CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 07

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 7. The Continuity And Differentiability questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Continuity And Differentiability concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Continuity And Differentiability worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Continuity And Differentiability chapter and other subjects too. Use them for better understanding of the subjects..

Continuity & Differentiability

Question. Find the value of 'a' and 'b' so that f(x) is continues at x = 4. \[ f(x) = \begin{cases} \frac{x-4}{|x-4|} + a & ; \ x < 4 \\ a + b & ; \ x = 4 \\ \frac{x-4}{|x-4|} + b & ; \ x > 4 \end{cases} \]
Answer: Redefining the given f(x) \[ f(x) = \begin{cases} \frac{x-4}{-(x-4)} + a & ; \ x < 4 \\ a+b & ; \ x = 4 \\ \frac{x-4}{x-4} + b & ; \ x > 4 \end{cases} \] since \( x < 4 \dots |x - 4| = -(x - 4) \)
and \( x > 4 \dots |x - 4| = (x - 4) \)
\( \Rightarrow f(x) = \begin{cases} -1 + a & ; \ x < 4 \\ a + b & ; \ x = 4 \\ 1 + b & ; \ x > 4 \end{cases} \)
LHL = \( (-1 + a) \)
\( \therefore \text{LHL} = -1 + a \)
RHL = \( (1 + b) \)
\( \therefore \text{RHL} = 1 + b \)
and \( f(4) = a + b \)
since f(x) is continuous at \( x = 4 \)
\( \therefore \text{LHL} = \text{RHL} = f(4) \)
\( \Rightarrow -1 + a = 1 + b = a + b \)
consider \( 1 + b = a + b \)
\( a = 1 \)
and \( -1 = a = a + b \)
\( b = -1 \)
\( \therefore f(x) \) is continuous at \( x = 4 \) for \( a = 1 \) & \( b = -1 \). Ans.

Question. Find value of 'k' so that f(x) is continues at x = 0. \[ f(x) = \begin{cases} \frac{1-\cos(kx)}{x\sin x} & ; \ x \neq 0 \\ \frac{1}{2} & ; \ x = 0 \end{cases} \]
Answer: LHL = \( \left[\frac{1-\cos(kx)}{x\sin x}\right] \)
put \( x = 0 - h = -h \) and \( h \to 0 \)
LHL = \( \left[\frac{1-\cos(-kh)}{(-h)\sin(-h)}\right] \)
LHL = \( \left[\frac{1-\cos(kh)}{h\sin h}\right] \dots\dots \{\because \cos(-x) = \cos x, \sin(-x) = \sin x\} \)
\( = \left[\frac{2\sin^2\left(\frac{kh}{2}\right)}{h\sin h}\right] \)
\[ = \left[ \frac{\frac{2\sin^2\left(\frac{kh}{2}\right)}{\frac{k^2h^2}{4}} \times \frac{k^2h^2}{4}}{\frac{\sin h}{h} \times h \times h} \right] \]
\[ = \left[ \frac{\frac{\sin^2(kh/2)}{\frac{k^2h^2}{4}}}{\frac{\sin h}{h}} \right] \times \frac{2k^2}{4} \dots\dots \left\{ \lim_{x \to 0}\left(\frac{\sin x}{x}\right) = 1 \left(\frac{\sin^2 x}{x^2} = 1\right) \right\} \]
\( = \frac{1}{1} \times \frac{k^2}{2} = \frac{k^2}{2} /> LHL = \( \frac{k^2}{2} \)
similarly RHL = \( \frac{k^2}{2} \)
\( f(0) = \frac{1}{2} \)
since f(x) is continuous at \( x = 0 \)
\( \therefore \text{LHL} = \text{RHL} = f(0) \)
\( \frac{k^2}{2} = \frac{k^2}{2} = \frac{1}{2} \)
\( \Rightarrow \frac{k^2}{2} = \frac{1}{2} \)
\( \Rightarrow k^2 = 1 \)
\( k = \pm 1 \)
\( \therefore f(x) \) is continuous at \( x = 0 \) for \( k = \pm 1 \) Ans.

