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y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{dy}{dx}$ 9y + 2 = 0

$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}$

For horizontal tangent $\frac{dy}{dx}=0\Rightarrow y=\frac{2}{9}$ which does not satisfy the equation so no horizontal

For vertical tangent $5y^4-9x=0$

m = 0, N = 2

$sin\left(2x^2\right).109_e\left(tan{x}^2\right)dy+4xydx=4\sqrt[]{2}.x.\left(sin{x}^{2}cos\frac{\pi }{4}-cos{x}^{2}sin\frac{\pi }{4}\right)dx$

$\Rightarrow \mathcal{l}n\left(tan{x}^2\right)dy+\frac{4x}{sin\left(2x^2\right)}ydx=\frac{4x\left(sin{x}^2-cos{x}^2\right)}{sin\left(2x^2\right)}dx$

Integrate

$\Rightarrow y.\mathcal{l}n\left(tan{x}^2\right)=2.\mathcal{l}n\left(\frac{sin{x}^2+cos{x}^2-1}{sin{x}^2+cos{x}^2+1}\right)+C$

$x=\sqrt[]{\frac{\pi }{6}},y=1$ calculate C.

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

9a + 1 + 10 = 0…… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$

$4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,  $\lambda$ =1

Equation of plane $\pi \equiv x+2y+3z-2=0$

$\Rightarrow$$7$$P$$+$$3$$-$$2$$P$$+$$4$$-$$1$$2$$P$$+$$9$$-$$2$$=$$0$$\Rightarrow$$P$$=2$$\mathrm{}$

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance = $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$

use $si{n}^{-1}x=co{s}^{-1}\sqrt[]{1-x^2}$

$ta{n}^{-1}\sqrt[]{1-x^2}=co{t}^{-1}\frac{1}{\sqrt[]{1-x^2}}$

$si{n}^{-1}\frac{x}{\sqrt[]{1-x^2}}=co{s}^{-1}\frac{\sqrt[]{1-2x^2}}{\sqrt[]{1-x^2}}$

Sum of roots b = 1 + 2 $\frac{\left(k^2-1\right)}{k^2-2}$

Product of roots 5 = 2 $\left(\frac{k^2-1}{k^2-2}\right)$

b = 4, k² = $\frac{1}{3}$

The requirements for Nursing admissions, including whether NEET is required for Nursing admission to top-most colleges, however not mandatory.

Candidates can check the table below to see the list of colleges offering Nursing admission 0224 without NEET.

Note: This information is sourced from the official website and may vary.

While a 48% score in Class 12 might pose a challenge for admission to some universities or colleges offering BSc Nursing due to the competitive nature of the programs, it doesn't necessarily eliminate all possibilities.

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Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 × 0.5 n = 0.3 n

So, overall % age yield of complete reaction = $\frac{0.3n}{n}\times 100=30\mathrm{\%}$

There are only 6 colleges in India that offer BSc in Nursing admission through the NEET exam. These are as follows:

Note: The information is sourced from the official/counselling body of the institute/exam. 

Candidates can check the BSc Nursing eligibility criteria below:

• For BSc Nursing Course – Candidates should pass Class 12 in Physics, Chemistry, and Biology and be medically fit.
• For BSc (Post-Basic) – Candidates should pass Class 12 in PCB (Physics, Chemistry, and Biology). One has to get a certificate in GNM (General Nursing Midwifery) and register him/herself as an RNRM (Registered Nurse Registered Midwife) from the State Nurses Registration Council.

 $\mathrm{\%}Opticalpurity=\frac{Observedrotationofmixture\times 100}{rotationofpureenautiomer}$

$=\frac{+12.6}{+30}\times 100=42\mathrm{\%}$

Candidates can check the Nursing eligibility criteria in the table given below.

Several colleges and universities offer BSc Nursing programs without a specific entrance exam requirement. 

Such colleges are mentioned below in the table.

Note: The information is sourced from the official/counselling body of the institute/exam. 

Yes, direct Nursing admission is possible. Several colleges accept Class 12 exam scores for Nursing admission. Candidates can check the table below:

Note: The information is sourced from the official/counselling body of the institute/exam. 

To get Nursing admission in Delhi University, candidates must make sure to fulfil the eligibility criteria. For Nursing admission 2025 at the UG level, candidates should pass Class 12 in Physics, Chemistry, and Biology and be medically fit.

DU colleges that offer Nursing admission 2025 are mentioned in the table below.

Note: The information is sourced from the official/counselling body of the institute/exam. 

There are 21 colleges in Delhi that offer Nursing admission 2025. Candidates can check the table below to know the admission process and the total fees for Nursing courses in these popular colleges.

Note: The information is sourced from the official/counselling body of the institute/exam.