Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(a)

Question 1.

Examine the continuity of the following functions at indicated points.

(i) f(x) = \(\left\{\begin{array}{cl}

\frac{x^2-a^2}{x-a} & \text { if } x \neq a \\

a & \text { if } x=a

\end{array}\right.\) at x = a

Solution:

(ii) f(x) = \(\left\{\begin{aligned}

\frac{\sin 2 x}{x} & \text { if } x \neq 0 \\

2 & \text { if } x=0

\end{aligned}\right.\) at x = 0

Solution:

(iii) f(x) = \(\begin{cases}(1+2 x)^{\frac{1}{x}} & \text { if } x \neq 0 \\ e^2 & \text { if } x=0\end{cases}\) at x = 0

Solution:

(iv) f(x) = \(\left\{\begin{array}{l}

x \sin \frac{1}{x} \text { if } x \neq 0 \\

0

\end{array}\right.\) at x = 0

Solution:

(v) f(x) = \(\left\{\begin{array}{l}

\frac{x^2-1}{x-1} \text { if } x \neq 1 \\

2

\end{array}\right.\) at x = 1

Solution:

(vi) f(x) = \(\begin{cases}\sin \frac{1}{x} & \text { if } x \neq a \\ 0 & \text { if } x=0\end{cases}\) at x = 0

Solution:

(vii) f(x) = [3x + 11] at x = –\(\frac{11}{3}\)

Solution:

(viii) f(x) = \(\left\{\begin{array}{l}

\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1} \text { if } x \neq 0 \\

0

\end{array}\right.\) at x= 0

Solution:

(ix) f(x) = \(\left\{\begin{array}{l}

\frac{1}{x+[x]} \text { if } x<0 \\

-1 \quad \text { if } x \geq 0

\end{array}\right.\) at x = 0

Solution:

because [- h]is the greatest integer not exceeding – h

and so [- h ] = – 1

As L.H.L. = R.H.L. = f(0)

f(x) is cntinuous at x = 0.

(x) f(x) = \(\begin{cases}\frac{|x|}{x} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}\) at x = 0

Solution:

(xi) f(x) = \(\left\{\begin{array}{l}

2 x+1 \text { if } x \leq 0 \\

x \quad \text { if } 0<x<1 \\

2 x-1 \text { if } x \geq 1

\end{array}\right.\) at x = 0, 1

Solution:

(xii) f(x) = \(\left\{\begin{array}{l}

\frac{1}{e^{\frac{1}{x}}-1} \text { if } x>0 \\

0

\end{array} \text { if } x \leq 0\right.\) at x = 0

Solution:

(xiii) f(x) = sin\(\frac{\pi[x]}{2}\) at x = 0

Solution:

(xiv) f(x) = \(\frac{g(x)-g(1)}{x-1}\) at x = 1

Solution:

g(x) = |x – 1|

Then g(1) = |1 – 11| = 0

Now f(1) = \(\frac{g(1)-f(1)}{1-1}\) = 0/0

which we cannot determine.

Hence f(x) is discontinuous at x = 1.

Question 2.

If a function is continuous at x = a, then find

(i) \(\lim _{h \rightarrow 0}+\frac{1}{2}\{f(a+h)+f(a-h)\}\)

(ii) \(\lim _{h \rightarrow 0}+\frac{1}{2}\{f(a+h)-f(a-h)\}\)

Solution:

Question 3.

Find the value ofa such that the function f defined by \(\begin{cases}\frac{\sin a x}{\sin x} & \text { if } x \neq 0 \\ \frac{1}{a} & \text { if } x=0\end{cases}\)

is continuous at x = 0.

Solution:

Question 4.

If f(x) = \(\left\{\begin{array}{l}

a x^2+b \text { if } x<1 \\

1 \quad \text { if } x=1 \\

2 a x-b \text { if } x>1

\end{array}\right.\)

is continuous at x = 1, then find a and b.

