117396
If \(\left|\frac{x^2+k x+1}{x^2+x+1}\right|\lt 3\) for all real number \(x\), then the range of the parameter \(k\) is
1 \((0,4)\)
2 \((-1,5)\)
3 \((-4,0)\)
4 \((-5,1)\)
Explanation:
B We have, \(\left|\frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\right|\lt 3\) \(-3\lt \frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\lt 3\) Case \(\mathrm{I}:-\mathrm{x}^2+\mathrm{kx}+1\lt 3 \mathrm{x}^2+3 \mathrm{x}+3\) \(2 x^2+x(3-k)+2>0\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((3-\mathrm{k})^2-4 \times 2 \times 2\lt 0\) \((3-\mathrm{k})^2-16\lt 0\) \((3-\mathrm{k})^2\lt 16\) \((3-\mathrm{k})\lt \pm 4\) \(-4\lt (3-\mathrm{k})\lt 4\) So, \(\mathrm{k}>-1, \mathrm{k}\lt 7\) \(\mathrm{k} \in(-1,7)\) Case II :- \(x^2+k x+1>-3 x^2-3 x-3\) \(\left.4 x^2+x(k+3)+4>0\right]\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((\mathrm{k}+3)^2-4 \times 4 \times 4\lt 0\) \((\mathrm{k}+3)^2-64\lt 0\) \((\mathrm{k}+3)^2\lt 64\) \(\mathrm{k}+3\lt \pm 8\) \(-8\lt \mathrm{k}+3\lt 8\) \(-11\lt \mathrm{k}\lt 5\) \(\mathrm{k} \in(-11,5)\) From equation (i) and (ii), we get- Hence, \(\mathrm{k} \in(-1,5)\).
AP EAMCET-23.04.2019
Sets, Relation and Function
117397
The domain of the function \(f(x)=\frac{\sqrt{9-x^2}}{\sin ^{-1}(3-x)}\) is
1 \((2,3)\)
2 \([2,3)\)
3 \((2,3]\)
4 None of these
Explanation:
B Given, \(\sqrt{9-x^2}\) is defined for - \(9-x^2 \geq 0 \Rightarrow(3-x)(3+x) \geq 0\) \(\Rightarrow \quad (x-3)(x+3) \leq 0\) \(\Rightarrow \quad -3 \leq x \leq 3\) \(\sin ^{-1}(3-\mathrm{x})\) defined for - \(-1 \leq 3-x \leq 1\) \(\Rightarrow \quad 2 \leq \mathrm{x} \leq 4\) Also, \(\sin ^{-1}(3-x) \neq 0\) \(\Rightarrow \quad 3-\mathrm{x} \neq 0 \text { or } \mathrm{x} \neq 3\) From equation (i), (ii) and (iii), we get The domain of \(f(x)\) is - \(([-3,3] \cap[2,4])-\{3\}=[2,3) .\)
Manipal UGET-2010
Sets, Relation and Function
117398
The range of the function \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) is
B Given that, Let, \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) It is open upward parabola then range, \(R \in\left(\frac{-D}{4 a}, \infty\right)\) \(R \in\left(-\frac{(-4)^2-4 \times 3 \times 5}{4 \times 3}, \infty\right) R \in\left(\frac{11}{3}, \infty\right)\) Hence, \(f(x)=\log _e[g(x)]\) Range of \(f(x)=\left[\log _e\left(\frac{11}{3}\right), \infty\right]\)
Manipal UGET-2010
Sets, Relation and Function
117399
The period of the function \(f(x)=\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) is
1 \(\pi\)
2 \(2 \pi\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given that, \(f(x)=\frac{\sin 8 x \times \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) \(f(x)=\frac{\frac{1}{2}[\sin (8 x+x)+\sin (8 x-x)]-\frac{1}{2}[\sin (6 x+3 x)+\sin (6 x-3 x)]}{\frac{1}{2}[\cos (2 x+x)+\cos (2 x-x)]+\frac{1}{2}[\cos (3 x+4 x)-\cos (4 x-3 x)]}\) \(f(x)=\frac{(\sin 9 x+\sin 7 x)-(\sin 9 x+\sin 3 x)}{(\cos 3 x+\cos x)+(\cos 7 x-\cos x)}\) \(f(x)=\frac{\sin 9 x+\sin 7 x-\sin 9 x-\sin 3 x}{\cos 3 x+\cos x+\cos 7 x-\cos x}\) \(f(x)=\frac{\sin 7 x-\sin 3 x}{\cos 3 x+\cos 7 x}\) \(f(x)=\frac{2 \cos \left(\frac{7 x+3 x}{2}\right) \sin \left(\frac{7 x-3 x}{2}\right)}{2 \cos \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)}\) \(f(x)=\frac{\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x}\) \(f(x)=\frac{\sin 2 x}{\cos 2 x}=\tan 2 x\) We know that, the general formula for the period tan \((\alpha x)\) is \(\frac{\pi}{\alpha}\) Hence, the period of \(\tan (2 x)\) is \(\frac{\pi}{2}\).
117396
If \(\left|\frac{x^2+k x+1}{x^2+x+1}\right|\lt 3\) for all real number \(x\), then the range of the parameter \(k\) is
1 \((0,4)\)
2 \((-1,5)\)
3 \((-4,0)\)
4 \((-5,1)\)
Explanation:
B We have, \(\left|\frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\right|\lt 3\) \(-3\lt \frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\lt 3\) Case \(\mathrm{I}:-\mathrm{x}^2+\mathrm{kx}+1\lt 3 \mathrm{x}^2+3 \mathrm{x}+3\) \(2 x^2+x(3-k)+2>0\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((3-\mathrm{k})^2-4 \times 2 \times 2\lt 0\) \((3-\mathrm{k})^2-16\lt 0\) \((3-\mathrm{k})^2\lt 16\) \((3-\mathrm{k})\lt \pm 4\) \(-4\lt (3-\mathrm{k})\lt 4\) So, \(\mathrm{k}>-1, \mathrm{k}\lt 7\) \(\mathrm{k} \in(-1,7)\) Case II :- \(x^2+k x+1>-3 x^2-3 x-3\) \(\left.4 x^2+x(k+3)+4>0\right]\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((\mathrm{k}+3)^2-4 \times 4 \times 4\lt 0\) \((\mathrm{k}+3)^2-64\lt 0\) \((\mathrm{k}+3)^2\lt 64\) \(\mathrm{k}+3\lt \pm 8\) \(-8\lt \mathrm{k}+3\lt 8\) \(-11\lt \mathrm{k}\lt 5\) \(\mathrm{k} \in(-11,5)\) From equation (i) and (ii), we get- Hence, \(\mathrm{k} \in(-1,5)\).
AP EAMCET-23.04.2019
Sets, Relation and Function
117397
The domain of the function \(f(x)=\frac{\sqrt{9-x^2}}{\sin ^{-1}(3-x)}\) is
1 \((2,3)\)
2 \([2,3)\)
3 \((2,3]\)
4 None of these
Explanation:
B Given, \(\sqrt{9-x^2}\) is defined for - \(9-x^2 \geq 0 \Rightarrow(3-x)(3+x) \geq 0\) \(\Rightarrow \quad (x-3)(x+3) \leq 0\) \(\Rightarrow \quad -3 \leq x \leq 3\) \(\sin ^{-1}(3-\mathrm{x})\) defined for - \(-1 \leq 3-x \leq 1\) \(\Rightarrow \quad 2 \leq \mathrm{x} \leq 4\) Also, \(\sin ^{-1}(3-x) \neq 0\) \(\Rightarrow \quad 3-\mathrm{x} \neq 0 \text { or } \mathrm{x} \neq 3\) From equation (i), (ii) and (iii), we get The domain of \(f(x)\) is - \(([-3,3] \cap[2,4])-\{3\}=[2,3) .\)
Manipal UGET-2010
Sets, Relation and Function
117398
The range of the function \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) is
B Given that, Let, \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) It is open upward parabola then range, \(R \in\left(\frac{-D}{4 a}, \infty\right)\) \(R \in\left(-\frac{(-4)^2-4 \times 3 \times 5}{4 \times 3}, \infty\right) R \in\left(\frac{11}{3}, \infty\right)\) Hence, \(f(x)=\log _e[g(x)]\) Range of \(f(x)=\left[\log _e\left(\frac{11}{3}\right), \infty\right]\)