Question. Find value of 'a' so that f(x) is continues at x = 0. \[ f(x) = \begin{cases} \frac{1-\cos(4x)}{x^2} & ; \ x < 0 \\ a & ; \ x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} & ; \ x > 0 \end{cases} \]
Answer: LHL = \( \left[\frac{1-\cos(4x)}{x^2}\right] \)
put \( x = 0 - h = -h \) and \( h \to 0 \)
LHL = \( \left[\frac{1-\cos(-4h)}{(-h)^2}\right] \)
\( = \left[\frac{1-\cos(4h)}{h^2}\right] \)
\( = \left(\frac{2\sin^2(2h)}{h^2}\right) \)
\( = \left[\frac{2\sin^2(2h)}{4h^2} \times 4\right] \)
\( = 8\left(\frac{\sin^2(2h)}{4h^2}\right) \)
\( = 8 \times 1 = 8 \dots\dots \left(\frac{\sin^2 x}{x^2} = 1\right) \)
\( \therefore \text{LHL} = 8 \)
RHL = \( \left(\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}\right) \)
\( \therefore \text{RHL} = \left[\frac{\sqrt{h}}{\sqrt{16+\sqrt{h}}-4}\right] \)
rationalize
RHL = \( \left[\frac{\sqrt{h}}{\sqrt{16+\sqrt{h}}-4} \times \frac{(\sqrt{16+\sqrt{h}}+4)}{(\sqrt{16+\sqrt{h}}+4)}\right] \)
\( = \left[\frac{\sqrt{h}(\sqrt{16+\sqrt{h}}+4)}{16+\sqrt{h}-16}\right] \)
\( = \left[\sqrt{16} + \sqrt{h} + 4\right] \)
\( = 4 + 4 \)
RHL = 8
Now \( f(0) = a \)
since f(x) is continuous at \( x = 0 \)
LHL = RHL = \( f(0) \)
\( \Rightarrow 8 = 8 = a \dots a = 8 \)
\( \therefore f(x) \) is cont. at \( x = 0 \) for \( a = 8 \) Ans.

Question. Determine the value of a, b and c so that the function is continues at x = 0. \[ f(x) = \begin{cases} \frac{\sin(a+1)x+\sin x}{x} & ; \ x < 0 \\ c & ; \ x = 0 \\ \frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}} & ; \ x > 0 \end{cases} \]
Answer: LHL = \( \left[\frac{\sin(a+1)x+\sin x}{x}\right] \)
put \( x = 0 - h = -h \) and \( h \to 0 \)
LHL = \( \left[\frac{\sin(a+1)(-h)+\sin(-h)}{-h}\right] \)
\( = \left[\frac{-\sin(a+1)h-\sin h}{-h}\right] \)
\( = \left[\frac{\sin(a+1)h+\sin h}{h}\right] \)
\( = \left(\frac{\sin(a+1)h}{h} + \frac{\sin h}{h}\right) \)
\( = \left(\frac{\sin(a+1)h}{h(a+1)} \times (a + 1) + \frac{\sin h}{h}\right) \)
\( = (a + 1)\left(\frac{\sin(a+1)h}{h(a+1)}\right) + \left(\frac{\sin h}{h}\right) \)
\( = (a + 1)1 + 1 \dots\dots \left\{\left(\frac{\sin x}{x}\right) = 1\right\} \)
LHL = \( a + 2 \)
RHL = \( \left(\frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}}\right) \)
put \( x = 0 + h = h \) and \( h \to 0 \)
RHL = \( \left(\frac{\sqrt{h+bh^2}-\sqrt{h}}{bh^{3/2}}\right) \)
\( = \left(\frac{\sqrt{h}\sqrt{1+bh}-\sqrt{h}}{bh\sqrt{h}}\right) \)
\( = \left(\frac{\sqrt{h}(\sqrt{1+bh}-1)}{bh\sqrt{h}}\right) \)
Rationalize
\( = \left(\frac{(\sqrt{1+bh}-1)(\sqrt{1+bh}+1)}{bh(\sqrt{1+bh}+1)}\right) \)
\( = \left(\frac{1+bh-1}{bh(\sqrt{1+bh}+1)}\right) \)
\( = \left(\frac{1}{\sqrt{1+bh}+1}\right) = \frac{1}{1+1} = \frac{1}{2}  /> RHL = \( \frac{1}{2} \)
\( f(0) = c \)
since f(x) is continuous at \( x = 0 \)
\( \therefore \text{LHL} = \text{RHL} = f(0) \)
\( \Rightarrow a + 2 = \frac{1}{2} = c \)
\( \Rightarrow a + 2 = \frac{1}{2} \) and \( c = \frac{1}{2} \)
\( a = -\frac{3}{2} \) and \( c = \frac{1}{2} \) and \( b = \mathbb{R} - \{0\} \dots\dots \{\text{for } b = 0 : f(x) \text{ does not exist}\} \). Ans.

