118096
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\) in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given equation - \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\text { Now, }\) \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(3 \sin ^2 \mathrm{x}-6 \sin \mathrm{x}-\sin \mathrm{x}+2=0\) \(3 \sin \mathrm{x}(\sin \mathrm{x}-2)-1(\sin \mathrm{x}-2)=0\) \((3 \sin \mathrm{x}-1)(\sin \mathrm{x}-2)=0\) \(\therefore \sin \mathrm{x}=-2(\text { not possible because) }(-1 \leq \sin \mathrm{x} \leq 1)\) \(\therefore 3 \sin \mathrm{x}-1=0\) \(\sin \mathrm{x}=\frac{1}{3}(\text { posible })\) \(\sin \mathrm{x}=\frac{1}{3} \text { in interval }(0,5 \pi)\) So, there are 6 solutions in intervals \((0,5 \pi)\)
BCECE-2015
Complex Numbers and Quadratic Equation
118097
The value of ' \(a\) ' for which \(a^2+\sin ^{-1}\left(x^2-2 x+2\right)\) \(+\cos ^{-1}\left(x^2-2 x+2\right)=0\) has a real solution, is
1 \(-\frac{2}{\pi}\)
2 \(\frac{2}{\pi}\)
3 \(-\frac{\pi}{2}\)
4 \(\frac{\pi}{2}\)
Explanation:
C Given equation is \(a x^2+\sin ^{-1}\left(x^2-2 x+2\right)+\cos ^{-1}\left(x^2-2 x+2\right)=0\) we know that \(\sin ^{-1}(x)+\cos ^{-1}(x)=\frac{\pi}{2}=0\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) Now let us check possible value of \(x^2\) \(\because-1 \leq \mathrm{x}^2-2 \mathrm{x}+2 \leq 1\) \(\therefore-1 \leq(\mathrm{x}-1)^2+\leq 1\) \(\therefore\) only possible value if \((\mathrm{x}-1)^2=0\) \(\mathrm{x}=1\) Put the value of \(x=1\) in equation (i) \(\mathrm{a}=\frac{-\pi}{2} \times \frac{1}{1^2}=\frac{-\pi}{2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118098
\(\frac{\mathrm{d}}{\mathrm{dx}} \operatorname{cosec}^{-1}\left(\frac{1+\mathrm{x}^2}{2 \mathrm{x}}\right)\) is equal to
1 \(\frac{-2}{\left(1+x^2\right)}, x \neq 0\)
2 \(\frac{2}{\left(1+x^2\right)}, x \neq 0\)
3 \(\frac{2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1,0\)
4 None of the above
Explanation:
C \( We know that-\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\left(\frac{1+x^2}{2 x}\right)^2-1}} \cdot \frac{d}{d x}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\frac{1+x^2+2 x^2-4 x^2}{4 x^2}} \cdot \frac{2 x(2 x)-\left(1+x^2\right) 2}{(2 x)^2}}\) \(=\frac{-1}{\frac{1+x^2}{2 x} \cdot \frac{\sqrt{1+x^4-2 x^2}}{2 x}} \cdot \frac{4 x^2-2-2 x^2}{4 x^2}\) \(=\frac{-4 x^2}{\left(1+x^2\right) \sqrt{\left(1-x^2\right)^2}} \cdot \frac{2 x^2-2}{4 x^2}\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)=\frac{+2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1\) For the next three (03) questions that follow: Consider the polynomial \(\mathbf{p}(\mathbf{x})=(\mathbf{x}-\boldsymbol{\alpha})(\mathbf{x}-\boldsymbol{\beta})(\mathbf{x}-\gamma)\) with \(\boldsymbol{\alpha}=\cos \mathbf{7 5}, \boldsymbol{\beta}=\cos 45^{\circ}\) and \(\gamma=\cos 165^{\circ}\)
