Choose the correct statement about two circles whose equations are given below :
Question: Choose the correct statement about two circles whose equations are given below : $x^{2}+y^{2}-10 x-10 y+41=0$ $x^{2}+y^{2}-22 x-10 y+137=0$ circles have same centrecircles have no meeting pointcircles have only one meeting pointcircles have two meeting pointsCorrect Option: , 3 Solution: $x^{2}+y^{2}-10 x-10 y+41=0$ $\mathrm{A}(5,5), \mathrm{R}_{1}=3$ $x^{2}+y^{2}-22 x-10 y+137=0$ $\mathrm{B}(11,5), \mathrm{R}_{2}=3$ $\mathrm{AB}=6=\mathrm{R}_{1}+\mathrm{R}_{2}$ Touch each other extern...
Read More →The value of
Question: The value of $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}}$, where $[\mathrm{x}]$ denotes the greatest integer $\leq \mathrm{x}$ is :$\pi$0$\frac{\pi}{4}$$\frac{\pi}{2}$Correct Option: , 4 Solution: $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1} x}{\left(1-x^{2}\right)} \times \frac{\sin ^{-1} x}{x}=\frac{\pi}{2}$...
Read More →Which of the following
Question: Which of the following is true for $\mathrm{y}(\mathrm{x})$ that satisfies the differential equation $\frac{d y}{d x}=x y-1+x-y ; y(0)=0$$y(1)=e^{\frac{1}{2}}-1$$y(1)=e^{\frac{1}{2}}-e^{-\frac{1}{2}}$$y(1)=1$$y(1)=e^{\frac{1}{2}}-1$Correct Option: 1 Solution: $\frac{d y}{d x}=(1+y)(x-1)$ $\frac{d y}{(y+1)}=(x-1) d x$ Integrate $\ln (y+1)=\frac{x^{2}}{2}-x+c$ $(0,0) \Rightarrow c=0 \Rightarrow y=e^{\left(\frac{x^{2}}{2}-x\right)}-1$...
Read More →If n is the number of irrational terms in the expansion of
Question: If $\mathrm{n}$ is the number of irrational terms in the expansion of $\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$, then $(n-1)$ is divisible by :263087Correct Option: 1 Solution: $\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$ ${ }^{60} C_{r}\left(3^{1 / 4}\right)^{60-r} \cdot\left(5^{1 / 8}\right)^{r}$ ${ }^{60} C_{r}(3)^{\frac{60-r}{4}} .5^{\frac{r}{8}}$ For rational terms. $\frac{\mathrm{r}}{8}=\mathrm{k} ; \quad 0 \leq \mathrm{r} \leq 60$ $0 \leq 8 \mathrm{k} \leq 60$ $0 \leq \mathrm{k} \leq \...
Read More →The solutions of the equation
Question: The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x \sin ^{2} x \sin ^{2} x \\ \cos ^{2} x 1+\cos ^{2} x \cos ^{2} x \\ 4 \sin 2 x 4 \sin 2 x 1+4 \sin 2 x\end{array}\right|=0,(0x\pi)$, are $\frac{\pi}{12}, \frac{\pi}{6}$$\frac{\pi}{6}, \frac{5 \pi}{6}$$\frac{5 \pi}{12}, \frac{7 \pi}{12}$$\frac{7 \pi}{12}, \frac{11 \pi}{12}$Correct Option: , 4 Solution: $\left|\begin{array}{ccc}1+\sin ^{2} x \sin ^{2} x \sin ^{2} x \\ \cos ^{2} x 1+\cos ^{2} x \cos ^{2} x \\ 4 \sin 2 x ...
Read More →The sum of solutions of the equation
Question: The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan 2 x|, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)-\left\{\frac{\pi}{4},-\frac{\pi}{4}\right\}$ is :(1) $-\frac{11 \pi}{30}$$\frac{\pi}{10}$$-\frac{7 \pi}{30}$$-\frac{\pi}{15}$Correct Option: 1, Solution: $\frac{\cos x}{1+\sin x}=|\tan 2 x|$ $\Rightarrow \frac{\cos ^{2} x / 2-\sin ^{2} x / 2}{(\cos x / 2+\sin x / 2}=1 \tan 2 x$ $\Rightarrow \tan ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan ^{2} 2 x$ $\Rightarrow ...
