MCQ
The matrix $\left( {\begin{array}{*{20}{c}}1&a&2\\1&2&5\\2&1&1\end{array}} \right)$ is not invertible, if $‘a’ $ has the value
- A$2$
- ✓$1$
- C$0$
- D$-1$
$ \Rightarrow $ $1\,(2 - 5) - a(1 - 10) + 2(1 - 4) = 0$
$ \Rightarrow $ $ - 3 + 9a - 6 = 0 $
$\Rightarrow a = 1$.
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Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
| $x=-1$ | $x=0$ | $x=2$ | |
| $f(x)$ | $3$ | $6$ | $0$ |
| $g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$