Manipal UGET-2010
Sets, Relation and Function
117399
The period of the function \(f(x)=\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) is
1 \(\pi\)
2 \(2 \pi\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given that, \(f(x)=\frac{\sin 8 x \times \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) \(f(x)=\frac{\frac{1}{2}[\sin (8 x+x)+\sin (8 x-x)]-\frac{1}{2}[\sin (6 x+3 x)+\sin (6 x-3 x)]}{\frac{1}{2}[\cos (2 x+x)+\cos (2 x-x)]+\frac{1}{2}[\cos (3 x+4 x)-\cos (4 x-3 x)]}\) \(f(x)=\frac{(\sin 9 x+\sin 7 x)-(\sin 9 x+\sin 3 x)}{(\cos 3 x+\cos x)+(\cos 7 x-\cos x)}\) \(f(x)=\frac{\sin 9 x+\sin 7 x-\sin 9 x-\sin 3 x}{\cos 3 x+\cos x+\cos 7 x-\cos x}\) \(f(x)=\frac{\sin 7 x-\sin 3 x}{\cos 3 x+\cos 7 x}\) \(f(x)=\frac{2 \cos \left(\frac{7 x+3 x}{2}\right) \sin \left(\frac{7 x-3 x}{2}\right)}{2 \cos \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)}\) \(f(x)=\frac{\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x}\) \(f(x)=\frac{\sin 2 x}{\cos 2 x}=\tan 2 x\) We know that, the general formula for the period tan \((\alpha x)\) is \(\frac{\pi}{\alpha}\) Hence, the period of \(\tan (2 x)\) is \(\frac{\pi}{2}\).
117396
If \(\left|\frac{x^2+k x+1}{x^2+x+1}\right|\lt 3\) for all real number \(x\), then the range of the parameter \(k\) is
1 \((0,4)\)
2 \((-1,5)\)
3 \((-4,0)\)
4 \((-5,1)\)
Explanation:
B We have, \(\left|\frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\right|\lt 3\) \(-3\lt \frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\lt 3\) Case \(\mathrm{I}:-\mathrm{x}^2+\mathrm{kx}+1\lt 3 \mathrm{x}^2+3 \mathrm{x}+3\) \(2 x^2+x(3-k)+2>0\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((3-\mathrm{k})^2-4 \times 2 \times 2\lt 0\) \((3-\mathrm{k})^2-16\lt 0\) \((3-\mathrm{k})^2\lt 16\) \((3-\mathrm{k})\lt \pm 4\) \(-4\lt (3-\mathrm{k})\lt 4\) So, \(\mathrm{k}>-1, \mathrm{k}\lt 7\) \(\mathrm{k} \in(-1,7)\) Case II :- \(x^2+k x+1>-3 x^2-3 x-3\) \(\left.4 x^2+x(k+3)+4>0\right]\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((\mathrm{k}+3)^2-4 \times 4 \times 4\lt 0\) \((\mathrm{k}+3)^2-64\lt 0\) \((\mathrm{k}+3)^2\lt 64\) \(\mathrm{k}+3\lt \pm 8\) \(-8\lt \mathrm{k}+3\lt 8\) \(-11\lt \mathrm{k}\lt 5\) \(\mathrm{k} \in(-11,5)\) From equation (i) and (ii), we get- Hence, \(\mathrm{k} \in(-1,5)\).
AP EAMCET-23.04.2019
Sets, Relation and Function
117397
The domain of the function \(f(x)=\frac{\sqrt{9-x^2}}{\sin ^{-1}(3-x)}\) is
1 \((2,3)\)
2 \([2,3)\)
3 \((2,3]\)
4 None of these
Explanation:
B Given, \(\sqrt{9-x^2}\) is defined for - \(9-x^2 \geq 0 \Rightarrow(3-x)(3+x) \geq 0\) \(\Rightarrow \quad (x-3)(x+3) \leq 0\) \(\Rightarrow \quad -3 \leq x \leq 3\) \(\sin ^{-1}(3-\mathrm{x})\) defined for - \(-1 \leq 3-x \leq 1\) \(\Rightarrow \quad 2 \leq \mathrm{x} \leq 4\) Also, \(\sin ^{-1}(3-x) \neq 0\) \(\Rightarrow \quad 3-\mathrm{x} \neq 0 \text { or } \mathrm{x} \neq 3\) From equation (i), (ii) and (iii), we get The domain of \(f(x)\) is - \(([-3,3] \cap[2,4])-\{3\}=[2,3) .\)
Manipal UGET-2010
Sets, Relation and Function
117398
The range of the function \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) is
B Given that, Let, \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) It is open upward parabola then range, \(R \in\left(\frac{-D}{4 a}, \infty\right)\) \(R \in\left(-\frac{(-4)^2-4 \times 3 \times 5}{4 \times 3}, \infty\right) R \in\left(\frac{11}{3}, \infty\right)\) Hence, \(f(x)=\log _e[g(x)]\) Range of \(f(x)=\left[\log _e\left(\frac{11}{3}\right), \infty\right]\)
Manipal UGET-2010
Sets, Relation and Function
117399
The period of the function \(f(x)=\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) is
1 \(\pi\)
2 \(2 \pi\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given that, \(f(x)=\frac{\sin 8 x \times \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) \(f(x)=\frac{\frac{1}{2}[\sin (8 x+x)+\sin (8 x-x)]-\frac{1}{2}[\sin (6 x+3 x)+\sin (6 x-3 x)]}{\frac{1}{2}[\cos (2 x+x)+\cos (2 x-x)]+\frac{1}{2}[\cos (3 x+4 x)-\cos (4 x-3 x)]}\) \(f(x)=\frac{(\sin 9 x+\sin 7 x)-(\sin 9 x+\sin 3 x)}{(\cos 3 x+\cos x)+(\cos 7 x-\cos x)}\) \(f(x)=\frac{\sin 9 x+\sin 7 x-\sin 9 x-\sin 3 x}{\cos 3 x+\cos x+\cos 7 x-\cos x}\) \(f(x)=\frac{\sin 7 x-\sin 3 x}{\cos 3 x+\cos 7 x}\) \(f(x)=\frac{2 \cos \left(\frac{7 x+3 x}{2}\right) \sin \left(\frac{7 x-3 x}{2}\right)}{2 \cos \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)}\) \(f(x)=\frac{\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x}\) \(f(x)=\frac{\sin 2 x}{\cos 2 x}=\tan 2 x\) We know that, the general formula for the period tan \((\alpha x)\) is \(\frac{\pi}{\alpha}\) Hence, the period of \(\tan (2 x)\) is \(\frac{\pi}{2}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117396
If \(\left|\frac{x^2+k x+1}{x^2+x+1}\right|\lt 3\) for all real number \(x\), then the range of the parameter \(k\) is
1 \((0,4)\)
2 \((-1,5)\)
3 \((-4,0)\)
4 \((-5,1)\)
Explanation:
B We have, \(\left|\frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\right|\lt 3\) \(-3\lt \frac{\mathrm{x}^2+\mathrm{kx}+1}{\mathrm{x}^2+\mathrm{x}+1}\lt 3\) Case \(\mathrm{I}:-\mathrm{x}^2+\mathrm{kx}+1\lt 3 \mathrm{x}^2+3 \mathrm{x}+3\) \(2 x^2+x(3-k)+2>0\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((3-\mathrm{k})^2-4 \times 2 \times 2\lt 0\) \((3-\mathrm{k})^2-16\lt 0\) \((3-\mathrm{k})^2\lt 16\) \((3-\mathrm{k})\lt \pm 4\) \(-4\lt (3-\mathrm{k})\lt 4\) So, \(\mathrm{k}>-1, \mathrm{k}\lt 7\) \(\mathrm{k} \in(-1,7)\) Case II :- \(x^2+k x+1>-3 x^2-3 x-3\) \(\left.4 x^2+x(k+3)+4>0\right]\) Therefore, \(\mathrm{D}\lt 0\) \(\mathrm{b}^2-4 \mathrm{ac}\lt 0\) \((\mathrm{k}+3)^2-4 \times 4 \times 4\lt 0\) \((\mathrm{k}+3)^2-64\lt 0\) \((\mathrm{k}+3)^2\lt 64\) \(\mathrm{k}+3\lt \pm 8\) \(-8\lt \mathrm{k}+3\lt 8\) \(-11\lt \mathrm{k}\lt 5\) \(\mathrm{k} \in(-11,5)\) From equation (i) and (ii), we get- Hence, \(\mathrm{k} \in(-1,5)\).
