- A$16$
- ✓$10$
- C$6$
- DNone of these
adj $A = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&4\end{array}} \right]$
$|adj\,A|\, = (4 \times 4) - ( - 3 \times - 2) = 16 - 6$
$|adj\,A|\,\, = 10.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $Face :$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| $P(F)$ | $0.2$ | $0.22$ | $0.11$ | $0.25$ | $0.05$ | $0.17$ |
The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is
$1.$ If $\mathrm{f}(-10 \sqrt{2})=2 \sqrt{2}$, then $\mathrm{f}^{\prime \prime}(-10 \sqrt{2})=$
$(A)$ $\frac{4 \sqrt{2}}{7^3 3^2}$ $(B)$ $-\frac{4 \sqrt{2}}{7^3 3^2}$ $(C)$ $\frac{4 \sqrt{2}}{7^3 3}$ $(D)$ $-\frac{4 \sqrt{2}}{7^3 3}$
$2.$ The area of the region bounded by the curves $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$, where $-\infty < \mathrm{a} < \mathrm{b} < -2$, is
$(A)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$
$(B)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$
$(C)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$
$(D)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$
$3.$ $\int_{-1}^1 g^{\prime}(x) d x=$
$(A)$ $2 g(-1)$ $(B)$ 0 $(C)$ $-2 g(1)$ $(D)$ $2 \mathrm{~g}(1)$
Give the answer question $1,2$ and $3.$
Define $S _i=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f( x ) \cdot g _i( x ) dx , i=1,2$
($1$) The value of $\frac{16 S _1}{\pi}$ is. . . . . .
($2$) The value of $\frac{48 S _2}{\pi^2}$ is. . . . .