Prove that the function
Question: Prove that the function $f: R \rightarrow R: f(x)=2 x$ is one-one and onto. Solution: To prove: function is one-one and onto Given: $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}: \mathrm{f}(\mathrm{x})=2 \mathrm{x}$ We have, $f(x)=2 x$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow 2 x_{1}=2 x_{2}$ $\Rightarrow x_{1}=x_{2}$ When, $f\left(x_{1}\right)=f\left(x_{2}\right)$ then $x_{1}=x_{2}$ f(x) is one-one $f(x)=2 x$ Let $f(x)=y$ such that $y \in R$ $\Rightarrow y=2 x$ $\R...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x$ Solution: Let $I=\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x$ We have, $\cos 2 x=1-2 \sin ^{2} x$ $I=e^{2 x}\left(\frac{1-\sin 2 x}{1-\left(1-2 \sin ^{2} x\right)}\right) d x$ $=\int e^{2 x}\left(\frac{1-\sin 2 x}{2 \sin ^{2} x}\right) d x$ $=\int \mathrm{e}^{2 \mathrm{x}}\left(\frac{\operatorname{cosec}^{2} \mathrm{x}}{2}-\frac{2 \sin \mathrm{x} \cos \mathrm{x}}{2 \sin ^{2} \m...
Read More →Prove that
Question: Let $f: R \rightarrow R: f(x)=x^{2}+2$ and $g: R \rightarrow R: g(x)=\frac{x}{x-1}, x \neq 1$. find $f$ o $g$ and $g$ o $f$ and hence find (f o $g$ ) (2) and (g o f) $(-3)$ Solution: To find: $f \circ g, g \circ f,(f \circ g)(2)$ and $(g \circ f)(-3)$ Formula used: (i) $f \circ g=f(g(x))$ (ii) $g \circ f=g(f(x))$ Given: (i) $f: R \rightarrow R: f(x)=x^{2}+2$ (ii) $g: R \rightarrow R: g(x)=\frac{x}{x-1}, x \neq 1$ f o g = f(g(x)) $\Rightarrow f\left(\frac{x}{x-1}\right)$ $\Rightarrow\le...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x$ Solution: Let $I=\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x$ $=\int e^{x} \frac{(x-2)-2)}{(x-2)^{3}} d x$ $=\int e^{x}\left\{\frac{1}{(x-2)^{2}}-\frac{2}{(x-2)^{2}}\right\} d x$ Let $f(x)=\frac{1}{(x-2)^{2}}$ and $f^{\prime}(x)=\frac{2}{(x-2)^{2}}$ We know that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\}=e^{x} f(x)+c$ $I=\frac{e^{x}}{(x-2)^{2}}+c$...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int\left\{\tan (\log x)+\sec ^{2}(\log x)\right\} d x$ Solution: Let $I=\int\left[\tan (\log x)+\sec ^{2}(\log x)\right] d x$ $\log x=z \Rightarrow x=e^{z} \Rightarrow d x=e^{z} d z$ $I=\int\left(\tan z+\sec ^{2} z\right) e^{z} d z$ $\mathrm{f}(\mathrm{z})=\tan \mathrm{z} ; \mathrm{f}^{\prime}(\mathrm{z})=\sec ^{2} \mathrm{z}$ We know that, $\int \mathrm{e}^{\mathrm{x}}\left\{\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right\}=\mathrm{e}^{...
Read More →Reactivity of transition elements decreases
Question: Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain. Solution: Effective nuclear charge increases as we move along the period from left to right. Due to the reason, the size also reduces. Therefore the electrons will be held more tightly and so that removing of the electron will be difficult from the outermost shell. So that ionization enthalpy also increases. Therefore, reactivity also decreases. Sc is more reactive than Cu....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\frac{\sin x \cos x-1}{\sin ^{2} x}\right) d x$ Solution: Let $I=\int e^{x}\left(\frac{\sin x \cos x-1}{\sin ^{2} x}\right) d x$ $=\int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x$ $=\int e^{x}\left(\cot x+-\operatorname{cosec}^{2} x\right) d x$ We know that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\}=e^{x} f(x)+c$ let $\mathrm{f}(\mathrm{x})=\cot \mathrm{x} ; \mathrm{f}^{\prime}(\mathrm{x})=-\operatorname{cosec}^{2} \mat...
