MCQ
The area enclosed between the parabolas ${y^2} = 4x$ and ${x^2} = 4y$ is
- A$\frac{{14}}{3}$ sq. unit
- B$\frac{3}{4}$ sq. unit
- C$\frac{3}{{16}}$ sq. unit
- ✓$\frac{{16}}{3}$ sq. unit
The given equations may be written as
$y = 2\sqrt x $ and $y = \frac{{{x^2}}}{4}.$
We know that area enclosed by the parabolas
$ = \int_{\,0}^{\,4} {2\sqrt x \,dx - } \int_{\,0}^{\,4} {\frac{{{x^2}}}{4}dx = \frac{{32}}{3} - \frac{{16}}{3} = \frac{{16}}{3}} \,\, sq. \,unit$.
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$f(x)=e^{x-1}-e^{-|x-1|} \text { and } g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right) \text {. }$ Then the area of the region in the first quadrant bounded by the curves $y=f(x), y=g(x)$ and $x=0$ is
$I.$ Adifferentiable function $' f '$ with maximum at $x = c$ ==> $ f "(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$. $\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx} $
$IV.$ If $f (x)$ has a maxima at $x = c$ , then $'f '$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \rightarrow 0$ for $h > 0.$ Now indicate the correct alternative.