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Question: $4^{(x+1)}+4^{(1-x)}=10$ Solution: Given : $4^{(x+1)}+4^{(1-x)}=10$ $\Rightarrow 4^{x} \cdot 4+4^{1} \cdot \frac{1}{4^{x}}=10$ Let $4^{x}$ be $y$. $\therefore 4 y+\frac{4}{y}=10$ $\Rightarrow 4 y^{2}-10 y+4=0$ $\Rightarrow 4 y^{2}-8 y-2 y+4=0$ $\Rightarrow 4 y(y-2)-2(y-2)=0$ $\Rightarrow y=2$ or $y=\frac{2}{4}=\frac{1}{2}$ $\Rightarrow 4^{x}=2$ or $\frac{1}{2}$ $\Rightarrow 4^{x}=2^{2 x}=2^{1}$ or $2^{2 x}=2^{-1}$ $\Rightarrow x=\frac{1}{2}$ or $x=\frac{-1}{2}$ Hence, $\frac{1}{2}$ and...
Read More →limx is equal to
Question: $\lim _{x \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to Solution: $\tan \left(\lim _{n \rightarrow \infty} \sum_{r=1}^{n}\left[\tan ^{-1}(r+1)-\tan ^{-1}(r)\right]\right)$ $=\tan \left(\lim _{n \rightarrow \infty}\left(\tan ^{-1}(n+1)-\frac{\pi}{4}\right)\right)$ $=\tan \left(\frac{\pi}{4}\right)=1$...
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Question: $3^{(x+2)}+3^{-x}=10$ Solution: $3^{(x+2)}+3^{-x}=10$ $3^{x} .9+\frac{1}{3^{x}}=10$ Let $3^{x}$ be equal to $y$. $\therefore 9 y+\frac{1}{y}=10$ $\Rightarrow 9 y^{2}+1=10 y$ $\Rightarrow 9 y^{2}-10 y+1=0$ $\Rightarrow(y-1)(9 y-1)=0$ $\Rightarrow y-1=0$ or $9 y-1=0$ $\Rightarrow y=1$ or $y=\frac{1}{9}$ $\Rightarrow 3^{x}=1$ or $3^{x}=\frac{1}{9}$ $\Rightarrow 3^{x}=3^{0}$ or $3^{x}=3^{-2}$ $\Rightarrow x=0$ or $x=-2$ Hence, 0 and $-2$ are the roots of the given equation....
Read More →The number of solutions of the equation
Question: The number of solutions of the equation $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$ for $\mathrm{x} \in[-1,1]$, and $[\mathrm{x}]$ denotes the greatest integer less than or equal to $\mathrm{x}$, is :(1) 2(2) 0(3) 4(4) InfiniteCorrect Option: , 2 Solution: Given equation $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$ Now, $\sin ^{-1}\left[\mathrm{x}^{2}+\frac{1}{3}\right]$ is defined if $-1 \leq x^...
Read More →A particle of charge q and mass m is subjected to an electric field
Question: A particle of charge $q$ and mass $m$ is subjected to an electric field $E=E_{0}\left(1-a x^{2}\right)$ in the $x$-direction, where $a$ and $E_{0}$ are constants. Initially the particle was at rest at $x=0$. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :(1) $a$(2) $\sqrt{\frac{2}{a}}$(3) $\sqrt{\frac{3}{a}}$(4) $\sqrt{\frac{1}{a}}$Correct Option: , 3 Solution: (3) Given, Electric field, $E=E_{0}\lef...
Read More →A particle of charge q and mass m is subjected to an electric field
Question: A particle of charge $q$ and mass $m$ is subjected to an electric field $E=E_{0}\left(1-a x^{2}\right)$ in the $x$-direction, where $a$ and $E_{0}$ are constants. Initially the particle was at rest at $x=0$. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :(1) $a$(2) $\sqrt{\frac{2}{a}}$(3) $\sqrt{\frac{3}{a}}$(4) $\sqrt{\frac{1}{a}}$Correct Option: , 3 Solution: (3) Given, Electric field, $E=E_{0}\lef...
