[mathjax]
Question 1
निम्नलिखित फलनों के सांतत्य (संतता) की जाँच दुए हुए बिन्दुओं पर कीजिए ।
[Examine the continuity of the following functions at indicated points]
[Note : f(x), x=a
f(a)𝜖R
$\lim _{x \rightarrow a} f(x)=f(a)$]
(i) f(x)=2x+3 , (at) x=1 पर
Sol :
x=1 पर ,
f(1)=2(1)+3
=5𝜖R
∴f(x) , x=1 पर परिभाषित है ।
$\lim _{x \rightarrow 1} f(x) \equiv \lim _{x \rightarrow 1}(2 x+3)$
=2(1)+3
=5
$\lim _{x \rightarrow 1} f(x)=f(1)$
∴ f(x) , x=1 पर संतता(continuous) है
(ii) f(x)=2x2-1 , (at) x=3 पर
Sol :
x=3 पर ,
f(3)=2(3)2-1
=17𝜖R
∴ f(x) , x=3 पर परिभाषित है ।
$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(2 x^{2}-1\right)$
=2(3)2-1=17
$\lim _{x \rightarrow 3} f(x)=f(3)$
∴ x=3 , f(x) संतता है ।
(iii) f(x)=5x-3 , (at) x=0 , -3 , 5 पर
Sol :
x=0 पर ,
f(0)=5(0)-3
=-3
f(x),x=0 पर परिभाषित है ।
$\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0}(5 x-3)$
=5(0)-3
=-3
$\lim _{x \rightarrow 0} f(x)=f(0)$
∴ f(x),x=0 पर संतता है ।
(b) x=-3 पर ,
f(-3)=5(-3)-3
=-18∈R
∴f(x),x=-3 पर संतता है ।
$=\lim _{x \rightarrow-3} f(x)=\lim_{x\rightarrow-3}(5x-3)$
=5(-3)-3
=-18
$\lim _{x \rightarrow-3} f(x)=f(-3)$
∴ f(x),x=-3 पर संतता है ।
(iv) $f(x)=x^{3}+1$. (at) x=1 पर
Sl :
(v) $f(x)=x^{2}$, (at) x=0 पर
Sol :
Question 2
साबित करें कि फलन $f(x)$, जहाँ
[Prove that the function f(x) where]
f(x)=4x+3 when x
=3x+7 , when x=4
x=4 पर संतता है । (is continuous at x=4)
Sol :
$f(x)=\left\{\begin{array}{l}4 x+3, \text {when } x \neq 4\left\{\begin{array}{l}x<4 \\ x>4\end{array}\right. \\ 3 x+7, \text{when }x=4\end{array}\right.$
x=4 ,
L.H.L
$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{-}}(4 x+3)$
$=\lim _{h \rightarrow 0}[4(4-h)+3]$
$=\lim _{h \rightarrow 0}[16-4 h+3]$
$=\lim _{h \rightarrow 0}(19-4h)$
=19-4(0)=19
R.H.L
$\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{+}}(4 x+3)$
$=\lim _{h \rightarrow 0}[4(4+h)+3]$
$=\lim_{h \rightarrow 0}(16+4 h+3]$
$=\lim_{h \rightarrow 0}(19+4h)$
=19+4(0)
=19
f(4)=3(4)+7
=19
∴$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(4)$
f(x), x=4 पर संतता(continuous) है
ALTERNATE METHOD
$f(x)=\left\{\begin{array}{l}4 x+3, \text {when } x \neq 4\left\{\begin{array}{l}x<4 \\ x>4\end{array}\right. \\ 3 x+7, \text{when }x=4\end{array}\right.$
L.H.L
$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4}(4 x+3)$
=4(4)+3=19
R.H.L
$\lim_{x\rightarrow4^{-}}f(x)=\lim _{x \rightarrow 4^+}(x)=f(
4)$
f(x),x=4 पर संतता(continuous) है
Question 3
दिखाएँ कि फलन
[Show that the function]
$f(x)=\left\{\begin{array}{r}x^{3}+3, \text { if } x \neq 0 \\ 1,\text { if } x=0\end{array}\right.$
x=4 पर असंतत है [is discontinuous at x=4]
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(x^{3}+3\right)$
$=\lim _{h \rightarrow 0}\left[(0-h)^{3}+3\right]$
$=\lim _{h \rightarrow 0}\left[-h^{3}+3\right]$
$=-0^{3}+3$
=3
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(x^{3}+3\right)$
$=\lim _{h \rightarrow 0}\left[(0+h)^{3}+3\right]$
f(0)=1
∴$\lim _{x \rightarrow 0^{-}} f(2)=\lim _{x \rightarrow 0^{+}} f(x)$≠f(0)
<to be added>
Question 4
यदि (If) f(x)=0 when x=0
$=\frac{1}{2}-x$ ,when $0<x<\frac{1}{2}$
$=\frac{1}{2}$ when $x=\frac{1}{2}$
तो x=0 तथा $x=\frac{1}{2}$ पर f(x) के सातत्य की जाँच करें ।
[then test the continuity of f(x) at x=0 and $x=\frac{1}{2}$. ]Sol :
At 2=0
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1}{2}-x\right)$