Read More →The halides of transition elements become
Question: The halides of transition elements become more covalent with increasing oxidation state of the metal. Why? Solution: Halides become more covalent with increasing oxidation state. As the oxidation state increases, the charge on the atom increases and the size of the ion of transition element decreases. Fajans rules states that greater the charge on an atom, greater the covalent character. 50. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d or...
Read More →E° of Cu is + 0.34V while
Question: E of Cu is + 0.34V while that of Zn is 0.76V. Explain. Solution: The reduced form of Cu2+ is more stable than the oxidized form of Cu. Therefore, the value of E is positive for Cu. Removing two electrons gives a stable configuration [Ar]3d10 with filled orbitals. So, the oxidized form is more stable than the reduced form. Therefore, the value of E is negative for Zn....
Read More →The second and third rows of transition elements
Question: The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why? Solution: Due to poor f orbital shielding, effective nuclear charge increases and there is a contraction in the size of the third-row elements. This contraction in size is called Lanthanide contraction. The second and third row of transition elements have similar atomic radii and therefore they resemble each other more....
Read More →If f be a greatest integer function and g be an absolute value function, find the value of
Question: If f be a greatest integer function and g be an absolute value function, find the value of $(f \circ g)\left(\frac{-3}{2}\right)+(g \circ f)\left(\frac{4}{3}\right)$ Solution: To find: (fog) $\left(\frac{-3}{2}\right)+($ gof $)\left(\frac{4}{3}\right)$ Formula used: (i) f o g = f(g(x)) (ii) g o f = g(f(x)) Given: (i) f is a greatest integer function (ii) g is an absolute value function $f(x)=[x]$ (greatest integer function) $g(x)=|x|$ (absolute value function) $f\left(\frac{4}{3}\right...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$ Solution: Let $I=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$ here, $\mathrm{f}(\mathrm{x})=\tan ^{-1} \mathrm{x}$ and $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}^{2}}$ and we know that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\}=e^{x} f(x)+c$ $\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x=e^{x} \tan ^{-1} x+c$...
Read More →A solution of KMnO4 on reduction yields either
Question: A solution of KMnO4 on reduction yields either colourless solution or a brown precipitate or a green solution depending on the pH of the solution. What different stages of the reduction do these represent and how are they carried out? Solution: In acidic medium, permanganate changes to manganous ion which is colourless. MnO4-+8H+ + 5e- Mn2+ + 4H2O (colourless) In alkaline medium, permanganate changes to manganate, which is a green solution MnO4-+ e- MnO42- (green) In a neutral medium, ...
Read More →When an orange solution containing Cr2O72– ion is treated with an alkali,
Question: When an orange solution containing Cr2O72 ion is treated with an alkali, a yellow solution is formed and when H+ ions are added to a yellow solution, an orange solution is obtained. Explain why does this happen? Solution: When Cr2O72is treated with an alkali: (orange) Cr2O72+ OH- 2CrO42-(yellow) When the yellow solution is treated with an acid, we get back the orange solution: (yellow)2CrO42- +2H+Cr2O72(orange)+ H2O...
Read More →Explain why the colour of KMnO4 disappears
Question: Explain why the colour of KMnO4 disappears when oxalic acid is added to its solution in acidic medium. Solution: This is a redox titration. The deep purple colour of KMnO4 disappears due to the formation of MnSO4. 5H2C2O4 + 2KMnO4 +3H2SO4 2MnSO4 + 8H2O + K2SO4 +10CO2...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{2 x}(-\sin x+2 \cos x) d x$ Solution: Let $I=\int e^{2 x}(-\sin x+2 \cos x) d x$ $I=\int e^{2 x}-\sin x d x+2 \int e^{2 x} \cos x d x$ Applying by parts in the second integral, $I=-\int e^{2 x} \sin x d x+2\left\{\frac{1}{2} e^{2 x} \cos x+\int \frac{1}{2} e^{2 x} \sin x d x\right\}$ $=-\int e^{2 x} \sin x d x+e^{2 x} \cos x+\int e^{2 x} \sin x d x+c$ $=e^{2 x} \cos x+c$...
Read More →Although +3 oxidation states are the characteristic
Question: Although +3 oxidation states are the characteristic oxidation state of lanthanoids cerium shows +4 oxidation state also. Why? Solution: Ce [Xe] 4f1 5d1 6s2. Usually, lanthanoids lose the 5d and 6s electrons and show +3 oxidation state, but Cerium loses the one 4f electron also to attain Xenons noble gas configuration which will make Ce4+ very stable....