Question. If the function f(x) is continues at x = 0. Find the value of k. \[ f(x) = \begin{cases} \frac{\log(1+ax)-\log(1-bx)}{x} & ; \ x \neq 0 \\ k & ; \ x = 0 \end{cases} \]
Answer: RHL = \( \left[\frac{\log(1+ax)-\log(1-bx)}{x}\right] \)
put \( x = 0 + h \) and \( h \to 0 \)
RHL = \( \left[\frac{\log(1+ah)-\log(1-bh)}{h}\right] \)
\( = \left[\frac{\log(1+ah)}{h} - \frac{\log(1-bh)}{h}\right] \)
\( = \left[\frac{\log(1+ah)}{ah} \times a - \frac{\log(1+(-bh))}{(-bh)} \times (-b)\right] \)
\( = a\left(\frac{\log(1+ah)}{ah}\right) + b\left(\frac{\log(1+(-bh))}{(-bh)}\right) \)
\( = a(1) + b(1) \dots\dots \left(\frac{\log(1+x)}{x} = 1\right) \)
RHL = \( a + b \)
\( f(0) = k \)
since f(x) is continuous at \( x = 0 \)
\( \therefore \text{RHL} = f(0) \Rightarrow a + b = k \)
\( \therefore k = a + b \) Ans.

Question. Prove that the greatest integer function [x] is discontinues at all integral points.
Answer: We have \( f(x) = [x] \)
let \( k \) be only integer i.e. \( k \in \mathbb{Z} \)
then \( f(x) = [x] = \begin{cases} k - 1 & ; \text{if } k - 1 \le x < k \\ k & ; \text{if } k \le x < k + 1 \end{cases} \)
LHL = \( (k - 1) \)
LHL = \( k - 1 \)
RHL = \( (k) \)
RHL = \( k \)
and \( f(k) = k \)
since LHL \( \neq \) RHL \( \therefore f(x) \) is discontinuous at 'k' i.e. all integral points \( (\because k \in \mathbb{Z}) \). Ans.

Question. Show that the function g(x) = x - [x] is discontinues at all integral points.
Answer: We have \( g(x) = x - [x] \)
let \( k \) be any integer i.e \( k \in \mathbb{Z} \)
\( g(x) = \begin{cases} x - (k - 1) & ; \text{if } k - 1 \le x < k \\ x - k & ; \text{if } k \le x < k + 1 \end{cases} \)
LHL = \( (x - (k - 1)) \)
put \( x = k - h \) and \( h \to 0 \)
\( \therefore \text{LHL} = (k - h - k + 1) \)
LHL = \( -h + 1 = 1 \)
RHL = \( (x - k) \)
put \( x = k + h \) & \( h \to 0 \)
\( \therefore \text{RHL} = (k + h - k) = 0 \)
\( f(x) = k - k = 0 \)
since LHL \( \neq \) RHL \( \dots f(x) \) is discontinuous at all integral points \( (\because k \in \mathbb{Z}) \). Ans.

