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$\begin{array}{l}Let\text{?}\text{?}\text{?}\text{?}I={\displaystyle {?}_0^{\raisebox{1ex}{$?$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}(1+tanx)dx?????(i)\\ ={\displaystyle {?}_0^{\raisebox{1ex}{$?$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}\left[1+tan\left(\frac{?}{4}?x\right)\right]dx?\{{\displaystyle {?}_0^{a}f(x)dx=}{\displaystyle {?}_0^{a}f(a?x)dx}\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}\left[1+\frac{tan\frac{?}{4}?tanx}{1+tan\frac{?}{4}tanx}\right]dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}·\left\{1+\frac{1?tanx}{1+tanx}\right\}dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}\left\{\frac{1+tanx+1?tanx}{1+tanx}\right\}dx\\ I={\displaystyle {?}_0^{\frac{?}{4}}log}\left(\frac{2}{1+tanx}\right)dx?????(ii)\\ Adding\left(i\right)\&\left(ii\right)weget,\\ I+I={\displaystyle {?}_0^{\frac{?}{4}}\left\{log(1+tanx)+log\left(\frac{2}{1+tanx}\right)\right\}}dx\\ I={\displaystyle {?}_0^{\frac{?}{4}}log}\left\{\underset{\_}{1+tanx}*\frac{2}{1+tanx}\right\}dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}2dx\\ ={[xlog2]}_0^{\frac{?}{4}}\\ =log2\left[\frac{?}{4}?0\right]=\frac{?}{4}log2\\ I=\frac{?}{8}log2.\end{array}$

33.  (c) JK mol^-1

$\begin{array}{l}LetI={\displaystyle {\int }_0^{1}x}\left(1-x^n\right)dx\\ ={\displaystyle {\int }_0^1(}1-x)\left[1-{\left(1-x\right)}^n\right]dx\left\{{\displaystyle {\int }_0^{a}f}(x)dx={\displaystyle {\int }_0^{a}f}(a-x)dx\right\}\\ ={\displaystyle {\int }_0^1(}1-x){(1-1+x)}^{n}dx\end{array}$

$\begin{array}{l}={\displaystyle {\int }_0^1(}1-x)x^{n}dx\\ ={\displaystyle {\int }_0^1\left(x^n-x^{n+1}\right)}dx\\ ={\left[\frac{x^{n+1}}{n+1}\right]}_0^1-{\left[\frac{x^{n+1+1}}{n+1+1}\right]}_0^1\\ =\left[\frac{1^{n+1}}{n+1}-\frac{0^{n+1}}{n+1}\right]-\left[\frac{1^{n+2}}{n+2}-\frac{0^{n+2}}{n+2}\right]\\ =\frac{1}{n+1}-\frac{1}{n+2}=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}\end{array}$

An abstract noun is an idea, state, quality, or feeling that is intangible in nature. By being intangible, abstract nouns cannot be seen, touched, or physically measured, but can be understood or felt emotionally.

For example, love, anger, beauty, anxiety, happiness, knowledge, wisdom, and hate.

$\begin{array}{l}LetI={\displaystyle {\int }_2^8|}x-5|dx.\left\{\begin{array}{l}x–5ifx–5>0,x>5\\ –\left(x–5\right)ifx–5<0,x<5\end{array}\right\}\\ ={\displaystyle {\int }_2^5-}(x-5)dx+{\displaystyle {\int }_5^8(}x-5)dx\\ =-{\left[\frac{x^2}{2}-5x\right]}_2^5+{\left[\frac{x^2}{2}-5x\right]}_5^8\\ =-\left[\left(\frac{5^2}{2}-5\times 5\right)-\left(\frac{2^2}{2}-5\times 2\right)\right]+\left[\left(\frac{8^2}{2}-5\times 8\right)-\left(\frac{5^2}{2}-5\times 5\right)\right]\\ =-\left[\frac{25}{2}-25-2+10\right]+[\left(32-40-\frac{25}{2}+25\right].\\ =-\frac{25}{2}+17+17-\frac{25}{2}\\ \begin{array}{ll}= 34 – 25\hfill & = 9\hfill \end{array}\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_{-5}^5|}x+2|dx\\ ={\displaystyle {\int }_{-5}^{-2}|}x+2|dx+{\displaystyle {\int }_{-2}^5|}x+2|dx\end{array}$

? |x + 2| = x + 2 if if x + 2 > 0 => x> –2

– (x + 2) if x + 2 < 0 => x< –2.

$\begin{array}{l}={\left[-\frac{x^2}{2}-2x\right]}_{-5}^{-2}+{\left[\frac{x^2}{2}+2x\right]}_{-2}^5\\ =-\left[\left(\frac{{(-2)}^2}{2}+2(-2)\right)-\left(\frac{{\left(-5\right)}^2}{2}+2(-5)\right)\right]+\left[\left(\frac{5^2}{2}+2\times 5\right)-\left(\frac{{(-2)}^2}{2}+2(-2)\right)\right]\\ =-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]\\ =-8+\frac{25}{2}+\frac{25}{2}+12\\ ==–8+25+12=29.\end{array}$

No, students cannot pursue the Shri Guru Ram Rai University DPharma course without completing Class 12. The admission committee has set the criteria that only those candidates who have successfully passed Class 12 examinations with Science as the stream are eligible to apply. Moreover, students with any other qualification approved by the Pharmacy Council of India are also considered for admissions.

The eligibility criteria required for the Creative Techno College is to pass Class 12 from a recognised board. The application process is conducted in online mode. Streams such as Science and Maths are compulsory in Class 12 for admission. Interested candidates can apply online on the official website for the desired course. 

Yes, the ARSD College BCom (Hons) course admissions do require candidates to have Maths as one of the compulsory subjects in their Class 12 board exams along with the minimum aggregate requirement. Candidates who fulfill the eligibility criteria need to clear the CUET-UG entrance exam to gain admissions in ARSD College, BCom (Hons) courses. 

$\begin{array}{l}Let,I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{5}xdx}{si{n}^{5}x+co{s}^{5}x}}-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^5\left(\frac{\pi }{2}-x\right)}{si{n}^5\left(\frac{\pi }{2}-x\right)+co{s}^5\left(\frac{\pi }{2}-x\right)}}dx\\ I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{5}x}{co{s}^{5}x+si{n}^{5}x}}dx-----(ii)\end{array}$

$\begin{array}{l}Adding\left(i\right)\&\left(ii\right)weget,\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{co{s}^{5}x}{si{n}^{5}x+co{s}^{5}x}+\frac{si{n}^{5}x}{co{s}^{5}x+si{n}^{5}x}\right\}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{5}x+si{n}^{5}x}{si{n}^{5}x+co{s}^{3}x}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x=\frac{\pi }{2}\\ I=\frac{\pi }{4}.\end{array}$

3.  (ii) zero

$\begin{array}{l}Let,I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}}dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}}dx-----(ii)\\ \left(i\right)+\left(ii\right)\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}+\frac{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}\right\}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}\right\}}dx\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x=\frac{\pi }{2}\\ I=\frac{\mathrm{\pi /4}}{}\end{array}$

The total tuition fee to pursue DPharma at Shri Guru Ram Rai University is INR 2.1 lakh. Other than this, the fee structure includes other charges such as admission fees, hostel fees, development & training fees, etc. Candidates who wish to know the complete fee structure can visit the official website of the institute.

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

2.  (iii) q = 0

Kindly go through the solution