Read More →Which of the following statements
Question: Which of the following statements is incorrect for the function $\mathrm{g}(\alpha)$ for $\alpha \in \mathrm{R}$ such that. $\mathrm{g}(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin ^{\alpha} \mathrm{x}}{\cos ^{\alpha} \mathrm{x}+\sin ^{\alpha} \mathrm{x}} \mathrm{dx}$$g(\alpha)$ is a strictly increasing function$\mathrm{g}(\alpha)$ has an inflection point at $\alpha=-\frac{1}{2}$$g(\alpha)$ is a strictly decreasing function$\mathrm{g}(\alpha)$ is an even functionCorrect Opti...
Read More →Solve this following
Question: Let $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}$ then $a_{1}+a_{3}+a_{5}+\ldots+a_{37}$ is equal to $2^{20}\left(2^{20}-21\right)$$2^{19}\left(2^{20}-21\right)$$2^{19}\left(2^{20}+21\right)$$2^{20}\left(2^{20}+21\right)$Correct Option: , 2 Solution: $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+\ldots .+a_{40} x^{40}$ put $x=$ $1,-1$ $\Rightarrow a_{0}+a_{1}+a_{2}+\ldots+a_{40}=2^{20}$ $a_{0}-a_{1}+a_{2}+\ldots+a_{40}=2^{20}$ $\Rightarrow a_{1}+a_{3}+\l...
Read More →Choose the incorrect statement
Question: Choose the incorrect statement about the two circles whose equations are given below : $x^{2}+y^{2}-10 x-10 y+41=0$ and $x^{2}+y^{2}-16 x-10 y+80=0$Distance between two centres is the average of radii of both the circles.Both circles' centres lie inside region of one another.Both circles pass through the centre of each other.Circles have two intersection points.Correct Option: , 2 Solution: $\mathrm{r}_{1}=3, \mathrm{c}_{1}(5,5)$ $\mathrm{r}_{2}=3, \mathrm{c}_{2}(8,5)$ $\mathrm{C}_{1} ...
Read More →The number of integral values of $m$ so that the abscissa of point of intersection of lines
Question: The number of integral values of $m$ so that the abscissa of point of intersection of lines $3 x+4 y=9$ and $y=m x+1$ is also an integer, is : 1230Correct Option: , 2 Solution: $3 x+4 y=9$ $y=m x+1$ $\Rightarrow 3 x+4 m x+4=9$ $\Rightarrow(3+4 m) x=5$ $\Rightarrow x$ will be an integer when $3+4 m=5,-5,1,-1$ $\Rightarrow \mathrm{m}=\frac{1}{2},-2,-\frac{1}{2},-1$ so, number of integral values of $m$ is 2...
Read More →The value of
Question: $2+\frac{2}{5} \sqrt{30}$$2+\frac{4}{\sqrt{5}} \sqrt{30}$$4+\frac{4}{\sqrt{5}} \sqrt{30}$$5+\frac{2}{5} \sqrt{30}$Correct Option: 1 Solution: $y=4+\frac{1}{\left(5+\frac{1}{y}\right)}$ $y-4=\frac{y}{(5 y+1)}$ $5 y^{2}-20 y-4=0$ $\mathrm{y}=\frac{20+\sqrt{480}}{10}$ $\mathrm{y}=\frac{20-\sqrt{480}}{10} \rightarrow$ rejected $y=2+\sqrt{\frac{480}{100}}$ Correct with Option (A)...
