- A$\frac{1}{3}(x + \log x) + c$
- B$\frac{1}{3}{(x + \log x)^2} + c$
- ✓$\frac{1}{3}{(x + \log x)^3} + c$
- DNone of these
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$f(x)=\left\{\begin{array}{ll} \min \left\{(x+6), x^{2}\right\}, & -3 \leq x \leq 0 \\ \max \left\{\sqrt{x}, x^{2}\right\}, & 0 \leq x \leq 1 \end{array}\right.$
If the area bounded by $y = f ( x )$ and $x$ -axis is $A,$ then the value of $6 A$ is equal to ....... .
$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
is $\frac{\mathrm{m}}{\sqrt{\mathrm{n}}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then the value of $\mathrm{m}+\mathrm{n}$ equals.
$(a-c) x^2+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix
$\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right]$
is singular. Then the value of
$\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$