MCQ
$\lambda $ ની કઈ કિંમત માટે નીચેના સમીકરણને ઉકેલ ના મળે.$\begin{vmatrix}x+y+z=6\\4x+\lambda y-\lambda z=0&\\3x+2y-4z=-8&\end{vmatrix}$
- ✓$3$
- B$ - 3$
- C$0$
- D$1$
$=\begin{vmatrix}1 & 1 & 1 \\4 & \lambda& -\lambda \\3 & 2 & -4\end{vmatrix}$
$\frac{C_{31}(-1)}{C_{32}(-1)}\begin{vmatrix}0 & 0 & 1 \\4+\lambda &2\lambda& -\lambda \\7&6&4\end{vmatrix}=0$
$\therefore 1(6(4+ \lambda)-14\lambda)=0$
$\therefore 24+6 \lambda-14\lambda=0$
$\therefore24-8\lambda=0$
$\therefore 24=8\lambda$
$\lambda=3$
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|
$X$ |
$0$ | $1$ | $2$ | $3$ | $4$ |
| $P(X)$ | $k$ | $2$ | $4k$ | $6k$ | $64$ |
નું મૂલ્ય....... $P (1< X <4 \mid X \leq 2)$ છે
$f(x) \rightarrow \frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)}, x<2$
$\quad\quad\quad\quad e^{\frac{\tan (x-2)}{x-[x]}}, \quad x>2$
$\quad\quad\quad\quad \mu \quad\quad\quad\quad x=2$
કે જ્યાં $[x]$ એ મહતમ પૃણાંક વિધેય છે. જો $f$ એ $x=2$ આગળ સતત હોય તો $\lambda+\mu$ ની કિમંત મેળવો.