Read More →Solve this following
Question: The differential equation satisfied by the system of parabolas $\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}+\mathrm{a})$ is : $y\left(\frac{d y}{d x}\right)^{2}-2 x\left(\frac{d y}{d x}\right)-y=0$ $y\left(\frac{d y}{d x}\right)^{2}-2 x\left(\frac{d y}{d x}\right)+y=0$$y\left(\frac{d y}{d x}\right)^{2}-2 x\left(\frac{d y}{d x}\right)+y=0$$y\left(\frac{d y}{d x}\right)+2 x\left(\frac{d y}{d x}\right)-y=0$Correct Option: , 3 Solution: $y^{2}=4 a x+4 a^{2}$ differentiate with respect to $x$ $\...
Read More →Team 'A' consists of
Question: Team 'A' consists of 7 boys and $n$ girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then $\mathrm{n}$ is equal to :5246Correct Option: , 3 Solution: Total matches between boys of both team $={ }^{7} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}=28$ Total matches between girls of both team $={ }^{n} \mathrm{C}_{1}{ }^{6} \mathrm{C}_{1}=6 \mathrm{n}$ Now, $28+6...
Read More →The line
Question: The line $2 x-y+1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2 y=4$. Then, the radius of the circle is:$3 \sqrt{5}$$5 \sqrt{3}$$5 \sqrt{4}$$4 \sqrt{5}$Correct Option: 1 Solution: $\left(\frac{h-\frac{(h-4)}{2}}{2-h}\right)(2)=-1$ $h=8$ center $(8,2)$...
Read More →The area of the triangle
Question: The area of the triangle with vertices $\mathrm{A}(\mathrm{z}), \mathrm{B}(\mathrm{iz})$ and $\mathrm{C}(\mathrm{z}+\mathrm{iz})$ is :1$\frac{1}{2}|z|^{2}$$\frac{1}{2}$$\frac{1}{2}|z+i z|^{2}$Correct Option: , 2 Solution: $\mathrm{A}=\frac{1}{2}|\mathrm{z}||\mathrm{iz}|$ $=\frac{|\mathrm{z}|^{2}}{2}$...
Read More →The sum of possible values
Question: The sum of possible values of $x$ for $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$ is :$-\frac{32}{4}$$-\frac{31}{4}$$-\frac{30}{4}$$-\frac{33}{4}$Correct Option: 1 Solution: $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1} \frac{8}{31}$ Taking tangent both sides :- $\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}$ $\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}$ $\Rightarrow 4 x^{2}+31 x-8=0$ But, if $x=\frac{1}{4}$ $...
Read More →In a school,
Question: In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement? $P$ and $Q$$\mathrm{P}$ and $\mathrm{R}$None of these$Q$ and $R$Correct Option: , 3 Solution: $A \cap B \cap C$ is visible in all three venn diagram Hence, Option (3)...
Read More →Let a complex number z,
Question: Let a complex number $\mathrm{z},|\mathrm{z}| \neq 1$, satisfy $\log _{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{(|z|-1)^{2}}\right) \leq 2$. Then, the largest value of $|z|$ is equal to______.8765Correct Option: Solution: $\log _{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{(|z|-1)^{2}}\right) \leq 2$ $\frac{|z|+11}{(|z|-1)^{2}} \geq \frac{1}{2}$ $2|z|+22 \geq(|z|-1)^{2}$ $2|z|+22 \geq|z|^{2}+1-2|z|$ $|z|^{2}-4|z|-21 \leq 0$ $\Rightarrow|z| \leq 7$ $\therefore \quad$ Largest value of $|z|$ is ...
Read More →If the fourth term in
Question: If the fourth term in the expansion of $\left(x+x^{\log _{2} x}\right)^{7}$ is 4480 , then the value of $x$ where $x \in N$ is equal to :2431Correct Option: 1 Solution: ${ }^{7} \mathrm{C}_{3} \mathrm{x}^{4} \mathrm{x}^{\left(3 \log _{2}^{\mathrm{x}}\right)}=4480$ $\Rightarrow \mathrm{x}^{\left(4+3 \log _{2}^{\mathrm{x}}\right)}=2^{7}$ $\Rightarrow(4+3 \mathrm{t}) \mathrm{t}=7 ; \mathrm{t}=\log _{2}^{\mathrm{x}}$ $\Rightarrow \mathrm{t}=1, \frac{-7}{3} \Rightarrow \mathrm{x}=2$...