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Question: $\frac{a}{(a x-1)}+\frac{b}{(b x-1)}=(a+b), \quad\left(x \neq \frac{1}{a}, \frac{1}{b}\right)$ Solution: $\frac{a}{(a x-1)}+\frac{b}{(b x-1)}=(a+b)$ $\Rightarrow\left[\frac{a}{(a x-1)}-b\right]+\left[\frac{b}{(b x-1)}-a\right]=0$ $\Rightarrow \frac{a-b(a x-1)}{a x-1}+\frac{b-a(b x-1)}{b x-1}=0$ $\Rightarrow \frac{a-a b x+b}{a x-1}+\frac{a-a b x+b}{b x-1}=0$ $\Rightarrow(a-a b x+b)\left[\frac{1}{(a x-1)}+\frac{1}{(b x-1)}\right]=0$ $\Rightarrow(a-a b x+b)\left[\frac{(b x-1)+(a x-1)}{(a ...
Read More →The sum of possible values of
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(1) $-\frac{32}{4}$(2) $-\frac{31}{4}$(3) $-\frac{30}{4}$(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 :- $\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}$ $\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}$ $\Rightarrow 4 x^{2}+31 x-8=0$ $\Rightarro...
Read More →Prove that
Question: $\frac{a}{(x-b)}+\frac{b}{(x-a)}=2,(x \neq b, a)$ Solution: $\frac{a}{(x-b)}+\frac{b}{(x-a)}=2$ $\Rightarrow\left[\frac{a}{(x-b)}-1\right]+\left[\frac{b}{(x-a)}-1\right]=0$ $\Rightarrow \frac{a-(x-b)}{x-b}+\frac{b-(x-a)}{x-a}=0$ $\Rightarrow \frac{a-x+b}{x-b}+\frac{a-x+b}{x-a}=0$ $\Rightarrow(a-x+b)\left[\frac{1}{(x-b)}+\frac{1}{(x-a)}\right]=0$ $\Rightarrow(a-x+b)\left[\frac{(x-a)+(x-b)}{(x-b)(x-a)}\right]=0$ $\Rightarrow(a-x+b)\left[\frac{2 x-(a+b)}{(x-b)(x-a)}\right]=0$ $\Rightarrow...
Read More →A two point charges $4 q$ and $-q$ are fixed on the $x$-axis at
Question: A two point charges $4 q$ and $-q$ are fixed on the $x$-axis at $x=-\frac{d}{2}$ and $x=\frac{d}{2}$, respectively. If a third point charge ' $q$ ' is taken from the origin to $x=d$ along the semicircle as shown in the figure, the energy of the charge will: (1) increase by $\frac{3 q^{2}}{4 \pi \varepsilon_{0} d}$(2) increase by $\frac{2 q^{2}}{3 \pi \varepsilon_{0} d}$(3) decrease by $\frac{q^{2}}{4 \pi \varepsilon_{0} d}$(4) decrease by $\frac{4 q^{2}}{3 \pi \varepsilon_{0} d}$Correc...
Read More →The increasing order of reactivity of the following
Question: The increasing order of reactivity of the following compound towards reaction with alkyl halides directly is: $(\mathrm{B})(\mathrm{A})(\mathrm{C})(\mathrm{D})$$(\mathrm{A})(\mathrm{B})(\mathrm{C})(\mathrm{D})$$(B)(A)(D)(C)$$(\mathrm{A})(\mathrm{C})(\mathrm{D})(\mathrm{B})$Correct Option: 1 Solution:...
Read More →if
Question: If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots$ upto 100 terms, then $\alpha$ is :(1) $1.01$(2) 1(3) $1.02$(4) $1.03$Correct Option: 1 Solution: $\operatorname{Cot}^{-1}(\alpha)=\cot ^{-1}(2)+\cot ^{-1}(8)+\cot ^{-1}(18)+\ldots \ldots$ $=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{2}{4 n^{2}}\right)$ $=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}\right)$ $=\sum_{n=1}^{100} \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$ $=\tan ^{...
