The integral
Question: The integral $\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$ is equal to (where $c$ is a constant of integration)(1) $\frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5}+c$(2) $\frac{1}{2} \cos \sqrt{(2 x+1)^{2}+5}+c$(3) $\frac{1}{2} \cos \sqrt{(2 x-1)^{2}+5}+c$(4) $\frac{1}{2} \sin \sqrt{(2 x+1)^{2}+5}+c$Correct Option: 1 Solution: $\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{(2 x-1)^{2}+5}} d x$ $(2 x-1)^{2}+5=t^{2}$ $2(2 x-1) 2 d x=2 t d t$ $2 \sqrt{t^{2}-5} d x=...
Read More →The set of elements that differ in mutual relationship from those of the other sets is :
Question: The set of elements that differ in mutual relationship from those of the other sets is :$\mathrm{L}_{\mathrm{i}}-\mathrm{Mg}$$\mathrm{B}-\mathrm{Si}$$\mathrm{Be}-\mathrm{Al}$$\mathrm{Li}-\mathrm{Na}$Correct Option: , 4 Solution: $\mathrm{Li}-\mathrm{Mg}, \mathrm{B}-\mathrm{Si}, \mathrm{Be}-\mathrm{A} 1$ show diagonal relationship but $\mathrm{Li}$ and $\mathrm{Na}$ do not show diagonal relationship as both belongs to same group and not placed diagonally....
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $x^{2}-4 a x-b^{2}+4 a^{2}=0$ Solution: We write, $-4 a x=-(b+2 a) x+(b-2 a) x$ as $x^{2} \times\left(-b^{2}+4 a^{2}\right)=\left(-b^{2}+4 a^{2}\right) x^{2}=-(b+2 a) x \times(b-2 a) x$ $\therefore x^{2}-4 a x-b^{2}+4 a^{2}=0$ $\Rightarrow x^{2}-(b+2 a) x+(b-2 a) x-(b-2 a)(b+2 a)=0$ $\Rightarrow x[x-(b+2 a)]+(b-2 a)[x-(b+2 a)]=0$ $\Rightarrow[x-(b+2 a)][x+(b-2 a)]=0$ $\Rightarrow x-(b+2 a)=0$ or $x+(b-2 a)=0$ $\Rightarrow x=2 a+b$ or $x=...
Read More →For real numbers
Question: For real numbers $\alpha, \beta, \gamma$ and $\delta$, if $\int \frac{\left(x^{2}-1\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)}{\left(x^{4}+3 x^{2}+1\right) \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)} d x$ $=\alpha \log _{e}\left(\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right)$ $+\beta \tan ^{-1}\left(\frac{\gamma\left(x^{2}-1\right)}{x}\right)+\delta \tan ^{-1}\left(\frac{x^{2}+1}{x}\right)+C$ where $\mathrm{C}$ is an arbitrary constant, then the value of $10(\alpha+\beta \gamma+\d...
Read More →Which of the following compound CANNOT
Question: Which of the following compound CANNOT act as a Lewis base?$\mathrm{NF}_{3}$$\mathrm{PCl}_{5}$$\mathrm{SF}_{4}$$\mathrm{ClF}_{3}$Correct Option: , 2 Solution: Lewis base : Chemical species which has capability to donate electron pair. In $\mathrm{NF}_{3}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}$ central atom (i.e. $\mathrm{N}, \mathrm{S}, \mathrm{Cl}$ )having lone pair therefore act as lewis base. In $\mathrm{PCl}_{5}$ central atom $(\mathrm{P})$ does not have lone pair therefore does not ac...
Read More →Solve this
Question: $a b x^{2}+\left(b^{2}-a c\right) x-b c=0$ Solution: Given: $a b x^{2}+\left(b^{2}-a c\right) x-b c=0$ $\Rightarrow a b x^{2}+b^{2} x-a c x-b c=0$ $\Rightarrow b x(a x+b)-c(a x+b)=0$ $\Rightarrow(b x-c)(a x+b)=0$ $\Rightarrow b x-c=0$ or $a x+b=0$ $\Rightarrow x=\frac{c}{b}$ or $x=\frac{-b}{a}$ Hence, the roots of the equation are $\frac{c}{b}$ and $\frac{-b}{a}$....
