MCQ
The matrix $\left[ {\begin{array}{*{20}{c}}\lambda &{ - 1}&4\\{ - 3}&0&1\\{ - 1}&1&2\end{array}} \right]$is invertible, if
- A$\lambda \ne - 15$
- ✓$\lambda \ne - 17$
- C$\lambda \ne - 16$
- D$\lambda \ne - 18$
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$x+y+\sqrt{3} z=0$
$-x+(\tan \theta) y+\sqrt{7} z=0$
$x+y+(\tan \theta) z=0$
has non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in s} \theta$ is equal to
$f(x)= \begin{cases}n(1-2 n x) \text { if } 0 \leq x \leq \frac{1}{2 n}2 n(2 n x-1) \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n}4 n(1-n x) \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(n x-1) \text { if } \frac{1}{n} \leq x \leq 1\end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$ , then the maximum value of the function $f$ is