MCQ
If the vectors $3i+\lambda \,j+k$ and $2i-j+8k$ are perpendicular, then $\lambda $ is
- A$-14$
- B$7$
- ✓$14$
- D$1/7$
$\because a \bot \,b$ $\therefore a\,.\,b = 0$
$(3i + \lambda \,j + k)\,.\,(2i - j + 8k) = 0$
$a,\,b,\,c$ $ \Rightarrow \lambda = 14.$
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$f\left( {x + a} \right) = \frac{1}{2} + \sqrt {f\left( x \right) - {f^2}\left( x \right)}$ a is a real constant, then $f(x)$ must be
Statement $1:$ $\mathop {\lim }\limits_{x \to {2^ - }} \,f(x)$ exists.
Statement $2:$ $f$ is continuous at $x = 2.$