BCECE-2011
Complex Numbers and Quadratic Equation
118100
If \(P\) is chosen at random in the closed interval \(\left[\begin{array}{ll}0,5\end{array}\right]\), the probability that the equation \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) will have real roots is equal to
1 \(1 / 2\)
2 \(1 / 5\)
3 \(2 / 3\)
4 \(3 / 5\)
Explanation:
D The quadratic equations \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) has real roots, if- \(D \geq 0\) \(\Rightarrow b^2-4 a c \geq 0\) \(\Rightarrow p^2-4\left(\frac{p+2}{4}\right) \geq 0\) \(\Rightarrow p^2-p-2 \geq 0\) \(\Rightarrow p^2-2 p+p-2 \geq 0\) \(\Rightarrow (p-2)(p+1) \geq 0\) Range of \(p \in(-\infty,-1] \cup[2, \infty)\) As \(\mathrm{p} \in[0,5]\), the required probability \(=\frac{3}{5}\)
SCRA-2012
Complex Numbers and Quadratic Equation
118101
The number of solutions of the equation \(\mathbf{x}^2+3|x|+2=0\) is
1 0
2 1
3 2
4 4
Explanation:
D Given, the equation is, \(x^2+3|x|+2=0\) Case- \(\mathrm{I}:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2+3 \mathrm{x}+2=0\) \(\Rightarrow \mathrm{x}^2+2 \mathrm{x}+\mathrm{x}+2=0\) \((x+2)(x+1)=0 \Rightarrow x=-2,-1\) Case II:- \(x\lt 0 \Rightarrow|x|=-x\) \(x^2-3 x+2=0\) \(x^2-2 x-x+2=0\) \(\Rightarrow(x-2)(x-1)=0 \Rightarrow x=1,2\)\(\therefore\) The number of solutions is 4 .
118096
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\) in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given equation - \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\text { Now, }\) \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(3 \sin ^2 \mathrm{x}-6 \sin \mathrm{x}-\sin \mathrm{x}+2=0\) \(3 \sin \mathrm{x}(\sin \mathrm{x}-2)-1(\sin \mathrm{x}-2)=0\) \((3 \sin \mathrm{x}-1)(\sin \mathrm{x}-2)=0\) \(\therefore \sin \mathrm{x}=-2(\text { not possible because) }(-1 \leq \sin \mathrm{x} \leq 1)\) \(\therefore 3 \sin \mathrm{x}-1=0\) \(\sin \mathrm{x}=\frac{1}{3}(\text { posible })\) \(\sin \mathrm{x}=\frac{1}{3} \text { in interval }(0,5 \pi)\) So, there are 6 solutions in intervals \((0,5 \pi)\)
BCECE-2015
Complex Numbers and Quadratic Equation
118097
The value of ' \(a\) ' for which \(a^2+\sin ^{-1}\left(x^2-2 x+2\right)\) \(+\cos ^{-1}\left(x^2-2 x+2\right)=0\) has a real solution, is
1 \(-\frac{2}{\pi}\)
2 \(\frac{2}{\pi}\)
3 \(-\frac{\pi}{2}\)
4 \(\frac{\pi}{2}\)
Explanation:
C Given equation is \(a x^2+\sin ^{-1}\left(x^2-2 x+2\right)+\cos ^{-1}\left(x^2-2 x+2\right)=0\) we know that \(\sin ^{-1}(x)+\cos ^{-1}(x)=\frac{\pi}{2}=0\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) Now let us check possible value of \(x^2\) \(\because-1 \leq \mathrm{x}^2-2 \mathrm{x}+2 \leq 1\) \(\therefore-1 \leq(\mathrm{x}-1)^2+\leq 1\) \(\therefore\) only possible value if \((\mathrm{x}-1)^2=0\) \(\mathrm{x}=1\) Put the value of \(x=1\) in equation (i) \(\mathrm{a}=\frac{-\pi}{2} \times \frac{1}{1^2}=\frac{-\pi}{2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118098
