- A$324$
- ✓$450$
- C$234$
- D$110$
$ 2 \mathrm{x}-5 \mathrm{y}=20 $
$ 3 \mathrm{x}+\mathrm{my}=\mathrm{m} $
$ \Rightarrow \mathrm{y}=\frac{2 \mathrm{~m}-60}{2 \mathrm{~m}+15} $
$ \mathrm{y}<0 \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 30\right) $
$ \mathrm{x}=\frac{25 \mathrm{~m}}{2 \mathrm{~m}+15} $
$ \mathrm{x}<0 \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) $
$ \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) $
$ |\mathrm{A}|=2 \mathrm{~m}+15 $
$ \text { Now, } $
$ 8 \int_{-15}^0(2 \mathrm{~m}+15) \mathrm{dm}=8\left\{\mathrm{~m}^2+15 \mathrm{~m}\right\}_{\frac{-15}{2}}^0 $
$ \Rightarrow 8\left\{-\left(\frac{225}{4}-\frac{225}{2}\right)\right\} $
$ =8 \times \frac{225}{4}=450$
Now,
$ 8 \int_{\frac{-15}{2}}^0(2 \mathrm{~m}+15) \mathrm{dm}=8\left\{\mathrm{~m}^2+15 \mathrm{~m}\right\}_{\frac{-15}{2}}^0 $
$ \Rightarrow 8\left\{-\left(\frac{225}{4}-\frac{225}{2}\right)\right\} $
$ =8 \times \frac{225}{4}=450$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ 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 $3.$