MCQ
$\int_{}^{} {{e^{{x^2}}}x\;dx} $ is equal to
- A${e^{{x^2}}}$
- ✓$\frac{1}{2}{e^{{x^2}}}$
- C$2{e^{{x^2}}}$
- D$\frac{{{e^{{x^2}}} - {x^2}}}{2}$
(Put ${x^2} = t \Rightarrow dt = 2x\,dx)$.
$ = \frac{1}{2}\int_{}^{} {{e^t}dt} = \frac{1}{2}{e^t} = \frac{1}{2}{e^{{x^2}}}$,
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Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
| $x=-1$ | $x=0$ | $x=2$ | |
| $f(x)$ | $3$ | $6$ | $0$ |
| $g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$
$x-2 y+3 z=9$
$2 x+y+z=b$
$x-7 y+a z=24$
has infinitely many solutions, then $a - b$ is equal to