)\) \(\frac{a^{-x}-1}{(-x)^n\left(a^{-x}+1\right)}=\frac{\left(a^x-1\right)}{\left(x^n\right)\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{\left(a^x-1\right)}{x^n\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{-\left(1-a^x\right)}{x^n\left(a^x+1\right)}\) \((-1)^n=-1\) So, it satisfies \(\mathrm{n}=3\) is odd. Hence, \(\mathrm{n}=3\)
116909
\(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\), then \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) is equal to
1 1
2 2
3 3
4 \(\frac{1}{2}\)
Explanation:
A Given, \(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\) Then, \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}\) \(=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{x^2-x^2+1}\) \(==\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)\) We have, \(\mathrm{x}=\frac{1}{2} \times \frac{4}{\sqrt{3}}=\frac{2}{\sqrt{3}}\) \(\therefore \mathrm{x}^2-1=\frac{4}{3}-1=\frac{1}{3}\) \(\therefore \sqrt{\mathrm{x}^2-1}\left(\mathrm{x}+\sqrt{\mathrm{x}^2-1}\right)=\frac{1}{\sqrt{3}}\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)=\frac{1 \times 3}{\sqrt{3} \times \sqrt{3}}=1\)
AP EAMCET-2005
Sets, Relation and Function
116910
If \(F\) is function such that \(F(0)=2, F(1)=3\), \(F(x+2)=2 F(x)-F(x+1)\) for \(x \geq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 17
4 13
Explanation:
D Given, \(F(0)=2, F(1)=3\) \(F(x+2)=2 F(x)-F(x-1)\) Putting \(\mathrm{x}=0\), we get - \(F(2)=2 F(0)-F(1)\) \(F(2)=2(2)-3\) \(F(2)=4-3\) \(F(2)=1\) Putting \(\mathrm{x}=1\), in equation (i) we get - \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2(3)-1 \{\because \mathrm{F}(1)=3, \mathrm{~F}(2)=1\}\) \(\mathrm{F}(3)=5\) Putting \(\mathrm{x}=2\), in equation (i) we get - \(\mathrm{F}(4)=2 \mathrm{~F}(2)-\mathrm{F}(3)\) \(\mathrm{F}(4)=2(1)-5 \quad\{\because \mathrm{F}(2)=1, \mathrm{~F}(3)=5\}\) \(F(4)=-3\) Putting \(\mathrm{x}=3\), in equation (i) we get - \(\mathrm{F}(5)=2 \mathrm{~F}(3)-\mathrm{F}(4)\) \(\mathrm{F}(5)=2(5)+3\) \(\mathrm{~F}(5)=13\)
VITEEE-2010
Sets, Relation and Function
116912
Let \(f\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\), for \(x \geq\) 0 then \(f(x)\) for \(x\lt 0\) is
1 \(-3 \sin x+4 \cos x\)
2 \(-3 \sin x-4 \cos x\)
3 \(3 \sin x+4 \cos x\)
4 \(3 \sin x-4 \cos x\)
Explanation:
D Given, \(\mathrm{f}\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\) for \(x \geq 0\). Then, \(f(x)=3 \sin x+4 \cos x\) Since, \(f\) is odd function. Then, \(f(-x)=-f(x), \quad x \geq 0\) \(f(-x)=3 \sin (-x)+4 \cos (-x)\) \(f(-x)=-3 \sin x+4 \cos x\left[\begin{array}{r} \because \cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta \\ \end{array}\right]\) \(f(-x)=-(3 \sin x-4 \cos x)\) So, comparing \(f(-x)=-f(x)\) \(-f(x)=-(3 \sin x-4 \cos x)\) Hence, for odd function \(f(x)\) for \(x\lt 0\) is \(3 \sin x-4 \cos\) \(\mathrm{x}\).