\(\frac{\mathrm{d}}{\mathrm{dx}} \operatorname{cosec}^{-1}\left(\frac{1+\mathrm{x}^2}{2 \mathrm{x}}\right)\) is equal to
1 \(\frac{-2}{\left(1+x^2\right)}, x \neq 0\)
2 \(\frac{2}{\left(1+x^2\right)}, x \neq 0\)
3 \(\frac{2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1,0\)
4 None of the above
Explanation:
C \( We know that-\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\left(\frac{1+x^2}{2 x}\right)^2-1}} \cdot \frac{d}{d x}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\frac{1+x^2+2 x^2-4 x^2}{4 x^2}} \cdot \frac{2 x(2 x)-\left(1+x^2\right) 2}{(2 x)^2}}\) \(=\frac{-1}{\frac{1+x^2}{2 x} \cdot \frac{\sqrt{1+x^4-2 x^2}}{2 x}} \cdot \frac{4 x^2-2-2 x^2}{4 x^2}\) \(=\frac{-4 x^2}{\left(1+x^2\right) \sqrt{\left(1-x^2\right)^2}} \cdot \frac{2 x^2-2}{4 x^2}\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)=\frac{+2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1\) For the next three (03) questions that follow: Consider the polynomial \(\mathbf{p}(\mathbf{x})=(\mathbf{x}-\boldsymbol{\alpha})(\mathbf{x}-\boldsymbol{\beta})(\mathbf{x}-\gamma)\) with \(\boldsymbol{\alpha}=\cos \mathbf{7 5}, \boldsymbol{\beta}=\cos 45^{\circ}\) and \(\gamma=\cos 165^{\circ}\)
BCECE-2011
Complex Numbers and Quadratic Equation
118100
If \(P\) is chosen at random in the closed interval \(\left[\begin{array}{ll}0,5\end{array}\right]\), the probability that the equation \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) will have real roots is equal to
1 \(1 / 2\)
2 \(1 / 5\)
3 \(2 / 3\)
4 \(3 / 5\)
Explanation:
D The quadratic equations \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) has real roots, if- \(D \geq 0\) \(\Rightarrow b^2-4 a c \geq 0\) \(\Rightarrow p^2-4\left(\frac{p+2}{4}\right) \geq 0\) \(\Rightarrow p^2-p-2 \geq 0\) \(\Rightarrow p^2-2 p+p-2 \geq 0\) \(\Rightarrow (p-2)(p+1) \geq 0\) Range of \(p \in(-\infty,-1] \cup[2, \infty)\) As \(\mathrm{p} \in[0,5]\), the required probability \(=\frac{3}{5}\)
SCRA-2012
Complex Numbers and Quadratic Equation
118101
The number of solutions of the equation \(\mathbf{x}^2+3|x|+2=0\) is
1 0
2 1
3 2
4 4
Explanation:
D Given, the equation is, \(x^2+3|x|+2=0\) Case- \(\mathrm{I}:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2+3 \mathrm{x}+2=0\) \(\Rightarrow \mathrm{x}^2+2 \mathrm{x}+\mathrm{x}+2=0\) \((x+2)(x+1)=0 \Rightarrow x=-2,-1\) Case II:- \(x\lt 0 \Rightarrow|x|=-x\) \(x^2-3 x+2=0\) \(x^2-2 x-x+2=0\) \(\Rightarrow(x-2)(x-1)=0 \Rightarrow x=1,2\)\(\therefore\) The number of solutions is 4 .
NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
118096
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\) in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given equation - \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\text { Now, }\) \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(3 \sin ^2 \mathrm{x}-6 \sin \mathrm{x}-\sin \mathrm{x}+2=0\) \(3 \sin \mathrm{x}(\sin \mathrm{x}-2)-1(\sin \mathrm{x}-2)=0\) \((3 \sin \mathrm{x}-1)(\sin \mathrm{x}-2)=0\) \(\therefore \sin \mathrm{x}=-2(\text { not possible because) }(-1 \leq \sin \mathrm{x} \leq 1)\) \(\therefore 3 \sin \mathrm{x}-1=0\) \(\sin \mathrm{x}=\frac{1}{3}(\text { posible })\) \(\sin \mathrm{x}=\frac{1}{3} \text { in interval }(0,5 \pi)\) So, there are 6 solutions in intervals \((0,5 \pi)\)
BCECE-2015
Complex Numbers and Quadratic Equation
118097
The value of ' \(a\) ' for which \(a^2+\sin ^{-1}\left(x^2-2 x+2\right)\) \(+\cos ^{-1}\left(x^2-2 x+2\right)=0\) has a real solution, is
1 \(-\frac{2}{\pi}\)
2 \(\frac{2}{\pi}\)
3 \(-\frac{\pi}{2}\)
4 \(\frac{\pi}{2}\)
Explanation:
C Given equation is \(a x^2+\sin ^{-1}\left(x^2-2 x+2\right)+\cos ^{-1}\left(x^2-2 x+2\right)=0\) we know that \(\sin ^{-1}(x)+\cos ^{-1}(x)=\frac{\pi}{2}=0\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) Now let us check possible value of \(x^2\) \(\because-1 \leq \mathrm{x}^2-2 \mathrm{x}+2 \leq 1\) \(\therefore-1 \leq(\mathrm{x}-1)^2+\leq 1\) \(\therefore\) only possible value if \((\mathrm{x}-1)^2=0\) \(\mathrm{x}=1\) Put the value of \(x=1\) in equation (i) \(\mathrm{a}=\frac{-\pi}{2} \times \frac{1}{1^2}=\frac{-\pi}{2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118098
\(\frac{\mathrm{d}}{\mathrm{dx}} \operatorname{cosec}^{-1}\left(\frac{1+\mathrm{x}^2}{2 \mathrm{x}}\right)\) is equal to
1 \(\frac{-2}{\left(1+x^2\right)}, x \neq 0\)
2 \(\frac{2}{\left(1+x^2\right)}, x \neq 0\)
3 \(\frac{2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1,0\)
4 None of the above
Explanation:
C \( We know that-\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\left(\frac{1+x^2}{2 x}\right)^2-1}} \cdot \frac{d}{d x}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\frac{1+x^2+2 x^2-4 x^2}{4 x^2}} \cdot \frac{2 x(2 x)-\left(1+x^2\right) 2}{(2 x)^2}}\) \(=\frac{-1}{\frac{1+x^2}{2 x} \cdot \frac{\sqrt{1+x^4-2 x^2}}{2 x}} \cdot \frac{4 x^2-2-2 x^2}{4 x^2}\) \(=\frac{-4 x^2}{\left(1+x^2\right) \sqrt{\left(1-x^2\right)^2}} \cdot \frac{2 x^2-2}{4 x^2}\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)=\frac{+2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1\) For the next three (03) questions that follow: Consider the polynomial \(\mathbf{p}(\mathbf{x})=(\mathbf{x}-\boldsymbol{\alpha})(\mathbf{x}-\boldsymbol{\beta})(\mathbf{x}-\gamma)\) with \(\boldsymbol{\alpha}=\cos \mathbf{7 5}, \boldsymbol{\beta}=\cos 45^{\circ}\) and \(\gamma=\cos 165^{\circ}\)
BCECE-2011
Complex Numbers and Quadratic Equation
118100
If \(P\) is chosen at random in the closed interval \(\left[\begin{array}{ll}0,5\end{array}\right]\), the probability that the equation \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) will have real roots is equal to
1 \(1 / 2\)
2 \(1 / 5\)
3 \(2 / 3\)
4 \(3 / 5\)
Explanation:
D The quadratic equations \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) has real roots, if- \(D \geq 0\) \(\Rightarrow b^2-4 a c \geq 0\) \(\Rightarrow p^2-4\left(\frac{p+2}{4}\right) \geq 0\) \(\Rightarrow p^2-p-2 \geq 0\) \(\Rightarrow p^2-2 p+p-2 \geq 0\) \(\Rightarrow (p-2)(p+1) \geq 0\) Range of \(p \in(-\infty,-1] \cup[2, \infty)\) As \(\mathrm{p} \in[0,5]\), the required probability \(=\frac{3}{5}\)
SCRA-2012
Complex Numbers and Quadratic Equation
118101
The number of solutions of the equation \(\mathbf{x}^2+3|x|+2=0\) is
1 0
2 1
3 2
4 4
Explanation:
D Given, the equation is, \(x^2+3|x|+2=0\) Case- \(\mathrm{I}:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2+3 \mathrm{x}+2=0\) \(\Rightarrow \mathrm{x}^2+2 \mathrm{x}+\mathrm{x}+2=0\) \((x+2)(x+1)=0 \Rightarrow x=-2,-1\) Case II:- \(x\lt 0 \Rightarrow|x|=-x\) \(x^2-3 x+2=0\) \(x^2-2 x-x+2=0\) \(\Rightarrow(x-2)(x-1)=0 \Rightarrow x=1,2\)\(\therefore\) The number of solutions is 4 .
