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14. Given f(x) = $\{\begin{array}{ccc}\hfill 2x, if x<0\hfill & \hfill 0, if 0\le x\le 1\hfill & \hfill 4x, if x>1.\hfill \end{array}$

For (c) = c < 0,

f(c) = 2c.

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 2x = 2c = f(c)

So, f is continuous at x $\left|<\right|$ 0

For x = c > 1,

f(c) = 4c

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 4x = 4c = f(c)

So, f is continuous at x> 1.

For x = 0

L.H.L. = $\underset{x\to 0^{-}}{lim}$ f(x) = $\underset{x\to 0^{-}}{lim}$ . 2x = 2 (0) = 0

R.H.L. = $\underset{x\to 0^{+}}{lim}$ f(x) = $\underset{x\to 0^{+}}{lim}$ . 0 = 0.

f(0) = 0.

∴ L.H.L. = R.H.L. = f(0).

So, f is continuous at x = 0.

For x = 1.

L.H.L. = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ . 0 = 0

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ . 4x = 4 (1) = 4.

∴ L.H.L. $\cancel{=}$ R.H.L.

So, f is discontinuous at x = 1.

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13. $$Given, f(x) = $\{\begin{array}{ccc}\hfill 3 \pi 0\le x\le 1\hfill & \hfill 4 \pi 1<x<3\hfill & \hfill 5 \pi 3\le x\le 10.\hfill \end{array}$

For x = c such that $0\le c<1$

f(c) = 3

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 3 = 3 = f(c)

So, f is continuous in [0, 1].

For x = c = 1,

L.H.L. = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ 3 = 3.

R.H.L. $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ 4 = 4

∴ L.H.L $\cancel{=}$ R.H.L.

f is discontinuity at x = 1

for x = c such that $1<c<3.$

f(c) = 4

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 4 = 4 = f(c)

So, f is continuous in $x\in (1,3)$

For x = c = 3

L.H.L. $\underset{x\to 3^{-}}{lim}$ f(x) = $\underset{x\to 3^{-}}{lim}$ 4 = 4

R.H.L. $\underset{x\to 3^{+}}{lim}$ f(x) = $\underset{x\to 3^{+}}{lim}$ 5 = 5.

So, f is discontinuous at x = 3.

For x = c such that $3<c\le 10$

f (c) = 5.

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 5 = 5 = f(c)

So, f is continuous in $x\in (3,10]$

12. Given, f(x) = $\{\begin{array}{ll}x+5 if x\left|?\right|1\hfill & x-5 if x>1.\hfill \end{array}$

For x = c < 1.

F (c) = c + 5

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ f x + 5 = c + 5

∴ $\underset{x\to c}{lim}$ f(x) = f(c)

So, f is continuous at x $\left|<\right|$ 1.

For x = c > 1

F (c) = c 5

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ x 5 = c 5.

$\underset{x\to c}{lim}$ f(x) = f(c)

So, f is continuous at x $\left|>\right|$ 1.

For x = 1

L.H.L. = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ x + 5 = 1 + 5 = 6.

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ x 5 = 1 5 = 4.

L.H.L. $\cancel{=}$ R.H.L.

f is not continuous at x = 1

So, point of discontinuity of f is at x = 1.

Discuss the continuity of the function $f$ , where $f$ is defined by

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11. Given, f (x) = $\{\begin{array}{ll}{x}^{10}-1, if x\le 1\hfill & x^2, if x>1.\hfill \end{array}$

For x = c < 1.

f (c) = $\underset{x\to c}{lim}$ f (x) = c10 1.

So, f is continuous for x $\left|>\right|$ 1.

For x = c > 1.

f (c) = $\underset{x\to c}{lim}$ f (x) = c2

So, f is continuous for x $\left|>\right|$ 1.

For x = c = 1,

L.H.L = $\underset{x\to 1^{-}}{lim}$ f (x) = $\underset{x\to 1^{-}}{lim}$ x10 1 110 1 = 0.

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f (x) = $\underset{x\to 1^{+}}{lim}$ x2 = 12 = 1.

∴ L.H.L $\cancel{=}$ R.H.L.

So, f is not continuous at x = 1.

Hence, f has point of discontinuity at x = 1.

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10. Given f (x) = $\{\begin{array}{cc}\hfill x^3-3if x\left|?\right|2\hfill & \hfill x^2+1 if x>2\hfill \end{array}$

For x = c < 2,

f (c) = c3 3

$\underset{x\to c}{lim}$ f (x) = $\underset{x\to c}{lim}$ x3 3 = c3 3.

So f is continuous at x $\left|<\right|$ 2.

For x = c > 2

f (c) = x2 + 1 = c2 + 1

$\underset{x\to 2}{lim}$ f (x) = $\underset{x\to 2}{lim}$ x2 + 1 = c2 + 1 = f (c)

So, f is continuous at x $\left|>\right|$ 2.

For x = c = 2, f (2) = 23 3 = 8 3 = 5.

L.H.L. $\underset{x\to 2^{-}}{lim}$ f (x) = $\underset{x\to 2^{-}}{lim}$ x3 3 = 23 3 = 5.

R.H.L. $\underset{x\to 2^{+}}{lim}$ f (x) = $\underset{x\to 2^{+}}{lim}$ x2 + 1 = 22 + 1 = 5

∴ R.H.L. = L.H.L. = f (2).

So, f is continuous at x = 2

Hence f has no point of discontinuity.

Only 1 student was placed during the IISER Pune MSc placement drive who was awarded a package of INR 8 LPA. The table below states the same:

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