MCQ
The set of all values of $\lambda $ for which the system of linear equations $x - 2y - 2z = \lambda x$ ; $x + 2y + z = \lambda y$ ; $-x - y = \lambda z$ has non zero solutions.
  • is a singleton
  • B
    contains exactly two elements
  • C
    is an empty set
  • D
    contains more than two elements

Answer

Correct option: A.
is a singleton
a
$\left| {\begin{array}{*{20}{c}}
{\lambda  - 1}&2&2\\
1&{2 - \lambda }&1\\
1&1&1
\end{array}} \right| = 0$

$ \Rightarrow {\left( {\lambda  - 1} \right)^3} = 0 \Rightarrow \lambda  = 1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The position vectors of the points  $ A, B, C $ are $(2i + j - k),$ $(3i - 2j + k)$ and $(i + 4j - 3k)$ respectively. These points
If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$
The area of the region bounded by the ellipse $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{16}=1$ is :
If $f(x)=\int_0^x e^{t^2}(t-2)(t-3) d t$ for all $x \in(0, \infty)$, then

$(A)$ $f$ has a local maximum at $x=2$

$(B)$ $f$ is decreasing on $(2,3)$

$(C)$ there exists some $c \in(0, \infty)$ such that $f ^{\prime \prime}( c )=0$

$(D)$ $f$ has a local minimum at $x=3$

Let $f(x) = (x^2 - 1)^n (x^2 + x + 1)$ then $f(x)$ has local extremum at $x = 1$ when
Let $f:[0,1] \rightarrow R$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)=f(1)=0$ and satisfies $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x, x \in[0,1]$

$1.$ 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 $2.$

Let $f (x) = ax^2 - b | x |$, where $a$ and $b$ are constants. Then at $x = 0, \,f (x) $has
The derivative of the function $\cot^{-1}\Big|(\cos2\text{x})^{\frac{1}{2}}\Big|\text{ at }\text{x}=\frac{\pi}{6}$ is:
Let $g _i:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R , i =1,2$, and $f:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R$ be functions such that $g_1(x)=1, g_2(x)=|4 x-\pi|$ and $f(x)=\sin ^2 x$, for all $x \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right]$

Define $S _i=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f( x ) \cdot g _i( x ) dx , i=1,2$

($1$) The value of $\frac{16 S _1}{\pi}$ is. . . . . .

($2$) The value of $\frac{48 S _2}{\pi^2}$ is. . . . .

The integral $\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right) \mathrm{dx}$ is equal to: