Ask & Answer: India's Largest Education Community

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

 ${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$

So, f (x) = αx – β

Now, $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$

f (x) = αx – β

$=\frac{18x-4}{13}$

option (D) satisfies

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$

option (C) is incorrect, there will be minima.

Candidates need to complete Class 12 to apply for any of the UG programmes at Munger University. The minimum aggregate required for most of the UG courses is 45%. Hence, aspirants with 61% in Class 12 can apply for the preferred UG course at Munger University. Aspirants are advised to keep the supporting documents with them for verification at the time of admission.

The university offers more than 71 undergraduate, 88 graduate, and vocational programs. There are a total of 8 faculties: From either of the faculty members above, international students may choose their dream specialisations given below:

 $x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

= 2

Candidates who have passed Class 12 with at least 45% aggregate with Mathematics or Statistics are eligible to apply for BCA at Munger University. Aspirants must hold the relevant documents proving their eligibility for verification at the time of admission. Besides, the selection for BCA programme is based on qualifying exam scores of the candidates.

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$

$\frac{25S}{36}=1+\frac{3/5}{1}$

$S=\frac{288}{125}$

B = (I – adjA)⁵

^{fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff}

$\left|z\right|=3\Rightarrow$ circle with radius = 3

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4},$ part of a circle (with radius $\sqrt[]{2}$ ). no common points

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

$\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$

= 1 + 1 – 1 = 1

Munger University admissions are primarily merit-based. The university considers the scores of the aspirants in the last qualifying exam for admission. Munger University conducts an entrance test to shortlist candidates for PhD programme. PAT is a written test comprising of 100 marks each. Paper-I consists of multiple-choice questions on qualitative, teaching and research aptitude. The paper-II includes subjective type questions in concerned subject.

Birla Institute of Technology Mesra considers CAT score for admission to MBA programme. However, it is not compulsory for the aspirants to have a valid CAT score. Aspirants can also get admission through MAT, XAT or CMAT scores. Selected candidates have to further pass the group discussion and personal interview round for final selection. 

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5\times 10^{-3}\right)\times \left(200\times 10^{-6}\right)}}$

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5\times 10^{2}Hz$ $\left[Taking\pi =\sqrt[]{10}\right]$

According to Young’s double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1\times 10^{-3}\times 3\times 10^{-5}}{5\times 10^{-2}}=60\times 10^{-8}m=600nm$

The admission portal for BTech courses at SRMCEM Lucknow started around February 2025. Moreover, the deadlines to apply for UPTAC counselling process are usually in the mid- to last week of July. The deadlines to apply for BTech courses are not explicitly mentioned on the official website of the college. However, candidates must track the entrance examination and counselling dates. 

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$

Time required = 4 × $t_{\frac{1}{2}}=20yrs$

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}\times \frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$