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  $f\left(g\left(x\right)\right)=x\forall x\in R$

$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)$      

For y = g (x)

x = y³ + y – 5

$\frac{dx}{dy}=3y^2+1\Rightarrow g\text{'}\left(63\right)=\frac{1}{3\left(16\right)+1}=\frac{1}{49}$

$g\text{'}\left(63\right)=\left(\frac{dy}{dx}\right)$

$\left(\begin{array}{l}x=63\\ y=4\end{array}\right)$                

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l}C{e}^{+4}+e^{-}\to C{e}^{+3}{E}^0=1.61V\\ C{e}^{+3}+3e^{-}\to Ce{E}^0=-2.336V\end{array}$

It exist as Ce+4 and acts like a strong oxidizing agent by gaining electrons

Yes, the B.Tech degree from LNCT University is valid since it is approved by UGC. For private universities, AICTE approval is not mandatory for individual courses if the university itself is recognised by UGC. So, your degree will be considered valid for jobs, higher studies, and other competitive exams, as long as the university remains UGC approved.

FMS Delhi MBA fee include different components, some of which have to be paid annually, semesterly, etc. The MBA total tuition fee is INR 2 lakh. The mentioned fee is taken from the official website/sanctioning body and is subject to change. Therefore, it is indicative.

$\left(p\Delta q\right)\Rightarrow \left(p\Delta \sim q\right)\vee \left(\sim p\Delta q\right)$

Case I

When  $\Delta$  is same as $\vee .$  

Then  $\left(p\Delta \sim q\right)\vee \left(-p\Delta q\right)$ becomes

$\left(p\vee \sim q\right)\vee \left(\sim p\vee q\right)$ which is always true, so x becomes tautology.

Case II

When  $\Delta$  is same as $\vee$  

Then  $\left(p\wedge q\right)\Rightarrow \left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$  becomes $p\wedge q$ is T, then $\left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$ is false, so x cannot be tautology.

Case III

When  $\Delta$  is same as $\Rightarrow$  

Then  $\left(p\Rightarrow \sim q\right)\vee \left(-p\Rightarrow q\right)$ is same as $\left(\sim p\vee \sim q\right)\vee \left(p\vee q\right)$ which is true, so x becomes tautology.

Case IV

When  $\Delta$ is same as $\iff$  

Then   $\left(p\iff q\right)\Rightarrow \left(p\iff \sim q\right)\vee \left(\sim p\iff q\right)$

$p\iff q$ is true when p and q have same truth values $p\iff \sim qand\sim p\iff q$  both are false. Hence x cannot be tautology.

$A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)$

$A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}-1\\ 0\\ 1\end{array}\right)$

$A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$

$\Rightarrow A\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$AB=C\Rightarrow A=C{B}^{-1}$

$\left|A-2l\right|=\left(-4\right)\left(-1\right)+3\left(-1\right)+1\left(-1\right)$

4 – 3 – 1 = 0

$\mathcal{l}\left(-1,0,1\right)+m\left(-1,1,0\right)=\left(-4,3,1\right)\Rightarrow -\mathcal{l}-m=-4$

$m=3,\mathcal{l}=1$

$AgC\mathcal{l}+2N{H}_3\to \underset{\left(Solublecomplex\right)}{\left[Ag{\left(N{H}_3\right)}_2\right]C\mathcal{l}}$

Size of hydrated ion $\propto$ $\frac{1}{Ionicmobility}$

Size of hydrated ions $\propto \frac{1}{Ionicsize}$

Size of hydrated ions ; $B{e}^{+2}>M{g}^{+2}>C{a}^{+2}>S{r}^{+2}$

Ionic mobility ; $B{e}^{+2}<M{g}^{+2}<C{a}^{+2}<S{r}^{+2}$

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

=4

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

  $\frac{sin\theta sin\theta }{cos\theta }+\frac{sin\theta }{cos\theta }=2sin\theta cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }\left[sin\theta +1\right]=2sin\theta cos\theta$            

$\therefore$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$  

$\theta =\frac{3\pi }{2},\theta =\frac{\pi }{6},\frac{5\pi }{6}$           

$\theta =\frac{3\pi }{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{\pi }{3}+cos\frac{5\pi }{3}=4$  

$\therefore$ T + n(s) = 4 + 5 = 9

  $l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$ ${\displaystyle \underset{}{\overset{}{}}\frac{{}^{}}{{}^{}{}^{}{}^{}}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

$\frac{x^2-5x+6}{x^2-9}\ge -1\&\frac{x^2-5x+6}{x^2-9}\le 1$

$\frac{2x+1}{x+3}\ge 0,x\ne -3\&\frac{1}{x+3}\ge 0,x\ne -3\Rightarrow x>-3$

$x\in \left[-\frac{1}{2},\infty \right]...........(i)$

$x^2-3x+2>0and{x}^2-3x+1\ne 0$

(x – 2) (x – 1) > 0 and $x\ne \frac{3\pm \sqrt[]{5}}{2}$

$x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

From (i) and (ii)   $x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

Number of electrons in

$P{H}_3=15+3=18$

$B_2{H}_6=5\times 2+6=16$

$\begin{array}{l}CC{l}_4=6+17\times 4=6+68=74\\ N{H}_3=7+1\times 3=10\\ LiH=3+1=4\\ BC{l}_3=5+17\times 3=56\end{array}$

$B_2{H}_6\&BC{l}_3$ are e^- deficient molecules. B₂H₆ is dimer of BH₃, both compound has 6e^- only.

Au + NaCN + O₂ → Na [Au (CN)₂]

$Zn+Na\left[Au{\left(CN\right)}_2\right]\to N{a}_2\left[Zn{\left(CN\right)}_4\right]+Au$

$Ais{\left[Au{\left(CN\right)}_2\right]}^{-}andBis{\left[Zn{\left(CN\right)}_4\right]}^{-2}$

n = 33, p = success, q = failure

3P (x = 0) = P (x = 1)

${\Rightarrow }^{33}{C}_0{p}^0{q}^{33}{=}^{33}{C}_{1}p{q}^{32}$            

$\Rightarrow p=\frac{1}{12},q=\frac{11}{12}\Rightarrow \frac{q}{p}=11$          

………. (i)

Subtracting, (ii) – (i), we get 1320