Read More →The absolute value of the electron gain enthalpy of
Question: The absolute value of the electron gain enthalpy of halogens satisfies:$\mathrm{I}\mathrm{Br}\mathrm{Cl}\mathrm{F}$$\mathrm{Cl}\mathrm{Br}\mathrm{F}\mathrm{I}$$\mathrm{Cl}\mathrm{F}\mathrm{Br}\mathrm{I}$$\mathrm{F}\mathrm{Cl}\mathrm{Br}\mathrm{I}$Correct Option: , 3 Solution: Order of electron gain enthalpy (Absolute value) $\mathrm{Cl}\mathrm{F}\mathrm{Br}\mathrm{I}$...
Read More →The electric field in a region is given by
Question: A plane electromagnetic wave of frequency $100 \mathrm{MHz}$ is travelling in vacuum along the $\mathrm{x}$ - direction. At a particular point in space and time, $\overrightarrow{\mathrm{B}}=2.0 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T} \cdot($ where,$\hat{\mathrm{k}}$ is unit vector along z-direction) What is $\overrightarrow{\mathrm{E}}$ at this point?(1) $0.6 \hat{\mathrm{j}} \mathrm{V} / \mathrm{m}$(2) $6.0 \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}$(3) $6.0 \hat{\mathrm{j}} \mathrm...
Read More →The characteristics of elements X,
Question: The characteristics of elements $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ with atomic numbers, respectively, 33,53 and 83$\mathrm{X}$ and $\mathrm{Y}$ are metalloids and $\mathrm{Z}$ is a metal.$\mathrm{X}$ is a metalloid, $\mathrm{Y}$ is a non-metal and $\mathrm{Z}$ is a metal.$\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are metals. $\mathrm{X}$ and $\mathrm{Z}$ are non-metals and $\mathrm{Y}$ is a metalloidCorrect Option: , 2 Solution: $\mathrm{X}={ }_{33} \mathrm{As} \rightarrow$ Metal...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$ Solution: We write, $6 x=(a+4) x-(a-2) x$ as $x^{2} \times\left[-\left(a^{2}+2 a-8\right)\right]=-\left(a^{2}+2 a-8\right) x^{2}=(a+4) x \times[-(a-2) x]$ $\therefore x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$ $\Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0$ $\Rightarrow x[x+(a+4)]-(a-2)[x+(a+4)]=0$ $\Rightarrow[x+(a+4)][x-(a-2)]=0$ $\Rightarrow x+(a+4)=0$ or $x-(a-2)=0$ $\Rightarrow x=-(a+4)$ or $x=a-2$...
Read More →The electric field in a region is given by
Question: The electric field in a region is given by $\overrightarrow{\mathrm{E}}=\frac{2}{5} \mathrm{E}_{0} \hat{\mathrm{i}}+\frac{3}{5} \mathrm{E}_{0} \hat{\mathrm{j}}$ with $\mathrm{E}_{0}=4.0 \times 10^{3} \frac{\mathrm{N}}{\mathrm{C}} \cdot$ The flux of this field through a rectangular surface area $0.4 \mathrm{~m}^{2}$ parallel to the $\mathrm{Y}-\mathrm{Z}$ plane is________ $\mathrm{Nm}^{2} \mathrm{C}^{-1}$ Solution: (640) $\phi=\mathrm{E}_{\mathrm{x}} \mathrm{A} \Rightarrow \frac{2}{5} \...
Read More →Identify the elements X and Y using the ionisation energy values given below:
Question: Identify the elements $X$ and $Y$ using the ionisation energy values given below: $\mathrm{X}=\mathrm{Na} ; \mathrm{Y}=\mathrm{Mg}$$\mathrm{X}=\mathrm{Mg} ; \mathrm{Y}=\mathrm{F}$$\mathrm{X}=\mathrm{Mg} ; \mathrm{Y}=\mathrm{Na}$$\mathrm{Y}=\mathrm{F} ; \mathrm{Y}=\mathrm{Mg}$Correct Option: 1 Solution: $\mathrm{Na} \rightarrow[\mathrm{Ne}] 3 \mathrm{~s}^{1} \mathrm{IE}_{1}$ is very low but $\mathrm{IE}_{2}$ is very high due to stable noble gas configuration of $\mathrm{Na}^{+}$ $\mathr...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $x^{2}-(2 b-1) x+\left(b^{2}-b-20\right)=0$ Solution: We write, $-(2 b-1) x=-(b-5) x-(b+4) x$ as $x^{2} \times\left(b^{2}-b-20\right)=\left(b^{2}-b-20\right) x^{2}=[-(b-5) x] \times[-(b+4) x]$ $\therefore x^{2}-(2 b-1) x+\left(b^{2}-b-20\right)=0$ $\Rightarrow x^{2}-(b-5) x-(b+4) x+(b-5)(b+4)=0$ $\Rightarrow x[x-(b-5)]-(b+4)[x-(b-5)]=0$ $\Rightarrow[x-(b-5)][x-(b+4)]=0$ $\Rightarrow x-(b-5)=0$ or $x-(b+4)=0$ $\Rightarrow x=b-5$ or $x=b+4...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $x^{2}-2 a x-\left(4 b^{2}-a^{2}\right)=0$ Solution: We write, $-2 a x=(2 b-a) x-(2 b+a) x$ as $x^{2} \times\left[-\left(4 b^{2}-a^{2}\right)\right]=-\left(4 b^{2}-a^{2}\right) x^{2}=(2 b-a) x \times[-(2 b+a) x]$ $\therefore x^{2}-2 a x-\left(4 b^{2}-a^{2}\right)=0$ $\Rightarrow x^{2}+(2 b-a) x-(2 b+a) x-(2 b-a)(2 b+a)=0$ $\Rightarrow x[x+(2 b-a)]-(2 b+a)[x+(2 b-a)]=0$ $\Rightarrow[x+(2 b-a)][x-(2 b+a)]=0$ $\Rightarrow x+(2 b-a)=0$ or $x...