UPSEE-2017
Sets, Relation and Function
116913
If the real valued function \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even, then \(n\) is equal to
1 2
2 \(\frac{2}{3}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
D Given function, \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even. We know that for even function. \(f(-x)=f(x)\) \(\frac{a^{-x}-1}{(-x)^n\left(a^{-x}+1\right)}=\frac{\left(a^x-1\right)}{\left(x^n\right)\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{\left(a^x-1\right)}{x^n\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{-\left(1-a^x\right)}{x^n\left(a^x+1\right)}\) \((-1)^n=-1\) So, it satisfies \(\mathrm{n}=3\) is odd. Hence, \(\mathrm{n}=3\)
116909
\(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\), then \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) is equal to
1 1
2 2
3 3
4 \(\frac{1}{2}\)
Explanation:
A Given, \(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\) Then, \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}\) \(=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{x^2-x^2+1}\) \(==\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)\) We have, \(\mathrm{x}=\frac{1}{2} \times \frac{4}{\sqrt{3}}=\frac{2}{\sqrt{3}}\) \(\therefore \mathrm{x}^2-1=\frac{4}{3}-1=\frac{1}{3}\) \(\therefore \sqrt{\mathrm{x}^2-1}\left(\mathrm{x}+\sqrt{\mathrm{x}^2-1}\right)=\frac{1}{\sqrt{3}}\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)=\frac{1 \times 3}{\sqrt{3} \times \sqrt{3}}=1\)
AP EAMCET-2005
Sets, Relation and Function
116910
If \(F\) is function such that \(F(0)=2, F(1)=3\), \(F(x+2)=2 F(x)-F(x+1)\) for \(x \geq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 17
4 13
Explanation:
D Given, \(F(0)=2, F(1)=3\) \(F(x+2)=2 F(x)-F(x-1)\) Putting \(\mathrm{x}=0\), we get - \(F(2)=2 F(0)-F(1)\) \(F(2)=2(2)-3\) \(F(2)=4-3\) \(F(2)=1\) Putting \(\mathrm{x}=1\), in equation (i) we get - \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2(3)-1 \{\because \mathrm{F}(1)=3, \mathrm{~F}(2)=1\}\) \(\mathrm{F}(3)=5\) Putting \(\mathrm{x}=2\), in equation (i) we get - \(\mathrm{F}(4)=2 \mathrm{~F}(2)-\mathrm{F}(3)\) \(\mathrm{F}(4)=2(1)-5 \quad\{\because \mathrm{F}(2)=1, \mathrm{~F}(3)=5\}\) \(F(4)=-3\) Putting \(\mathrm{x}=3\), in equation (i) we get - \(\mathrm{F}(5)=2 \mathrm{~F}(3)-\mathrm{F}(4)\) \(\mathrm{F}(5)=2(5)+3\) \(\mathrm{~F}(5)=13\)
VITEEE-2010
Sets, Relation and Function
116912
Let \(f\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\), for \(x \geq\) 0 then \(f(x)\) for \(x\lt 0\) is
1 \(-3 \sin x+4 \cos x\)
2 \(-3 \sin x-4 \cos x\)
3 \(3 \sin x+4 \cos x\)
4 \(3 \sin x-4 \cos x\)
Explanation:
D Given, \(\mathrm{f}\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\) for \(x \geq 0\). Then, \(f(x)=3 \sin x+4 \cos x\) Since, \(f\) is odd function. Then, \(f(-x)=-f(x), \quad x \geq 0\) \(f(-x)=3 \sin (-x)+4 \cos (-x)\) \(f(-x)=-3 \sin x+4 \cos x\left[\begin{array}{r} \because \cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta \\ \end{array}\right]\) \(f(-x)=-(3 \sin x-4 \cos x)\) So, comparing \(f(-x)=-f(x)\) \(-f(x)=-(3 \sin x-4 \cos x)\) Hence, for odd function \(f(x)\) for \(x\lt 0\) is \(3 \sin x-4 \cos\) \(\mathrm{x}\).
