- A$A. P.$
- ✓$G. P.$
- C$H. P.$
- DNone of these
= $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\0&0&{ - (a{\alpha ^2} + 2b\alpha + c)}\end{array}\,} \right|$, by ${R_3} \to {R_3} - \alpha {R_1} - {R_2}$
= $a\,\{ - c(a{\alpha ^2} + 2b\alpha + c) - 0\} - b\{ - b(a{\alpha ^2} + 2b\alpha + c) - 0\} $
by expanding along ${C_1}$
$ = ({b^2} - ac)\,(a{\alpha ^2} + 2b\alpha + c)$
Thus, $\Delta = 0$, if either ${b^2} - ac = 0$ or $a{\alpha ^2} + 2b\alpha + c = 0$
i.e., $a,b,c$ in $G.P.$ or $a{\alpha ^2} + 2b\alpha + c = 0$.
Trick: Put $\alpha = 0$, then the determinant
$\left| {\,\begin{array}{*{20}{c}}a&b&b\\b&c&c\\b&c&0\end{array}\,} \right|\, = \,\left| {\,\begin{array}{*{20}{c}}a&b&0\\b&c&0\\b&c&{ - c}\end{array}\,} \right|\, = \, - c(ac - {b^2}) = 0$.
Hence the result.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x)=\left\{\begin{array}{rc}x^5+5 x^4+10 x^3+10 x^2+3 x+1, & x<0 \\ x^2-x+1, & 0 \leq x<1 \\ \frac{2}{3} x^3-4 x^2+7 x-\frac{8}{3}, & 1 \leq x<3 \\ (x-2) \log _e(x-2)-x+\frac{10}{3}, & x \geq 3\end{array}\right.$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x =1$ $(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$ $(4)$ $f^{\prime}$ is $NOT$ differentiable at $x =1$