Read More →If the vertices of a hyperbola be at
Question: If the vertices of a hyperbola be at $(-2,0)$ and $(2,0)$ and one of its foci be at $(-3,0)$, then which one of the following points does not lie on this hyperbola?(1) $(-6,2 \sqrt{10})$(2) $(2 \sqrt{6}, 5)$(3) $(4, \sqrt{15})$(4) $(6,5 \sqrt{2})$Correct Option: , 4 Solution: Let the points are, $A(2,0), A^{\prime}(-2,0)$ and $S(-3,0)$ $\Rightarrow$ Centre of hyperbola is $O(0,0)$ $A A^{\prime}=2 a \Rightarrow 4=2 a \Rightarrow a=2$ $\because \quad$ Distance between the centre and foci...
Read More →Find out the surface charge density at the intersection
Question: Find out the surface charge density at the intersection of point $\mathrm{x}=3 \mathrm{~m}$ plane and $\mathrm{x}$-axis, in the region of uniform line charge of $8 \mathrm{n} \mathrm{C} / \mathrm{m}$ lying along the $\mathrm{z}$-axis in free space.(1) $0.424 \mathrm{nCm}^{-2}$(2) $47.88 \mathrm{C} / \mathrm{m}$(3) $0.07 \mathrm{nCm}^{-2}$(4) $4.0 \mathrm{nCm}^{-2}$Correct Option: 1 Solution: (1) $\frac{2 \mathrm{~K} \lambda}{\mathrm{r}}=\frac{\sigma}{\varepsilon_{0}} \quad(\mathrm{x}=3...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $x^{2}+5 x-\left(a^{2}+a-6\right)=0$ Solution: We write, $5 x=(a+3) x-(a-2) x$ as $x^{2} \times\left[-\left(a^{2}+a-6\right)\right]=-\left(a^{2}+a-6\right) x^{2}=(a+3) x \times[-(a-2) x]$ $\therefore x^{2}+5 x-\left(a^{2}+a-6\right)=0$ $\Rightarrow x^{2}+(a+3) x-(a-2) x-(a+3)(a-2)=0$ $\Rightarrow x[x+(a+3)]-(a-2)[x+(a+3)]=0$ $\Rightarrow[x+(a+3)][x-(a-2)]=0$ $\Rightarrow x+(a+3)=0$ or $x-(a-2)=0$ $\Rightarrow x=-(a+3)$ or $x=a-2$ Hence, ...