$=\frac{1}{2}-0=\frac{1}{2}$
f(0)=0
$\therefore \lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$
Hence , f(x) is discontinuous at x=0
ALTERNATE METHOD
At $x=\frac{1}{2}$
L.H.L
$\lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}-x\right)$
$=\frac{1}{2}-\frac{1}{2}=0$
$f\left(\frac{1}{2}\right)=\frac{1}{2}$
∴$\lim _{x \rightarrow \frac{1}{2}^{-}}f(x)\neq f\left(\frac{1}{2}\right)$
Question 5
साचित करें कि f(x)=|x|, x=0 पर संतत है।
[Prove that f(x)=|x| is continuous at x=0]
Sol :
$f(x)=\left\{\begin{array}{l}-x \text{ if x<0} \\ 0\text{ if x=0}\\x\text{ if x>0}\end{array}\right.$
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim_{x\rightarrow0}(-x)$
=-0
=0
R.H.L
$\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0}(x)$
=0
f(0)=0
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
=f(0)
f(x), x=0 पर संतता(continuous) है
Question 6
(क्या)Is $f(x)=(1+x)^{\frac{1}{x}}$ , when x≠0
f(x)=e , when x=0
x=0 पर संतत हैं ? (continuous at x=0 ?)
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim_{x\rightarrow0}(1+x)^{\frac{1}{x}}$
$=\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}$
$=\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
f(0)=e
∴$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
x=0 पर f(x) संतता(continuous) है
Question 7
क्या f(x) , x=0 पर संतत है? जहाँ
[If f(x) continuous at x=0 ? where]
$f(x)=\frac{\cos a x-\cos b x}{x^{2}}, \text{where }x\neq0 \left\{\begin{matrix}x<0\\x>0\end{matrix}\right.$
$f(x)=\frac{b^2-a^2}{2}, \text{where }x=0$
Sol :
At x=0 ,
L.H.L
=$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$
[$\cos C-\cos D=2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$]
$=\lim _{x \rightarrow 0} \frac{2 \sin \frac{a x+b x}{2} \operatorname{sin} \frac{b x-a x}{2}}{x^{2}}$
$=2 \lim _{x \rightarrow 0} \frac{\sin \left(\frac{a+b}{2}\right) x}{x} \cdot \lim _{x \rightarrow 0} \frac{\sin \left(\frac{b-9}{2}\right) x}{x}$
$=2 \lim _{x \rightarrow 0} \frac{\operatorname{sin}\left(\frac{a+b}{2}\right) x}{\left(\frac{a+b}{2}\right)x} \times\left(\frac{a+b}{2}\right) \cdot \lim _{x \rightarrow 0} \frac{\sin \left(\frac{b-a}{2}\right)x}{\left(\frac{b-a}{2}\right) x}\times \left(\frac{b-a}{2}\right)$
$=2 \times\left(\frac{a+b}{x}\right) \times \frac{b-a}{2}$
$=\frac{b^{2}-a^{2}}{2}$
R.H.L
=$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\cos a x-\cos bx}{x^{2}}$
$=\frac{b^{2}-a^{2}}{2}$
$f(0)=\frac{b^{2}-a^{2}}{2}$
∴$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
f(x),x=0 संतता(continuous) है
Question 8
(यदि) If $f(x)=\left\{\begin{array}{ll}\frac{|x^3|}{x},\text{when }x\neq0;\left\{\begin{matrix}x<0\\x>0\end{matrix}\right. \\ 0,\text{when }x=0\end{array}\right.$
तो f(x) का x=0 पर सांतत्य की जाँच करें ।
[then test the continuity of f(x) at x=0]
Sol :
At x=0 ,
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\left|x^{3}\right|}{x}$
$=\lim _{x \rightarrow 0} \frac{-x^{2}}{x}=0^2$
=0
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{| x^{-3}|}{x}$
$=\lim _{x \rightarrow 0} \frac{x^{3}}{x}=0^{2}$
=0
f(0)=0
∴ $\lim _{x \rightarrow 0^{-}} f(x)-\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
f(x) ,x=0 संतता(continuous) है
Question 9
सिद्ध करें कि $f(x)=x^{n}, x=0$ पर संतत है, जहाँ n एक धन पूर्णांक है।
[Prove that the function $f(x)=x^{n}, x=0$ is continuous at x=0 , where n is a positive integer.]
Sol :
x=0 ,
f(0)=$=0^{n}$=0
f(x), x=0 पर परिभाषित है ।
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{n}$
$=0^{n}$
=0
$\lim _{x \rightarrow 0} f(x)=f(0)$
f(x) ,x=0 पर संतता(continuous) है
Question 10
क्या $f(x)=x^{2}-\sin x+5$ से परिभाषित फलन $x=\pi$ पर सतत है ?
Is the function defined by $f(x)=x^{2}-\sin x+5$ continuous at $x=\pi ?$
Sol:
At x = 𝜋
L.H.S
$\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi}\left(x^{2}-\sin x+5\right)$
$=\lim _{h \rightarrow 0}\left[(\pi-h)^{2}-\sin(\pi-h)+5\right]$
$=\lim _{h \rightarrow 0}\left[x^{2}-2 \pi h+h^{2}-\sin h+5\right]$
$= \pi^{2}-2 \pi(0)+0^{2}-\sin \theta+5$
$=\pi^{2}+5$
R.H.L
$\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi}\left(x^{2}-\sin x+5\right)$
$=\lim _{h \rightarrow 0}\left[(\pi+h)^{2}-\sin (\pi+h)+5\right]$
$=\lim _{h \rightarrow 0}\left[(\pi+h)^{2}-\sin (\pi+h)+5\right]$
$=\lim _{h \rightarrow 0}\left(\pi^{2}+2 \pi h+h^{2}+\operatorname{sin} h+5\right)$
$= \pi^{2}+2 \pi(0)+0^{2}+\sin 0+5$
$=\pi^{2}+5$
$f(\pi)=\pi^{2}-\sin \pi+5$
$=\pi^{2}+5$
∴$\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)=f(x)$
f(x) , x=π पर संतता(continuous) है
Question 11
(i) f(x) का x=0 पर सांतत्य की जाँच करें, जहाँ
[Test the continuity of f(x) at x=0 where]
$\begin{matrix}f(x)=\frac{x e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right.\\=0,x=0\end{matrix}$
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{x e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}\right)$
$=\lim _{h \rightarrow 0}\left(\frac{(0-h) \cdot e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}\right)$
$=\lim _{h \rightarrow 0} \frac{-h \cdot e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}$
$=-0 \cdot \frac{e^{-\infty}}{1+e^{-\infty}}=0$
R.H.S
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{x \cdot e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}\right)$
$=\lim _{h \rightarrow 0}\left(\frac{(0+h) \cdot e^{\frac{1}{0+h}}}{1+e^{\frac{1}{0+h}}}\right)$
$={\lim}_{h \rightarrow 0} \frac{h \cdot e^{\frac{1}{h}}}{1+e^{\frac{1}{h}}}$
$=0 \cdot \frac{e^{\infty}}{1+e^{\infty}}=0$
f(0)=0
∵$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}}f(x)=f(0)$
f(x),x=0 पर संतता(continuous) है
(ii) माना कि [Let ]$\begin{aligned} f(x) &=\frac{e^{\frac{1}{x}}-1}{1+e^{\frac{1}{x}}}\text{when }x\neq0 \left\{\begin{matrix}x<0\\x>0\end{matrix}\right. \\ &=0 ,\text{when }x=0 \end{aligned}$
दिखाएँ कि $f(x), x=0$ पर असंतत है।
[Show that f(x) is discontinuous at x=0]
Sol :
At x=0 ,
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}}-1}{1+e^{\frac{1}{x}}}\right)$
$=\lim _{h \rightarrow 0}\left(\frac{e^{\frac{1}{0-h}-1}}{1+e^{\frac{1}{0-h}}}\right)$
$=\lim _{h \rightarrow 0} \frac{e^{\frac{-1}{h}}-1}{1+e^{-\frac{1}{h}}}$
$=\frac{e^{-\infty}-1}{1+e^{-\infty}}$
$=\frac{\frac{1}{e^{\infty}}-1}{1+\frac{1}{e^{\infty}}}$
$=\frac{\frac{1}{\infty}-1}{1+\frac{1}{\infty}}$
$=\frac{0-1}{1+0}=\frac{-1}{1}$
=-1
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}-1}}{1+e^{\frac{1}{x}}}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}-1}}{e^{\frac{1}{x}+1}}\right)$