118096
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\) in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given equation - \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\text { Now, }\) \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(3 \sin ^2 \mathrm{x}-6 \sin \mathrm{x}-\sin \mathrm{x}+2=0\) \(3 \sin \mathrm{x}(\sin \mathrm{x}-2)-1(\sin \mathrm{x}-2)=0\) \((3 \sin \mathrm{x}-1)(\sin \mathrm{x}-2)=0\) \(\therefore \sin \mathrm{x}=-2(\text { not possible because) }(-1 \leq \sin \mathrm{x} \leq 1)\) \(\therefore 3 \sin \mathrm{x}-1=0\) \(\sin \mathrm{x}=\frac{1}{3}(\text { posible })\) \(\sin \mathrm{x}=\frac{1}{3} \text { in interval }(0,5 \pi)\) So, there are 6 solutions in intervals \((0,5 \pi)\)
BCECE-2015
Complex Numbers and Quadratic Equation
118097
The value of ' \(a\) ' for which \(a^2+\sin ^{-1}\left(x^2-2 x+2\right)\) \(+\cos ^{-1}\left(x^2-2 x+2\right)=0\) has a real solution, is
1 \(-\frac{2}{\pi}\)
2 \(\frac{2}{\pi}\)
3 \(-\frac{\pi}{2}\)
4 \(\frac{\pi}{2}\)
Explanation:
C Given equation is \(a x^2+\sin ^{-1}\left(x^2-2 x+2\right)+\cos ^{-1}\left(x^2-2 x+2\right)=0\) we know that \(\sin ^{-1}(x)+\cos ^{-1}(x)=\frac{\pi}{2}=0\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) Now let us check possible value of \(x^2\) \(\because-1 \leq \mathrm{x}^2-2 \mathrm{x}+2 \leq 1\) \(\therefore-1 \leq(\mathrm{x}-1)^2+\leq 1\) \(\therefore\) only possible value if \((\mathrm{x}-1)^2=0\) \(\mathrm{x}=1\) Put the value of \(x=1\) in equation (i) \(\mathrm{a}=\frac{-\pi}{2} \times \frac{1}{1^2}=\frac{-\pi}{2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118098
\(\frac{\mathrm{d}}{\mathrm{dx}} \operatorname{cosec}^{-1}\left(\frac{1+\mathrm{x}^2}{2 \mathrm{x}}\right)\) is equal to
1 \(\frac{-2}{\left(1+x^2\right)}, x \neq 0\)
2 \(\frac{2}{\left(1+x^2\right)}, x \neq 0\)
3 \(\frac{2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1,0\)
4 None of the above
Explanation:
C \( We know that-\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\left(\frac{1+x^2}{2 x}\right)^2-1}} \cdot \frac{d}{d x}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\frac{1+x^2+2 x^2-4 x^2}{4 x^2}} \cdot \frac{2 x(2 x)-\left(1+x^2\right) 2}{(2 x)^2}}\) \(=\frac{-1}{\frac{1+x^2}{2 x} \cdot \frac{\sqrt{1+x^4-2 x^2}}{2 x}} \cdot \frac{4 x^2-2-2 x^2}{4 x^2}\) \(=\frac{-4 x^2}{\left(1+x^2\right) \sqrt{\left(1-x^2\right)^2}} \cdot \frac{2 x^2-2}{4 x^2}\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)=\frac{+2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1\) For the next three (03) questions that follow: Consider the polynomial \(\mathbf{p}(\mathbf{x})=(\mathbf{x}-\boldsymbol{\alpha})(\mathbf{x}-\boldsymbol{\beta})(\mathbf{x}-\gamma)\) with \(\boldsymbol{\alpha}=\cos \mathbf{7 5}, \boldsymbol{\beta}=\cos 45^{\circ}\) and \(\gamma=\cos 165^{\circ}\)
