$\vec{\beta}=(4 \lambda-2) \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}$
$\bar{\alpha}$ and $\bar{\beta}$ are collinear
$\left|\begin{array}{cc}{\lambda-2} & {1} \\ {4 \lambda-2} & {3}\end{array}\right|=0$
$3 \lambda-6-4 \lambda+2=0$
$-\lambda-4=0$
$\lambda=-4$
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$1.$ 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 $3.$