Read More →If a hyperbola has length of its conjugate axis
Question: If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13 , then the eccentricity of the hyperbola is :(1) $\frac{13}{12}$(2) 2(3) $\frac{13}{6}$(4) $\frac{13}{8}$Correct Option: 1 Solution: $\therefore$ Conjugate axis $=5$ $\therefore \quad 2 b=5$ Distance between foci $=13$ $2 a e=13$ Then, $b^{2}=a^{2}\left(e^{2}-1\right)$ $\Rightarrow a^{2}=36$ $\therefore \quad a=6$ $a e=\frac{13}{2} \Rightarrow e=\frac{13}{12}$...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$ Solution: We write, $-4 a^{2} x=-2\left(a^{2}+b^{2}\right) x-2\left(a^{2}-b^{2}\right) x$ as $4 x^{2} \times\left(a^{4}-b^{4}\right)=4\left(a^{4}-b^{4}\right) x^{2}=\left[-2\left(a^{2}+b^{2}\right)\right] x \times\left[-2\left(a^{2}-b^{2}\right)\right] x$ $\therefore 4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$ $\Rightarrow 4 x^{2}-2\left(a^{2}+b^{2}\right) x-2\left(a^{2}-b^{2}\right) x+\le...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0$ Solution: We write, $4 b x=2(a+b) x-2(a-b) x$ as $4 x^{2} \times\left[-\left(a^{2}-b^{2}\right)\right]=-4\left(a^{2}-b^{2}\right) x^{2}=2(a+b) x \times[-2(a-b) x]$ $\therefore 4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0$ $\Rightarrow 4 x^{2}+2(a+b) x-2(a-b) x-(a-b)(a+b)=0$ $\Rightarrow 2 x[2 x+(a+b)]-(a-b)[2 x+(a+b)]=0$ $\Rightarrow[2 x+(a+b)][2 x-(a-b)]=0$ $\Rightarrow 2 x+(a+b)=0$ or $2 x-(a-b)=0...
Read More →The correct match between item (I) and item (II) is :
Question: The correct match between item (I) and item (II) is : $(\mathrm{A}) \rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{S})$$(\mathrm{A}) \rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{R})$$(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{S})$$(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rig...
Read More →Let
Question: Let $\mathrm{S}=\left\{(x, y) \in \mathbf{R}^{2}: \frac{y^{2}}{1+\mathrm{r}}-\frac{x^{2}}{1-\mathrm{r}}=1\right\}$ where $r \neq \pm 1$ Then $S$ represents: (1) a hyperbola whose eccentricity is $\frac{2}{\sqrt{1-r}}$, when $0r1$(2) an ellipse whose eccentricity is $\sqrt{\frac{2}{r+1}}$, when $r1$(3) a hyperbola whose eccentricity is $\frac{2}{\sqrt{\mathrm{r}+1}}$, when $0\mathrm{r}1$(4) an ellipse whose eccentricity is $\frac{1}{\sqrt{r+1}}$, when $r1$Correct Option: , 2 Solution: S...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $2 x^{2}+a x-a^{2}=0$ Solution: We write, $a x=2 a x-a x$ as $2 x^{2} \times\left(-a^{2}\right)=-2 a^{2} x^{2}=2 a x \times(-a x)$ $\therefore 2 x^{2}+a x-a^{2}=0$ $\Rightarrow 2 x^{2}+2 a x-a x-a^{2}=0$ $\Rightarrow 2 x(x+a)-a(x+a)=0$ $\Rightarrow(x+a)(2 x-a)=0$ $\Rightarrow x+a=0$ or $2 x-a=0$ $\Rightarrow x=-a$ or $x=\frac{a}{2}$ Hence, $-a$ and $\frac{a}{2}$ are the roots of the given equation....
Read More →A light wave is incident normally on a glass slab of refractive index 1.5.
Question: A light wave is incident normally on a glass slab of refractive index 1.5. If $4 \%$ of light gets reflected and the amplitude of the electric field of the incident light is $30 \mathrm{~V} / \mathrm{m}$, then the amplitude of the electric field for the wave propogating in the glass medium will be:(1) $30 \mathrm{~V} / \mathrm{m}$(2) $10 \mathrm{~V} / \mathrm{m}$(3) $24 \mathrm{~V} / \mathrm{m}$(4) $6 \mathrm{~V} / \mathrm{m}$Correct Option: , 3 Solution: (3) As $4 \%$ of light gets re...
Read More →Solve this
Question: $\frac{2}{x^{2}}-\frac{5}{x}+2=0$ Solution: Given: $\frac{2}{x^{2}}-\frac{5}{x}+2=0$ $\Rightarrow 2-5 x+2 x^{2}=0 \quad$ [Multiplying both side by $\left.x^{2}\right]$ $\Rightarrow 2 x^{2}-5 x+2=0$ $\Rightarrow 2 x^{2}-(4 x+x)+2=0$ $\Rightarrow 2 x^{2}-4 x-x+2=0$ $\Rightarrow 2 x(x-2)-1(x-2)=0$ $\Rightarrow(2 x-1)(x-2)=0$ $\Rightarrow 2 x-1=0$ or $x-2=0$ $\Rightarrow x=\frac{1}{2}$ or $x=2$ Hence, the roots of the equation are $\frac{1}{2}$ and 2 ....
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