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Question: $\left(\frac{x}{x+1}\right)^{2}-5\left(\frac{x}{x+1}\right)+6=0,(x \neq-1)$ Solution: Given : $\left(\frac{x}{x+1}\right)^{2}-5\left(\frac{x}{x+1}\right)+6=0$ Putting $\frac{x}{x+1}=y$, we get: $y^{2}-5 y+6=0$ $\Rightarrow y^{2}-5 y+6=0$ $\Rightarrow y^{2}-(3+2) y+6=0$ $\Rightarrow y^{2}-3 y-2 y+6=0$ $\Rightarrow y(y-3)-2(y-3)=0$ $\Rightarrow(y-3)(y-2)=0$ $\Rightarrow y-3=0$ or $y-2=0$ $\Rightarrow y=3$ or $y=2$ Case I If $y=3$, we get: $\frac{x}{x+1}=3$ $\Rightarrow x=3(x+1)[$ On cros...
Read More →The correct order of atomic radii is :
Question: The correct order of atomic radii is :$\mathrm{N}\mathrm{Ce}\mathrm{Eu}\mathrm{Ho}$$\mathrm{Ho}\mathrm{N}\mathrm{Eu}\mathrm{Ce}$$\mathrm{Ce}\mathrm{Eu}\mathrm{Ho}\mathrm{N}$$\mathrm{Eu}\mathrm{Ce}\mathrm{Ho}\mathrm{N}$Correct Option: , 4 Solution: Atomic radii follows the order...
Read More →Given that the inverse trigonometric functions take principal values only.
Question: Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $\mathrm{x}$ which satisfy $\sin ^{-1}\left(\frac{3 \mathrm{x}}{5}\right)+\sin ^{-1}\left(\frac{4 \mathrm{x}}{5}\right)=\sin ^{-1} \mathrm{x}$ is equal to:(1) 2(2) 1(3) 3(4) 0Correct Option: , 3 Solution: $\sin ^{-1} \frac{3 x}{5}+\sin ^{-1} \frac{4 x}{5}=\sin ^{-1} x$ $\sin ^{-1}\left(\frac{3 \mathrm{x}}{5} \sqrt{1-\frac{16 \mathrm{x}^{2}}{25}}+\frac{4 \mathrm{x}}{5} \sqrt{1-\...
Read More →Two charged thin infinite plane sheets of uniform surface charge density
Question: Two charged thin infinite plane sheets of uniform surface charge density $\sigma_{+}$and $\sigma_{-}$, where $\left|\sigma_{+}\right|\left|\sigma_{-}\right|$, intersect at right angle. Which of the following best represents the electric field lines for this system ?Correct Option: , 3 Solution: (3) The electric field produced due to uniformly charged infinite plane is uniform. So option (2) and (4) are wrong. And +ve charge density $\sigma_{+}$is bigger in magnitude so its field along ...
Read More →The correct option with respect to the Pauling electronegativity values of the elements is:
Question: The correct option with respect to the Pauling electronegativity values of the elements is:$\mathrm{Te}\mathrm{Se}$$\mathrm{Ga}\mathrm{Ge}$$\mathrm{Si}\mathrm{Al}$$PS$Correct Option: , 2 Solution: Correct order of electronegativity values of the elements is $\mathrm{Si}\mathrm{Al} ; \mathrm{S}\mathrm{P} ; \mathrm{Se}\mathrm{Te} ; \mathrm{Ge}\mathrm{Ga}$...
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Question: $\left(\frac{4 x-3}{2 x+1}\right)-10\left(\frac{2 x+1}{4 x-3}\right)=3,\left(x \neq \frac{-1}{2}, \frac{3}{4}\right)$ Solution: Given : $\left(\frac{4 x-3}{2 x+1}\right)-10\left(\frac{2 x+1}{4 x-3}\right)=3$ Putting $\frac{4 x-3}{2 x+1}=y$, we get: $y-\frac{10}{y}=3$ $\Rightarrow \frac{y^{2}-10}{y}=3$ $\Rightarrow y^{2}-10=3 y[$ On cross multiplying $]$ $\Rightarrow y^{2}-3 y-10=0$ $\Rightarrow y^{2}-(5-2) y-10=0$ $\Rightarrow y^{2}-5 y+2 y-10=0$ $\Rightarrow y(y-5)+2(y-5)=0$ $\Rightar...