BCECE-2011
Complex Numbers and Quadratic Equation
118100
If \(P\) is chosen at random in the closed interval \(\left[\begin{array}{ll}0,5\end{array}\right]\), the probability that the equation \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) will have real roots is equal to
1 \(1 / 2\)
2 \(1 / 5\)
3 \(2 / 3\)
4 \(3 / 5\)
Explanation:
D The quadratic equations \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) has real roots, if- \(D \geq 0\) \(\Rightarrow b^2-4 a c \geq 0\) \(\Rightarrow p^2-4\left(\frac{p+2}{4}\right) \geq 0\) \(\Rightarrow p^2-p-2 \geq 0\) \(\Rightarrow p^2-2 p+p-2 \geq 0\) \(\Rightarrow (p-2)(p+1) \geq 0\) Range of \(p \in(-\infty,-1] \cup[2, \infty)\) As \(\mathrm{p} \in[0,5]\), the required probability \(=\frac{3}{5}\)
SCRA-2012
Complex Numbers and Quadratic Equation
118101
The number of solutions of the equation \(\mathbf{x}^2+3|x|+2=0\) is
1 0
2 1
3 2
4 4
Explanation:
D Given, the equation is, \(x^2+3|x|+2=0\) Case- \(\mathrm{I}:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2+3 \mathrm{x}+2=0\) \(\Rightarrow \mathrm{x}^2+2 \mathrm{x}+\mathrm{x}+2=0\) \((x+2)(x+1)=0 \Rightarrow x=-2,-1\) Case II:- \(x\lt 0 \Rightarrow|x|=-x\) \(x^2-3 x+2=0\) \(x^2-2 x-x+2=0\) \(\Rightarrow(x-2)(x-1)=0 \Rightarrow x=1,2\)\(\therefore\) The number of solutions is 4 .
118096
The number of solutions of the equation \(3 \sin ^2 x-7 \sin x+2=0\) in the interval \([0,5 \pi]\) is
1 0
2 5
3 6
4 10
Explanation:
C Given equation - \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(\text { Now, }\) \(3 \sin ^2 \mathrm{x}-7 \sin \mathrm{x}+2=0\) \(3 \sin ^2 \mathrm{x}-6 \sin \mathrm{x}-\sin \mathrm{x}+2=0\) \(3 \sin \mathrm{x}(\sin \mathrm{x}-2)-1(\sin \mathrm{x}-2)=0\) \((3 \sin \mathrm{x}-1)(\sin \mathrm{x}-2)=0\) \(\therefore \sin \mathrm{x}=-2(\text { not possible because) }(-1 \leq \sin \mathrm{x} \leq 1)\) \(\therefore 3 \sin \mathrm{x}-1=0\) \(\sin \mathrm{x}=\frac{1}{3}(\text { posible })\) \(\sin \mathrm{x}=\frac{1}{3} \text { in interval }(0,5 \pi)\) So, there are 6 solutions in intervals \((0,5 \pi)\)
BCECE-2015
Complex Numbers and Quadratic Equation
118097
The value of ' \(a\) ' for which \(a^2+\sin ^{-1}\left(x^2-2 x+2\right)\) \(+\cos ^{-1}\left(x^2-2 x+2\right)=0\) has a real solution, is
1 \(-\frac{2}{\pi}\)
2 \(\frac{2}{\pi}\)
3 \(-\frac{\pi}{2}\)
4 \(\frac{\pi}{2}\)
Explanation:
C Given equation is \(a x^2+\sin ^{-1}\left(x^2-2 x+2\right)+\cos ^{-1}\left(x^2-2 x+2\right)=0\) we know that \(\sin ^{-1}(x)+\cos ^{-1}(x)=\frac{\pi}{2}=0\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) \(\mathrm{ax}^2=\frac{-\pi}{2}\) Now let us check possible value of \(x^2\) \(\because-1 \leq \mathrm{x}^2-2 \mathrm{x}+2 \leq 1\) \(\therefore-1 \leq(\mathrm{x}-1)^2+\leq 1\) \(\therefore\) only possible value if \((\mathrm{x}-1)^2=0\) \(\mathrm{x}=1\) Put the value of \(x=1\) in equation (i) \(\mathrm{a}=\frac{-\pi}{2} \times \frac{1}{1^2}=\frac{-\pi}{2}\)