UPSEE-2017
Sets, Relation and Function
116913
If the real valued function \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even, then \(n\) is equal to
1 2
2 \(\frac{2}{3}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
D Given function, \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even. We know that for even function. \(f(-x)=f(x)\) \(\frac{a^{-x}-1}{(-x)^n\left(a^{-x}+1\right)}=\frac{\left(a^x-1\right)}{\left(x^n\right)\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{\left(a^x-1\right)}{x^n\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{-\left(1-a^x\right)}{x^n\left(a^x+1\right)}\) \((-1)^n=-1\) So, it satisfies \(\mathrm{n}=3\) is odd. Hence, \(\mathrm{n}=3\)
116909
\(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\), then \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) is equal to
1 1
2 2
3 3
4 \(\frac{1}{2}\)
Explanation:
A Given, \(x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\) Then, \(\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}\) \(=\frac{\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{x^2-x^2+1}\) \(==\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)\) We have, \(\mathrm{x}=\frac{1}{2} \times \frac{4}{\sqrt{3}}=\frac{2}{\sqrt{3}}\) \(\therefore \mathrm{x}^2-1=\frac{4}{3}-1=\frac{1}{3}\) \(\therefore \sqrt{\mathrm{x}^2-1}\left(\mathrm{x}+\sqrt{\mathrm{x}^2-1}\right)=\frac{1}{\sqrt{3}}\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)=\frac{1 \times 3}{\sqrt{3} \times \sqrt{3}}=1\)
AP EAMCET-2005
Sets, Relation and Function
116910
If \(F\) is function such that \(F(0)=2, F(1)=3\), \(F(x+2)=2 F(x)-F(x+1)\) for \(x \geq 0\), then \(F(5)\) is equal to
1 -7
2 -3
3 17
4 13
Explanation:
D Given, \(F(0)=2, F(1)=3\) \(F(x+2)=2 F(x)-F(x-1)\) Putting \(\mathrm{x}=0\), we get - \(F(2)=2 F(0)-F(1)\) \(F(2)=2(2)-3\) \(F(2)=4-3\) \(F(2)=1\) Putting \(\mathrm{x}=1\), in equation (i) we get - \(\mathrm{F}(3)=2 \mathrm{~F}(1)-\mathrm{F}(2)\) \(\mathrm{F}(3)=2(3)-1 \{\because \mathrm{F}(1)=3, \mathrm{~F}(2)=1\}\) \(\mathrm{F}(3)=5\) Putting \(\mathrm{x}=2\), in equation (i) we get - \(\mathrm{F}(4)=2 \mathrm{~F}(2)-\mathrm{F}(3)\) \(\mathrm{F}(4)=2(1)-5 \quad\{\because \mathrm{F}(2)=1, \mathrm{~F}(3)=5\}\) \(F(4)=-3\) Putting \(\mathrm{x}=3\), in equation (i) we get - \(\mathrm{F}(5)=2 \mathrm{~F}(3)-\mathrm{F}(4)\) \(\mathrm{F}(5)=2(5)+3\) \(\mathrm{~F}(5)=13\)
VITEEE-2010
Sets, Relation and Function
116912
Let \(f\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\), for \(x \geq\) 0 then \(f(x)\) for \(x\lt 0\) is
1 \(-3 \sin x+4 \cos x\)
2 \(-3 \sin x-4 \cos x\)
3 \(3 \sin x+4 \cos x\)
4 \(3 \sin x-4 \cos x\)
Explanation:
D Given, \(\mathrm{f}\) be an odd function defined on the real number such that \(f(x)=3 \sin x+4 \cos x\) for \(x \geq 0\). Then, \(f(x)=3 \sin x+4 \cos x\) Since, \(f\) is odd function. Then, \(f(-x)=-f(x), \quad x \geq 0\) \(f(-x)=3 \sin (-x)+4 \cos (-x)\) \(f(-x)=-3 \sin x+4 \cos x\left[\begin{array}{r} \because \cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta \\ \end{array}\right]\) \(f(-x)=-(3 \sin x-4 \cos x)\) So, comparing \(f(-x)=-f(x)\) \(-f(x)=-(3 \sin x-4 \cos x)\) Hence, for odd function \(f(x)\) for \(x\lt 0\) is \(3 \sin x-4 \cos\) \(\mathrm{x}\).
UPSEE-2017
Sets, Relation and Function
116913
If the real valued function \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even, then \(n\) is equal to
1 2
2 \(\frac{2}{3}\)
3 \(\frac{1}{4}\)
4 3
Explanation:
D Given function, \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\) is even. We know that for even function. \(f(-x)=f(x)\) \(\frac{a^{-x}-1}{(-x)^n\left(a^{-x}+1\right)}=\frac{\left(a^x-1\right)}{\left(x^n\right)\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{\left(a^x-1\right)}{x^n\left(a^x+1\right)}\) \(\frac{1-a^x}{(-1)^n x^n\left(a^x+1\right)}=\frac{-\left(1-a^x\right)}{x^n\left(a^x+1\right)}\) \((-1)^n=-1\) So, it satisfies \(\mathrm{n}=3\) is odd. Hence, \(\mathrm{n}=3\)