AP EAMCET-23.04.2019
Sets, Relation and Function
117397
The domain of the function \(f(x)=\frac{\sqrt{9-x^2}}{\sin ^{-1}(3-x)}\) is
1 \((2,3)\)
2 \([2,3)\)
3 \((2,3]\)
4 None of these
Explanation:
B Given, \(\sqrt{9-x^2}\) is defined for - \(9-x^2 \geq 0 \Rightarrow(3-x)(3+x) \geq 0\) \(\Rightarrow \quad (x-3)(x+3) \leq 0\) \(\Rightarrow \quad -3 \leq x \leq 3\) \(\sin ^{-1}(3-\mathrm{x})\) defined for - \(-1 \leq 3-x \leq 1\) \(\Rightarrow \quad 2 \leq \mathrm{x} \leq 4\) Also, \(\sin ^{-1}(3-x) \neq 0\) \(\Rightarrow \quad 3-\mathrm{x} \neq 0 \text { or } \mathrm{x} \neq 3\) From equation (i), (ii) and (iii), we get The domain of \(f(x)\) is - \(([-3,3] \cap[2,4])-\{3\}=[2,3) .\)
Manipal UGET-2010
Sets, Relation and Function
117398
The range of the function \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) is
B Given that, Let, \(f(x)=\log _e\left(3 x^2-4 x+5\right)\) It is open upward parabola then range, \(R \in\left(\frac{-D}{4 a}, \infty\right)\) \(R \in\left(-\frac{(-4)^2-4 \times 3 \times 5}{4 \times 3}, \infty\right) R \in\left(\frac{11}{3}, \infty\right)\) Hence, \(f(x)=\log _e[g(x)]\) Range of \(f(x)=\left[\log _e\left(\frac{11}{3}\right), \infty\right]\)
Manipal UGET-2010
Sets, Relation and Function
117399
The period of the function \(f(x)=\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) is
1 \(\pi\)
2 \(2 \pi\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given that, \(f(x)=\frac{\sin 8 x \times \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) \(f(x)=\frac{\frac{1}{2}[\sin (8 x+x)+\sin (8 x-x)]-\frac{1}{2}[\sin (6 x+3 x)+\sin (6 x-3 x)]}{\frac{1}{2}[\cos (2 x+x)+\cos (2 x-x)]+\frac{1}{2}[\cos (3 x+4 x)-\cos (4 x-3 x)]}\) \(f(x)=\frac{(\sin 9 x+\sin 7 x)-(\sin 9 x+\sin 3 x)}{(\cos 3 x+\cos x)+(\cos 7 x-\cos x)}\) \(f(x)=\frac{\sin 9 x+\sin 7 x-\sin 9 x-\sin 3 x}{\cos 3 x+\cos x+\cos 7 x-\cos x}\) \(f(x)=\frac{\sin 7 x-\sin 3 x}{\cos 3 x+\cos 7 x}\) \(f(x)=\frac{2 \cos \left(\frac{7 x+3 x}{2}\right) \sin \left(\frac{7 x-3 x}{2}\right)}{2 \cos \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)}\) \(f(x)=\frac{\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x}\) \(f(x)=\frac{\sin 2 x}{\cos 2 x}=\tan 2 x\) We know that, the general formula for the period tan \((\alpha x)\) is \(\frac{\pi}{\alpha}\) Hence, the period of \(\tan (2 x)\) is \(\frac{\pi}{2}\).
