MCQ
If $F(x) = \int_{{x^2}}^{{x^3}} {\log t\,dt,\,\,(x > 0),} $ then $F'(x) = $
- ✓$(9{x^2} - 4x)\log x$
- B$(4x - 9{x^2})\log x$
- C$(9{x^2} + 4x)\log x$
- DNone of these
Applying Leibnitzaes theorem,
$F\,'(x) = \log {x^3}.\frac{d}{{dx}}{x^3} - \log {x^2}.\frac{d}{{dx}}{x^2}$
$ = 3\log x \cdot 3{x^2} - 2\log x \cdot 2x$
$ = (9{x^2} - 4x)\log x$.
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Statement $-I$ : Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{r}}=0$ is of magnitude $\sqrt{10}$.
Statement $-II$ : In a triangle $A B C, \cos 2 A+\cos 2 B$ $+\cos 2 \mathrm{C} \geq-\frac{3}{2}$
$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to