BCECE-2013
Complex Numbers and Quadratic Equation
118098
\(\frac{\mathrm{d}}{\mathrm{dx}} \operatorname{cosec}^{-1}\left(\frac{1+\mathrm{x}^2}{2 \mathrm{x}}\right)\) is equal to
1 \(\frac{-2}{\left(1+x^2\right)}, x \neq 0\)
2 \(\frac{2}{\left(1+x^2\right)}, x \neq 0\)
3 \(\frac{2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1,0\)
4 None of the above
Explanation:
C \( We know that-\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\left(\frac{1+x^2}{2 x}\right)^2-1}} \cdot \frac{d}{d x}\left(\frac{1+x^2}{2 x}\right)\) \(=\frac{-1}{\frac{1+x^2}{2 x} \sqrt{\frac{1+x^2+2 x^2-4 x^2}{4 x^2}} \cdot \frac{2 x(2 x)-\left(1+x^2\right) 2}{(2 x)^2}}\) \(=\frac{-1}{\frac{1+x^2}{2 x} \cdot \frac{\sqrt{1+x^4-2 x^2}}{2 x}} \cdot \frac{4 x^2-2-2 x^2}{4 x^2}\) \(=\frac{-4 x^2}{\left(1+x^2\right) \sqrt{\left(1-x^2\right)^2}} \cdot \frac{2 x^2-2}{4 x^2}\) \(\frac{d}{d x} \operatorname{cosec}^{-1}\left(\frac{1+x^2}{2 x}\right)=\frac{+2\left(1-x^2\right)}{\left(1+x^2\right)\left|1-x^2\right|}, x \neq \pm 1\) For the next three (03) questions that follow: Consider the polynomial \(\mathbf{p}(\mathbf{x})=(\mathbf{x}-\boldsymbol{\alpha})(\mathbf{x}-\boldsymbol{\beta})(\mathbf{x}-\gamma)\) with \(\boldsymbol{\alpha}=\cos \mathbf{7 5}, \boldsymbol{\beta}=\cos 45^{\circ}\) and \(\gamma=\cos 165^{\circ}\)
BCECE-2011
Complex Numbers and Quadratic Equation
118100
If \(P\) is chosen at random in the closed interval \(\left[\begin{array}{ll}0,5\end{array}\right]\), the probability that the equation \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) will have real roots is equal to
1 \(1 / 2\)
2 \(1 / 5\)
3 \(2 / 3\)
4 \(3 / 5\)
Explanation:
D The quadratic equations \(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}+2}{4}=0\) has real roots, if- \(D \geq 0\) \(\Rightarrow b^2-4 a c \geq 0\) \(\Rightarrow p^2-4\left(\frac{p+2}{4}\right) \geq 0\) \(\Rightarrow p^2-p-2 \geq 0\) \(\Rightarrow p^2-2 p+p-2 \geq 0\) \(\Rightarrow (p-2)(p+1) \geq 0\) Range of \(p \in(-\infty,-1] \cup[2, \infty)\) As \(\mathrm{p} \in[0,5]\), the required probability \(=\frac{3}{5}\)
SCRA-2012
Complex Numbers and Quadratic Equation
118101
The number of solutions of the equation \(\mathbf{x}^2+3|x|+2=0\) is
1 0
2 1
3 2
4 4
Explanation:
D Given, the equation is, \(x^2+3|x|+2=0\) Case- \(\mathrm{I}:-\mathrm{x}>0 \Rightarrow|\mathrm{x}|=\mathrm{x}\) \(\therefore \mathrm{x}^2+3 \mathrm{x}+2=0\) \(\Rightarrow \mathrm{x}^2+2 \mathrm{x}+\mathrm{x}+2=0\) \((x+2)(x+1)=0 \Rightarrow x=-2,-1\) Case II:- \(x\lt 0 \Rightarrow|x|=-x\) \(x^2-3 x+2=0\) \(x^2-2 x-x+2=0\) \(\Rightarrow(x-2)(x-1)=0 \Rightarrow x=1,2\)\(\therefore\) The number of solutions is 4 .