Read More →Let Sk
Question: Let $S_{\mathrm{k}}=\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{6^{\mathrm{r}}}{2^{2 \mathrm{r}+1}+3^{2 \mathrm{r}+1}}\right) .$ Then $\lim _{\mathrm{k} \rightarrow \infty} \mathrm{S}_{\mathrm{k}}$ equal to :(1) $\tan ^{-1}\left(\frac{3}{2}\right)$(2) $\frac{\pi}{2}$(3) $\cot ^{-1}\left(\frac{3}{2}\right)$(4) $\tan ^{-1}(3)$Correct Option: , 3 Solution: $\mathrm{S}_{\mathrm{k}}=\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{6^{\mathrm{r}}}{2^{2 \mathrm{r}+1}+3^{2 \ma...
Read More →The correct order of the atomic radii of
Question: The correct order of the atomic radii of $\mathrm{C}, \mathrm{Cs}, \mathrm{Al}$, and $\mathrm{S}$ is :$\mathrm{C}\mathrm{S}\mathrm{A} 1\mathrm{Cs}$$\mathrm{S}\mathrm{C}\mathrm{Cs}\mathrm{Al}$$\mathrm{S}\mathrm{C}\mathrm{Al}\mathrm{Cs}$$\mathrm{C}\mathrm{S}\mathrm{Cs}\mathrm{Al}$Correct Option: Solution: On going down the group, size increases while going from left to right in a period size decreases, so order is $\mathrm{C}\mathrm{S}\mathrm{Al}\mathrm{Cs}$....
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Question: The $71^{\text {st }}$ electron of an element $X$ with an atomic number of 71 enters into the orbital:$6 p$$4 f$$5 d$$6 s$Correct Option: , 3 Solution: ${ }_{71} \mathrm{X}=[\mathrm{Xe}] 6 s^{2} 4 f^{14} 5 d^{1}$ $\therefore$ Orbital occupied by last $e^{-}$is $5 d$....
Read More →The electronegativity of aluminium is similar to:
Question: The electronegativity of aluminium is similar to:CarbonBerylliumBoronLithiumCorrect Option: , 2 Solution: Be and $\mathrm{Al}$ show diagonal relationship due to which these two elements have similar electronegativity....
Read More →Concentric metallic hollow spheres of radii R and 4 R hold charges
Question: Concentric metallic hollow spheres of radii $R$ and $4 R$ hold charges $Q_{1}$ and $Q_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V(R)-V(4 R)$ is :(1) $\frac{3 Q_{1}}{16 \pi \varepsilon_{0} R}$(2) $\frac{3 Q_{2}}{4 \pi \varepsilon_{0} R}$(3) $\frac{Q_{2}}{4 \pi \varepsilon_{0} R}$(4) $\frac{3 Q_{1}}{4 \pi \varepsilon_{0} R}$Correct Option: 1 Solution: (1) We have given two metallic hollow spheres of radii $R$ and...
Read More →When the first electron gain enthalpy
Question: When the first electron gain enthalpy $\left(\Delta_{\mathrm{eg}} \mathrm{H}\right)$ of oxygen is - $141 \mathrm{~kJ} / \mathrm{mol}$, its second electron gain enthalpy is:a more negative value than the firstalmost the same as that of the firstnegative, but less negative than the firsta positive valueCorrect Option: , 4 Solution: The second electron gain enthalpy of oxygen is positive as energy has to be added for the addition of another electron....
Read More →In general, the properties that decrease and increase down a group in the periodic table,
Question: In general, the properties that decrease and increase down a group in the periodic table, respectively, are:atomic radius and electronegativity.electron gain enthalpy and electronegativity.electronegativity and atomic radius.electronegativity and electron gain enthalpy.Correct Option: Solution: Generally, electronegativity decreases down the group as the size increases. This can also be formulated as: Electronegativity $=\frac{1}{\text { size }}$...
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