Class 12th

Candidates seeking admission to the GNM, BSc and MSc programme can enroll for admission with Class 12 and graduation marks. Candidates must complete Class 12 to enroll for UG course. Similarly, for PG course, candidates are selected based on BSc Nursing. Apollo School and College of Nursing, Hyderabad admissions are merit + entrance-based.

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This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\frac{{\left(a^2+1\right)}^2}{2a-i}=x+yi\dots \left(i\right)\\ Takingconjugateonbothsides\\ \frac{{\left(a^2+1\right)}^2}{2a+i}=x-yi\dots \left(ii\right)\\ Multiplyingeqn.\left(i\right)and\left(ii\right)wehave\\ \frac{{\left(a^2+1\right)}^2{\left(a^2+1\right)}^2}{\left(2a-i\right)\left(2a+i\right)}=x^2+y^2\\ \Rightarrow \frac{{\left(a^2+1\right)}^4}{4a^2-i^2}=x^2+y^2\\ \Rightarrow \frac{{\left(a^2+1\right)}^4}{4a^2+1}=x^2+y^2\\ Hence,thevalueof{x}^2+y^2=\frac{{\left(a^2+1\right)}^4}{4a^2+1}.\end{array}$

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l}Let{z}_1=x_1+y_{1}iand{z}_2=x_2+y_{2}i\\ \therefore \left|x_1+y_{1}i\right|=\left|x_2+y_{2}i\right|\\ \Rightarrow \sqrt[]{x_1^2+y_1^2}=\sqrt[]{x_2^2+y_2^2}\\ \Rightarrow x_1^2+y_1^2=x_2^2+y_2^2\Rightarrow x_1^2=x_2^{2}and{y}_1^2=y_2^2\\ \Rightarrow x_1=\pm x_{2}and{y}_1=\pm y_2\\ So,z_1=x_1+y_{1}iand{z}_2=\pm x_2\pm y_{2}i\\ \therefore z_1\ne z_2\\ Hence,itisnotnecessarythat{z}_1=z_2.\end{array}$

$\bar{x}=10$

$\Rightarrow \stackrel{?}{\mathrm{x}}=\frac{63+\mathrm{a}+\mathrm{b}}{8}=10$

$\Rightarrow a+b=17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So,  ${\sigma }^2=13.5=\frac{79+(a-10{)}^2+(b-10{)}^2}{8}$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2-20(\mathrm{a}+\mathrm{b})=-171$

$\Rightarrow a^2+b^2=169$

From (1) and (2) ; $a=12$ and $b=5$

Candidates seeking admission to various programme can enrol for admission with Class 12/graduation/PG marks as per course requirement. The college offers various UG courses in various streams such as Management Studies, Computer Science, etc. ISBMA, Kolkata admissions are based on merit.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhateachedgeofthecuboidis2units.\\ \therefore Coordinatesoftheverticesare\\ A\left(2,0,0\right),B\left(2,2,0\right),C\left(0,2,0\right),D\left(0,2,2\right),E\left(0,0,2\right),F\left(2,0,2\right),G\left(2,2,2\right)andO\left(0,0,0\right).\end{array}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab)\frac{2b^2}{a}=10\Rightarrow b^2=5a$

Now,  $?\left(t\right)=\frac{5}{12}+t-t^2=\frac{8}{12}-{\left(t-\frac{1}{2}\right)}^2$
$?(\mathrm{t}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{4}{9}$

$\Rightarrow {\mathrm{a}}^2=81\mathrm{}$  (From (i) and (ii)

So,  $a^2+b^2=81+45=126$

$f\left(2\right)=8,f^{\mathrm{\text{'}}}\left(2\right)=5,f^{\mathrm{\text{'}}}\left(x\right)\ge 1,f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)\ge 4,\forall x\in \left(\mathrm{1,6}\right)$

Using LMVT

$f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)=\frac{f^{\mathrm{\text{'}}}\left(5\right)-f^{\mathrm{\text{'}}}\left(2\right)}{5-2}\ge 4\Rightarrow f^{\mathrm{\text{'}}}\left(5\right)\ge 17$

$f^{\mathrm{\text{'}}}\left(x\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\ge 1\Rightarrow f\left(5\right)\ge 11$

Therefore $f^{\mathrm{\text{'}}}\left(5\right)+f\left(5\right)\ge 28$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Letthegiven\text{\hspace{0.17em}points}areA\left(0,-1,-7\right),B\left(2,1,-9\right),C\left(6,5,-13\right)\\ AB=\sqrt[]{{\left(2-0\right)}^2+{\left(1+1\right)}^2+{\left(-9+7\right)}^2}=\sqrt[]{4+4+4}=\sqrt[]{12}=2\sqrt[]{3}\\ BC=\sqrt[]{{\left(6-2\right)}^2+{\left(5-1\right)}^2+{\left(-13+9\right)}^2}=\sqrt[]{16+16+16}=\sqrt[]{48}=4\sqrt[]{3}\\ AC=\sqrt[]{{\left(6-0\right)}^2+{\left(5+1\right)}^2+{\left(-13+7\right)}^2}=\sqrt[]{36+36+36}=\sqrt[]{108}=6\sqrt[]{3}\\ 2\sqrt[]{3}+4\sqrt[]{3}=6\sqrt[]{3}\\ i.e.,AB+BC=AC\\ \therefore AB:AC=2\sqrt[]{3}:6\sqrt[]{3}=1:3\\ Hence,\text{\hspace{0.17em}point}AdividesBandCin1:3externally.\end{array}$