- A$343$
- B$125$
- C$64$
- ✓$216$
$ \left(2 \alpha^2-3 \alpha\right)=\alpha $
$ \alpha=0,2 \text { (accept) }$
Now, $2 \alpha^2-\alpha \beta=3 \alpha$
$ \alpha=2 \quad \beta=1 $
$ |A B|=|A \operatorname{cof}(A)|=|A|^3$
$A=\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4\end{array}\right|=6-2(9)+3(6)=6$
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$1.$ Which of the following is true for $0 < x < 1$ ?
$(A)$ $0 < $ f(x) $ < \infty$
$(B)$ $-\frac{1}{2} < f(x) < \frac{1}{2}$
$(C)$ $-\frac{1}{4} < f(x) < 1$
$(D)$ $-\infty < $ f $($ x $) < 0$
$2.$ If the function $e^{-x} f(x)$ assumes its minimum in the interval $[0,1]$ at $x=\frac{1}{4}$, which of the following is true?
$(A)$ $f^{\prime}(x)$
$(B)$ $f^{\prime}(x)>f(x), 0$
$(C)$ f $^{\prime}(x)$
$(D)$ $f^{\prime}(x)$
Give the answer question $1$ and $2.$
$\text{f(x)}=1,\ \text{f(y)}\neq1,\ \text{f(z)}\neq2.$
The value of f-1(1)is: