Class 12th

$\begin{array}{l}f\left(x\right)={\displaystyle \int 4x^3}–\frac{3}{x^4}dx\\ =4\cdot \frac{x^4}{4}-3\cdot \frac{x^{-3}}{-3}+\; C\\ =x^4+\frac{1}{x^3}+\; C\end{array}$

Now,  $f\left(2\right)=0 \therefore f\left(2\right) =2^4+\frac{1}{2^3}+C=0$

$\begin{array}{l}=16+\frac{1}{8}+c=0\\ =c=-\left(16+\frac{1}{8}\right)\\ =c=-\frac{129}{8}\end{array}$

Therefore, correct answer is A.

$\frac{x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{x^{1/2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C$

$=\frac{2}{3}{x}^{3/2}+2x^{1/2}+$ C

? The correct Answer is (C)

$\begin{array}{l}\int \left(\frac{2}{co{s}^{2}x}-\frac{3sinx}{co{s}^{2}x}\right)dx\\ \begin{array}{l}={\displaystyle \int \left(2se{c}^{2}x–3tanx.secx\right)dx}\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 2se{c}^{2}x}.dx–{\displaystyle \int 3tan}x.secx.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=2tanx– 3 secx+C\hfill \end{array}\end{array}$

$\begin{array}{l}\int \frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}dx\\ =\int \frac{si{n}^{2}x}{co{s}^{2}x}dx\\ \begin{array}{l}={{\displaystyle \int tan}}^{2}xdx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(se{c}^{2}x–1\right)}dx={\displaystyle \int se{c}^2}xdx–{\displaystyle \int 1}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=tanx–x+C\hfill \end{array}\end{array}$

$={\displaystyle \int \left(se{c}^{2}x+secxtanx\right)dx}$

$={\displaystyle \int se{c}^{2}x.dx}+{\displaystyle \int secxtanxdx}$

$=tanx+secx+C$

$\begin{array}{l}={\displaystyle \int 2x^2}.dx–{\displaystyle \int 3sinx}.dx+{\displaystyle \int \mathrm{5\surd x}}dx\\ =\frac{2x^3}{3}+3cosx+\frac{5x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C\\ =\frac{2x^3}{3}+3cosx+\frac{10x^{3/2}}{3}+\; C\end{array}$

$={\displaystyle \int 2x}dx–{\displaystyle \int 3cosx}+e^{x}dx$

$\begin{array}{l}=\frac{2x^2}{2}-3sinx+e^x+\; C\\ =x^2-3sinx+e^x+C\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(3x^2.x^{1/2}+2x.x^{1/2}+3x^{1/2}\right)}.dz\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(3x^{5/2}+2x^{3/2}+3x^{4/2}\right)}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 3x^{5/2}}.dx+{\displaystyle \int 2x^{3/2}}.dx+{\displaystyle \int 3x^{1/2}}.dx\hfill \end{array}\\ =3·\frac{x^{7/2}}{7/2}+\frac{2·x^{5/2}}{5/2}+\frac{3·x^{3/2}}{3/2}+\; C\\ =\frac{6·x^{7/2}}{7}+\frac{4x^{5/2}}{5}+\frac{6}{3}{x}^{3/2}+\; C\\ =\frac{6x^{7/2}}{7}+\frac{4·x^{5/2}}{5}+2x^{3/2}+\; C.\end{array}$

$\begin{array}{l}\int (\surd x-x·x^{1/2})dx\\ \int (\surd x-x^{3/2})·dx\\ \int x^{1/2}·dx-\int x^{3/2}·dx\\ \frac{x^{3/2}}{3/2}-\frac{x^{5/2}}{5/2}+C\\ \frac{2}{3}{x}^{3/2}-\frac{2}{5}{x}^{5/2}+\end{array}$ c

As per the National Law University Kochi admission guidelines, the eligibility criteria for BA LLB (Hons) requires candidates must have a 45% in their Higher Secondary School Examination (10+2) system or an equivalent examination. In the case of SC/ST/Person with Disability, 42% is also eligible. Therefore, candidates can get admission to NLU Kochi BA LLB (Hons) with 60& in Class 12th.