Question. Discus the continuity of the function f(x) = |x - 3| - |x - 1|
Answer: We have \( f(x) = |x - 3| - |x - 1| \)
first arrange moduli so that their critical points are in ascending order.
i.e \( f(x) = -|x - 1| + |x - 3| \)
\( f(x) = \begin{cases} +(x - 1) - (x - 3) & ; \ x < 1 \\ -1(x - 1) - (x - 3) & ; \ 1 \le x < 3 \\ -(x - 1) + (x - 3) & ; \ x \ge 3 \end{cases} \)
\( f(x) = \begin{cases} +2 & ; \ x < 1 \\ -2x + 4 & ; \ 1 \le x < 3 \\ -2 & ; \ x \ge 3 \end{cases} \)
when \( x < 1 \)
\( f(x) = 2 \) which is a constant function, which is everywhere continuous. \( \therefore f(x) \) is continuous when \( x < 1 \).
when \( 1 < x < 3 \)
\( f(x) = -2x + 4 \) which is a polynomial function, which is everywhere continuous. \( f(x) \) is continuous when \( 1 < x < 3 \).
when \( x > 3 \)
\( f(x) = -2 \) which is a constant function, which is everywhere continuous. \( \therefore f(x) \) is continuous when \( x \ge 3 \).
Now Continuity at \( x = 1 \)
LHL = \( (2) \Rightarrow \text{LHL} = 2 \)
RHL = \( (-2x + 4) \)
put \( x = 1 + h \) & \( h \to 0 \)
\( \therefore \text{RHL} = (-2(1 + h) + 4) = 2 \)
RHL = \( -2 + 4 = 2 \)
\( f(1) = -2(1) + 4 = 2 \)
LHL = RHL = \( f(1) = 2 \)
\( \therefore f(x) \) is also continuous at \( x = 1 \)
Now continuity at \( x = 3 \)
LHL = \( (-2x + 4) \)
put \( x = 3 - h \) and \( h \to 0 \)
LHL = \( (-2(3 - h) + 4) \)
\( \Rightarrow \text{LHL} = -6 + 4 = -2 \)
RHL = \( (-2) \)
\( f(3) = -2 \)
LHL = RHL = \( f(3) = -2 \)
\( \therefore f(x) \) is also continuous at \( x = 3 \)
\( \therefore f(x) \) is continuous everywhere (or) there is no point of discontinuity. (Ans.)

Question. Show that the function f(x) = | 1 - x + |x| | is a continues function.
Answer: Let \( g(x) = | -x + |x| | \)
and \( h(x) = |x| \)
Now \( (hog)(x) = h(g(x)) \)
\( = h(1 - x + |x|) \)
\( (hog)(x) = |1 - x + |x||  /> \( \Rightarrow (hog)(x) = f(x) \dots\dots(1) \)
Now \( h(x) = |x| \) is sum of polynomial and modulus function and sum of two continuous functions is also continuous. \( \therefore g(x) \) is continuous everywhere
and composite function of two continuous functions is also continuous
here \( f(x) \) being a composite function of \( h(x) \) and \( g(x) \dots\dots\{\text{from (1) is also continuous}\} \) Ans.

Question. Examine that sin | x | is a continues function.
Answer: Let \( f(x) = \sin|x| \)
again let \( g(x) = |x| \) and \( h(x) = \sin x \)
Now, \( (hog)(x) = h(g(x)) \)
\( = h(|x|) \)
\( = \sin|x| \)
\( hog(x) = f(x) \dots\dots(1) \)
\( g(x) = |x| \); which is a modulus function and it is continuous everywhere.
\( h(x) = \sin x \); is a sine function and it is continuous everywhere.
and composite function of two continuous functions is also continuous.
here, \( f(x) \) being a composite function of \( g(x) \) and \( h(x) \) is also continuous. (Ans)

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