Read More →Although Zr belongs to 4d and Hf belongs
Question: Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why? Solution: Due to the lanthanoid contraction poor f, orbital shielding leads to an increase in effective nuclear charge, which reduces the size of Hf. So the atomic radii of both Zr and Hf are similar which means they have similar physical and chemical properties and hence separation becomes difficult....
Read More →Ionisation enthalpies of Ce,
Question: Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why? Solution: Th, Pa and U, 5f electrons start filling and they have penetration lower than 4f electrons for Ce, Pr and Nd. Removing 4f electrons will be difficult, so ionization enthalpy for Th, Pa and U are lower than Ce, Pr and Nd....
Read More →Although Cr3+ and Co2+ ions have the same number
Question: Although Cr3+ and Co2+ ions have the same number of unpaired electrons the magnetic moment of Cr3+ is 3.87 B.M. and that of Co2+ is 4.87 B.M. Why? Solution: Cr3+ has a symmetrical electron distribution and will only have spin magnetic moment contribution whereas Co2+ has no symmetrical distribution of electrons so it will have an orbital magnetic moment and spin magnetic moment contribution. Therefore, the total magnetic moment for Co2+ will be higher than the Cr3+....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x} \cdot \frac{\sqrt{1-x^{2}} \sin ^{-1} x+1}{\sqrt{1-x^{2}}} d x$ Solution: Let $I=\int e^{x} \frac{\sqrt{1-x^{2}} \sin ^{-1} x+1}{\sqrt{1-x^{2}}} d x$ $I=\int e^{x} \sin ^{-1} x+\int e^{x} \frac{1}{\sqrt{1-x^{2}}} d x$ Integrating by parts $=e^{x} \sin ^{-1} x-\int e^{x}\left(\frac{d}{d x}\left(\sin ^{-1} x\right)\right) d x+\int e^{x} \frac{1}{\sqrt{1-x^{2}}} d x$ $=e^{x} \sin ^{-1} x-\int e^{x} \frac{1}{\sqrt{1-x^{2}}} d x+\int e^{x} \frac...
Read More →Although fluorine is more electronegative than oxygen,
Question: Although fluorine is more electronegative than oxygen, the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why? Solution: Fluorine has one unpaired electron and forms a single bond, while Oxygen has two unpaired electrons and can form multiple bonds thereby stabilizing higher oxidation states....
Read More →When a brown compound of manganese
Question: When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess reacts with NH3 to give an explosive compound (C). Identify compounds A, B and C. Solution: When brown co pound of manganese (A) is treated with HCl it gives chlorine gas. MnO2+ 4HCl MnCl2+ Cl2+2H2O (A) (B) The chlorine gas reacts with NH3 to given NCl3 3Cl2+ NH3NCl3+ 3HCl (B) (C) The brown compound A = MnO2 Gas B = Cl2 Explosive compound C= NCl3...
Read More →Solve this
Question: Let $f: N \rightarrow N: f(x)=2 x, g: N \rightarrow N: g(y)$ $=3 y+4$ and $h: N \rightarrow N: h(z)$ $=\sin z .$ Show that $h \circ(g \circ f)$ $=(\mathrm{h} \circ \mathrm{g}) \mathrm{o}$ f. Solution: To show: h o (g o f ) = (h o g) o f Formula used: (i) f o g = f(g(x)) (ii) $g \circ f=g(f(x))$ Given: (i) $f: N \rightarrow N: f(x)=2 x$ (ii) $g: N \rightarrow N: g(y)=3 y+4$ (iii) $h: N \rightarrow N: h(z)=\sin z$ Solution: We have, $\mathrm{LHS}=\mathrm{h} \mathrm{o}(\mathrm{g} \mathrm{...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}}\left\{\mathrm{x}(\log \mathrm{x})^{2}+2 \log \mathrm{x}\right\} \mathrm{dx}$ Solution: Let $I=\int \frac{e^{x}}{x}\left\{x(\log x)^{2}+2 \log x\right\} d x$ $=\int e^{x}(\log x)^{2} d x+2 \int \frac{e^{x}}{x} \log x d x$ Using integration by parts, $=e^{x}(\log x)^{2}-\int e^{x} \frac{d}{d x}(\log x)^{2}+2 \int \frac{e^{x}}{x} \log x d x$ $=e^{x}(\log x)^{2}-2 \int \frac{e^{x}}{x} \log x d x+2 \int \frac...
Read More →