$=\lim _{x \rightarrow 0} \frac{e^{\frac{1}{x}}\left(1-e^{-\frac{1}{x}}\right)}{e^{\frac{1}{x}}\left(1+e^{\frac{-1}{x}}\right)}$
$=\lim _{h \rightarrow 0}\left(\frac{1-e^{-\frac{1}{0+h}}}{1+e^{-\frac{1}{0+h}}}\right)$
$=\lim _{h \rightarrow 0}\left(\frac{1-e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}\right)$
$=\frac{1-e^{-\infty}}{1+e^{-\infty}}$
$=\frac{1-0}{1+0}=\frac{1}{1}=1$
$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x)$
f(x), x=0 पर असंतता(discontinuous) है
Question 12
निम्नीलिखत फलनों के निदिष्ट बिन्दुओं पर सांतत्य की जाँच करें ।
[Test the continuity of the following functions at indicated points]
(i) $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, x \neq 1\left\{\begin{array}{l}x<1 \\ x>1\\\end{array}\right.\\ 2, x=1\end{array}\right.$
Sol :
At x=1
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)$
$=\lim _{x \rightarrow 1}\left(\frac{x^{2}-1^{2}}{x-1}\right)$
$={\lim}_{x\rightarrow1} \frac{(x-1)(x+1)}{x-1}$
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)$
=1+1
=2
f(1)=2
∴$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
f(x) ,x=1 संतता(continuous) है
(ii) $f(x)=\left\{\begin{array}{cc}\frac{x^{2}-x-6}{x-3}, x \neq 3 \left\{\begin{array}{c}x<3 \\x>3\end{array}\right. \\ 5, x=3\end{array}\right.$
Sol :
x=3
L.H.L
$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(\frac{x^{2}-x-6}{x-3}\right)$
$=\lim _{x \rightarrow 3}\left(\frac{x^{2}-3 x+2 x-6}{x-3}\right)$
$=\lim _{x \rightarrow 3}\left(\frac{x(x-3)+2(x-3)}{x-3}\right)$
$=\lim _{x \rightarrow 3} \frac{(x-3)(x+2)}{x-3}$
=3+2
=5
R.H.L
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}\left(\frac{x^{2}-x-6}{x-3}\right)$
=5
f(3)=5
∴$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3)$
f(x) ,x=3 संतता(continuous) है
(iii) $f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}},x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right. \\ 1, x=0\end{array}\right.$
Sol :
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4}} \times \frac{1}{4}$
$=\frac{1}{4}\times {2}\lim_{x\rightarrow0} \frac{\sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4}}$
$=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$
$=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$
$=\frac{1}{2}(1)^{2}=\frac{1}{2}$
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)=\frac{1}{2}$
f(0)=1
$\therefore \lim _{x \rightarrow 0^{-}} f(x)\neq\lim _{x \rightarrow 0^{+}} f(x)\neq f(0)$
f(x) ,x=0 असंतता(discontinuous) है
(iv) $f(x)=\left\{\begin{array}{cl}\frac{|x-a|}{x-a},x \neq a\left\{\begin{array}{l}x<a \\ x>a\end{array}\right. \\ 1, x=a\end{array}\right.$
Sol :
At x=a
L.H.L
$\lim _{x \rightarrow a^{-}} f(x)=\lim_{x\rightarrow a^{-}} \frac{|x-a|}{x-a}$
$=\lim _{x \rightarrow 0} \frac{-(x-a)}{x-a}$
=-1
R.H.L
$\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}} \frac{|x-a|}{x-a}$
$=\lim_{x\rightarrow a} \frac{x-a}{x-a}$
=1
∴$\lim _{x \rightarrow a^{-}} f(x) \neq \lim _{x\rightarrow a^+} f(x)$
f(x) ,x=a असंतता (discontinuous) है