Solution:

Let f(x) be continuous at x = 1

Then L.H.L. = R.H.L. = f(1)

Question 5.

Show that sin x is continuous for every real x.

Solution:

Let f(x) = sin x

Consider the point x = a, where ‘a’ is any real number.

Then f(a) = sin a

Thus L.H.L. = R.H.L. = f(a)

Hence f(x) = sin x is continuous for every real x.

(Proved)

Question 6.

Show that the function f defined by \(\left\{\begin{array}{l}

1 \text { if } x \text { is rational } \\

0 \text { if } x \text { is irrational }

\end{array}\right.\) is discontinuous ∀ ≠ 0 ∈ R.

Solution:

Consider any real point x = a

If a is rational then f(a) = 1.

Again lim_{x→a+}f(x) = lim_{h→0}f(a + h)

which does not exist because a + h may be rational or irrational

Similarly lim_{x→a-}f(x) does not exist.

Thus f(x) is discontinuous at any rational point. Similarly we can show that f(x) is discontinuous at any irrational point.

Hence f(x) is discontinuous for all x ∈ R

(Proved)

Question 7.

Show that the function f defined by f(x) = \(f(x)=\left\{\begin{array}{l}

x \text { if } x \text { is rational } \\

-x \text { if } x \text { is irrational }

\end{array}\right.\)

is continuous at x = 0 and discontinuous ∀ x ≠ 0 ∈ R.

Solution:

f(0) = 0

L.H.L. = lim_{x→0}f(x) = lim_{h→0}f(-h)

= \(\lim _{h \rightarrow 0} \begin{cases}-h & \text { if } h \text { is rational } \\ h & \text { if } h \text { is irrational }\end{cases}\) = 0

Similarly R.H.L. = 0

Thus L.H.L. = R.H.L. = f(0)

Hence f(x) is continuous at x = 0.

We can easily show that f(x) is discontinuous at all real points x ≠ 0.

Question 8.

Show that the function f defined by

f(x) = \(\left\{\begin{array}{l}

x \text { if } x \text { is rational } \\

0 \text { if } x \text { is irrational }

\end{array}\right.\)

is discontinuous everywhere except at x = 0.

Solution:

f(0) = 0

L.H.L. = lim_{h→0}f(-h)

= lim_{h→0}\(\begin{cases}-h & \text { if }-h \text { is rational } \\ 0 & \text { if }-h \text { is irrational }\end{cases}\)

Similarly R.H.L. = 0

Thus L.H.L. = R.H.L = f(0)

Hence f(x) is continuous at x = 0.

Let a be any real number except 0.

If a is rational then f(a) = a.

L.H.L. = lim_{h→0}f(a – h) which does not exist because a – h may be rational or may be irrational.

Similarly R.H.L. does not exist.

Thus f(x) is discontinuous at any rational point x – a ≠ 0.

Similarly f(x) is discontinuous at any irrational point.

Hence f(x) is discontinuous everywhere except at x = 0.

(Proved)

Question 9.

Show that f(x) = \(\begin{cases}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}\) is continuous at x = 0.

Solution:

Refer to No. 1(iv) of Exercise – 7(a).

Question 10.

Prove that e^{x} – 2 = 0 has a solution between 0 and 1. [Hints: Use continuity of e^{x}– 2 and fact – 2]

Solution:

∴ f(x) is continuous in [0, 1]

f(0). f(1) = (-1) (e – 2) < 0

∴ f(x) has a zero between 0 and 1

i.e. e^{x} – 2 = 0 has a solution between 0 and 1

Question 11.

So that x^{5} + x +1 = 0 for some value of x between -1 and 0.

Solution:

Let f(x) = x^{5} + x + 1 and any a ∈ (-1, 0)

f(a) = a^{5} + a + 1

= -1 = f(-1)

∴ f is continuous on [-1, 0]

But f(-1) f(0) = 1 × -1 < 0

∴ f has a zero between -1 and 0

⇒ x5 + x + 1 = 0 for some value of x between -1 and 0.