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Let $f(x)=x^{-\left(\frac{1}{3}\right)}$ and $A$ denote the area of region bounded by $f(x)$ and the X-axis, when $x$ varies from $-1$ to $1$. Which of the following statements is/are TRUE?

  1. $f$ is continuous in $[-1, 1]$
  2. $f$ is not bounded in $[-1, 1]$
  3. $A$ is nonzero and finite
  1. II only
  2. III only
  3. II and III only
  4. I, II and III
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7 Answers

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49 votes
  1. $f$ is continuous in $[-1,1]$
    Given, $f(x) = \frac{1}{\sqrt[3]x}$
    We need to check continuity at $x = 0.$
    Left hand limit $= \displaystyle \lim_{x \to 0^-} \frac{1}{\sqrt[3]x} = \lim_{h \to 0^-}\frac{1}{\sqrt[3]{0-h}} = \lim_{h \to 0^-}\frac{-1}{\sqrt[3]h} = – \infty$
    Right hand limit $= \displaystyle \lim_{x \to 0^+} \frac{1}{\sqrt[3]x} = \lim_{h \to 0^+}\frac{1}{\sqrt[3]{0+h}} = \lim_{h \to 0^+}\frac{1}{\sqrt[3]h} =  +\infty$
    $\therefore \textsf{LHL} \neq \textsf{RHL}$
    So, $f$ is not continuous in $[-1,1].$ Statement I is FALSE.
  2. $f$ is not bounded in $[-1,1]$
    Since at $x=0,f(x)$ goes from $-\infty$ to $+\infty,$ $f$ is not bounded at $[-1,1].$ Statement II is TRUE.
  3. $A$ is non zero and finite.
    $A = \int_{-1}^0 x^{\frac{-1}{3}} dx +  \int_{0}^1 x^{\frac{-1}{3}} dx$
    $\qquad =  \dfrac{3}{2} [x^{\frac{2}{3}}]_{-1}^0 + \dfrac{3}{2} [x^{\frac{2}{3}}]_{0}^1$
    $\qquad = \frac{3}{2} + \frac{3}{2} = 3.$
    $\therefore A$ is non-zero and finite. Statement III is TRUE.

Answer: C

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Answer: C

I. False.
II. True.
III. True. An area is always positive, while the definite integral might be composed of several regions, some positive and some negative. A definite integral gets you the net area, because any part of the graph that is below the $x$-axis will give you a negative area. So, a definite integral is not necessarily the area under the curve, but the value of the area above the $x$-axis less the area under the $x$-axis. So, A is non-zero and finite.
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S1 : is False

S2: is True

S3: Area is finite but don't know whether we have to take signed Area or unsigned Area.

because signed Area is 0 and unsigned Area is 3 sq unit

Ans A or C.
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$f(x)=\frac{1}{x^{\frac{1}{3}}}$

Clearly it is visible that f(x) won't be defined at x=0($-1 \leq0 \leq 1$).

A function $f(x)$ is said to be continuous on a closed interval [a,b] iff

(i)f is continuous on the open interval (a,b)

(ii)$\lim_{x \to a^+}f(x)=f(a)$

(iii)$lim_{x \to b^-}f(x)=f(b)$

In words, f(x) is continous on [a,b] iff it is continous on (a,b) and it is continous at $a$ from the right and at $b$ from the left.

Since, the function f(x) here fails to satisfy point (1), it is not continous in closed interval [-1,1].For other intervals, you might need to check it.

A bounded function is one, which can be bounded if its set of values is bounded.

In short, it's graph can be bounded by using two lines y=k1 and y=k2 for some constants K1 and K2.

Here no need to check with graph, as $x \to 0$, $f(x) \to \infty$. Not Bounded ofcourse.

From question description it is given that

$A=\int_{-1}^{1}f(x)dx.$

Since, $f(x)$ is not defined at x=0, you need to integrate it piecewise.

$A=\int_{-1}^{0}f(x)dx+\int_{0}^{1}f(x)dx=3$

So, III is true.A is non-zero and finite.

So, option II and III true. (C)

 

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