(v) $f(x)=\left\{\begin{array}{cc}\frac{\sin 3 x}{x}, & x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right. \\ 1, & x=0\end{array}\right.$
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin3 x}{x}$
$=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \times 3$
=1×3
=3
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}$
$=\lim_{x\rightarrow {0}} \frac{\sin 3 x}{3 x} \times 3$
=1×3
=3
f(0)=1
∴$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$
f(x) ,x=0 असंतता (discontinuous) है
Question 13
(a) सिद्ध करें कि वास्तविक संख्याओं पर तत्समक फलन f(x)=x संतत होता है।
[Prove that the identity function f(x)=x is a continuous function on real numbers]
Sol :
Let c∈R <to be added>
At x=c
f(c)=c∈R
∴f(x), x=c पर परिभाषित है ।
$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x)=c$
∴$\lim _{x \rightarrow c} f(x)=f(c)$
$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(x)=c$
<to be added>
(b) सिद्ध करें कि निम्नलिखित फलन अपने परिभाषा प्रान्त में संतत है।
[Prove that the following functions are continuous function in their domains of definition]
(i) tan x
Sol :
Let f(x)=tan x
$f(x)=R-(2 n+1) \frac{\pi}{2}, n \in z$
$f(x)=\tan x=\frac{\sin x}{\cos x}$
∵sin x तथा(and) cos x , x∈R पर संतता (continuous) होता है
∴f(x)=tan x , $x \in R-(2 n+1) \frac{\pi}{2}, n\in z$ पर संतता (continuous) है
(ii) sec x
Sol :
Let f(x)=sec x
$=R-(2 n+1) \frac{\pi}{2}, n \in z$
$f(x)=\sec x=\frac{1}{\cos x}$
∵ cos x , x∈R पर संतता (continuous) होता है
∵f(x)=sec x, x∈R पर संतता (continuous) होगा लेकिन cosx≠0
$x \in R-(2 n+1) \frac{\pi}{2}, n \in z$
(iii) cosec x
Sol :
Let f(x)=cosec x
=R-nπ , n∈z
f(x)=cosecx$=\frac{1}{\sin x}$
∵sinx, x∈R पर संतता (continuous) होता है
∴f(1)=cosecx, x∈R पर संतता (continuous) होगा लेकिन sinx≠0
x∈R-nπ , n∈z
Question 14
निम्नलिखित फलनों के सांतत्य पर विचार करें।
[Discuss the continuity of the following functions]
(i) f(x)=x-5
Sol :
(ii) $f(x)=x^{3}+x^{2}-1$
Sol :
(iii) f(x)=sinxcosx
Sol :
f(x)=sinxcosx
$f(x)=\frac{1}{2} \cdot 2 \sin x \cos x$
$f(x)=\frac{1}{2} \sin 2 x$
Let c∈R <to be added>
$f(c)=\frac{1}{2} \sin 2 c \in R$
∴f(x), x=c पर परिभाषित है ।
$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} \frac{1}{2} \sin 2 x$
$=\frac{1}{2} \sin 2 c$
∴$\lim _{x \rightarrow c} f(x)=f(c)$
f(x) ; x=c∈R पर संतता (continuous) है
(iv) f(x)=sinx+cosx
Sol :
f(x)=sinx+cosx
$=\sqrt{2} \cdot \frac{1}{\sqrt{2}}(\sin x+\cos x)$
$=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)$
$=\sqrt{2}\left(\cos \frac{\pi}{4} \tan x+\sin \frac{\pi}{4} \cos x\right)$
$f(x)=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$
<to be added>
$f(c)=\sqrt{2} \sin \left(c+\frac{\pi}{4}\right) \in R$
∴f(x), x=c पर परिभाषित है ।
$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} \sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$
$=\sqrt{2} \sin \left(c+\frac{\pi}{4}\right)$
∴$\lim _{x \rightarrow c}f(x)=f(x)$
f(x), x=c∈R पर संतता (continuous) है
Question 15
दिखाएँ कि निम्नलिखित फलन संतत फलन है ।
[Show that the following functions are continuous functions]
(i) sin |x|
Sol :
Let h(x)=sin |x|
