- $\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
- $\frac{\text{x}^2}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
- $2\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
- none of these.
Solution:
$\text{I}=\int\text{x}\sec\text{x}^2\text{ dx}$
Put $\text{x}^2=\text{t}$
$=\text{x}=\sqrt{\text{t}}$
$2\text{xdx}=\text{dt}$
$\text{xdx}=\frac{\text{dt}}{2}$
$\text{I}=\int\sec\text{t}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\log(\sec\text{t}+\tan\text{t})+\text{C}$
$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$-x+2 y+5 z=b_1$
$2 x-4 y+3 z=b_2$
$x-2 y+2 z=b_3$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each$\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ $\in$ $S$ ?
$(A)$ $x+2 y+3 z=b_1, 4 y+5 z=b_2$ and $x+2 y+6 z=b_3$
$(B)$ $x+y+3 z=b_1, 5 x+2 y+6 z=b_2$ and $-2 x-y-3 z=b_3$
$(C)$ $-x+2 y-5 z=b_1, 2 x-4 y+10 z=b_2$ and $x-2 y+5 z=b_3$
$(D)$ $x+2 y+5 z=b_1, 2 x+3 z=b_2$ and $x+4 y-5 z=b_3$
Which one is not a requirement of a binomial dstribution?