Read More →Two dices are rolled.
Question: Two dices are rolled. If both dices have six faces numbered $1,2,3,5,7$ and 11 , then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :$\frac{4}{9}$$\frac{17}{36}$$\frac{5}{12}$$\frac{1}{2}$Correct Option: , 2 Solution: $n(E)=5+4+4+3+1=17$ So, $P(E)=\frac{17}{36}$...
Read More →Which of the following Boolean expression is a tautology?
Question: Which of the following Boolean expression is a tautology?$(p \wedge q) \vee(p \vee q)$$(\mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{p} \rightarrow \mathrm{q})$$(\mathrm{p} \wedge \mathrm{q}) \wedge(\mathrm{p} \rightarrow \mathrm{q})$$(p \wedge q) \rightarrow(p \rightarrow q)$Correct Option: , 4 Solution: $(\mathrm{p} \wedge \mathrm{q}) \rightarrow(\mathrm{p} \rightarrow \mathrm{q})$ is tautology...
Read More →If the Boolean expression
Question: If the Boolean expression $(\mathrm{p} \Rightarrow \mathrm{q}) \Leftrightarrow(\mathrm{q} *(\sim \mathrm{p}))$ is a tautology, then the Boolean expression $p *(\sim q)$ is equivalent to :$\mathrm{q} \Rightarrow \mathrm{p}$$\sim \mathrm{q} \Rightarrow \mathrm{p}$$\mathrm{p} \Rightarrow \sim \mathrm{q}$$p \Rightarrow q$Correct Option: 1 Solution: $\because \mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \vee \mathrm{q}$ $\mathrm{SO}, * \mathrm{~V}$ Thus, $p^{*}(\sim q) \equiv p ...
Read More →Let the functions f : R → R and g : R → R be
Question: Let the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined as : $f(x)=\left\{\begin{array}{ll}x+2, x0 \\ x^{2}, x \geq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{lr}x^{3}, x1 \\ 3 x-2, x \geq 1\end{array}\right.$ Then, the number of points in $\mathbb{R}$ where $(f \circ g)(x)$ is NOT differentiable is equal to :3100Correct Option: , 2 Solution: $f(g(x))= \begin{cases}g(x)+2, g(x)0 \\ (g(x))^{2}, g(x) \geq 0\end{cases}$ $= \b...
Read More →The number of the real roots of the equation
Question: The number of the real roots of the equation $(x+1)^{2}+|x-5|=\frac{27}{4}$ is Solution: $x \leq 5$ $(x+1)^{2}-(x-5)=\frac{27}{4}$ $(x+1)^{2}-(x+1)-\frac{3}{4}=0$ $x+1=\frac{3}{2},-\frac{1}{2}$ $x=\frac{1}{2},-\frac{3}{2}$ Case-II $x5$ $(x+1)+(x-5)=\frac{27}{4}$ $(x+1)^{2}+(x+1)-\frac{51}{4}=0$ $x=\frac{-1 \pm \sqrt{52}}{2}($ rejected as $x5)$ So, the equation have two real root....
Read More →Solve the Following Questions
Question: If $A=\left(\begin{array}{cc}0 \sin \alpha \\ \sin \alpha 0\end{array}\right)$ and $\operatorname{det}\left(A^{2}-\frac{1}{2} I\right)=0$, then a possible value of $\alpha$ is$\frac{\pi}{2}$$\frac{\pi}{3}$$\frac{\pi}{4}$$\frac{\pi}{6}$Correct Option: , 3 Solution: $\mathrm{A}^{2}=\sin ^{2} \alpha \mathrm{I}$ So, $\left|A^{2}-\frac{I}{2}\right|=\left(\sin ^{2} \alpha-\frac{1}{2}\right)^{2}=0$ $\Rightarrow \sin \alpha=\pm \frac{1}{\sqrt{2}}$...
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