Let f(x)=sin x , g(x)=|x|
<to be added>
(fog)(x)=f(g(x))
=f(|x|)
=sin |x|
=h(x)
∴ f(1)=sin x तथा g(2)=|x| संतता फलन है ।
∴fog भी एक संतता फलन होगा।
h(x)=sin|x| एक संतता फलन है ।
(ii) |cos x|
Sol :
Let h(x)=|cosx|
Let f(x)=|x| , g(x)=cosx
domain(f)=R ,range(g)=[-1,1]
range(g)≤domain(f)
∴fog परिभाषित है ।
fog(x)=f(g(x))
=f(cos x)
=|cos x|=h(x)
∵f(x)=|x|, तथा g(x)=cos x संतता फलन हैं ।
∴fog भी संतता फलन होगा ।
h(x)=|cos x| एक संतता फलन है ।
(iii) $\sin x^{2}$
Sol :
Question 16
निम्नलिखित फलनों के सांतत्य पर विचार करें ।
[Discuss the continuity of the following functions]
(i) $f(x)=\left\{\begin{array}{l}x+2, \text{if }x\leq 1 \\ x-2,\text{if }x>1\end{array}\right.$
Sol :
At x=1
L.H.L
$\lim_{x\rightarrow1^-} f\left(x\right)=\lim _{x \rightarrow 1}(x+2)$
=1+1
=2
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x-2)$
=1-2
=-1
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
f(x),x=1 पर असंतता हैं
(ii) f(x)=|x-5|
Sol :
∵|x| एक संतता फलन होता है , x𝜖R
|x-5| भी सभी बिन्दुओ पर संतता होगा ।
(iii) $f(x)=\left\{\begin{array}{l}x+5,\text{if }x\leq1 \\ x-5, (if) x>1\end{array}\right.$
Sol :
At x=1
L.H.L
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow 1}(x+5)$
=1+5
=6
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x-5)$
=1-5
=-4
∴$\lim _{x \rightarrow 1^{-}} f(x)\neq\lim_{x\rightarrow1^+}f(x)$
f(x), x=1 पर असंतता है
(iv) $f(x)=\left\{\begin{array}{l}x+2,\text{if }x\leq1 \\ -x+2, (if) x>1\end{array}\right.$
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(x+2)$
=0+2
=2
R.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(-x+2)$
=-0+2
=2
f(0)=0+2=2
∴f(x),x=0 पर संतता है ।
(v) $f(x)=\left\{\begin{array}{c}-2,\text{if }x\leq -1 \\ 2 x,\text{if }-1<x\leq1 \\ 2,\text{if }x>1\end{array}\right.$
Sol :
At x=1
L.H.L
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1}(-2)=-2$
R.H.L
$=\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1}(2x)$
=2(-1)
=-2
f(-1)=-2
∴$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)=f(-1)$
f(x),x=-1 पर संतता है ।
(v) $f(x)=\left\{\begin{array}{c}-2,\text{if }x\leq -1\\ 2 x ,\text{if }-1<x\leq 1\\ 2,\text{if }x>1\end{array}\right.$
Sol :
At x=1 ,
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(2 x)$
=2×1
=2
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(2)=2$
f(1)=2(1)=2
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
f(x),x=1 पर संतता है ।
(vi) $f(x)=\left\{\begin{array}{l}\sin x-\cos x, \text {if } x \neq 0\left\{\begin{array}{l}x<0 \\ x>0\end{array}\right. \\-1, \text{if }x=0\end{array}\right.$
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)$
=sin0-cos0
=0-1
=-1
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\sin x-\cos x\right)$
=sin0-cos0
=0-1
=-1
f(0)=-1
∴$\lim _{x \rightarrow 0^{-}} f(x)=\lim_{x\rightarrow 0^{+}}f(x)=f(0)$
f(x), x=0 पर संतता हैं ।
Question 17
(यदि) If $\begin{aligned} f(x) &=\frac{1}{2}-x, 0 \leq x<\frac{1}{2} \\ &=\frac{1}{2}, x=\frac{1}{2} \\ &=\frac{3}{2}-x, \frac{1}{2}<x \leq 1 \end{aligned}$
तो f(x) के सांतत्य की जाँच करें ।
[examine the continuity of f(x)]
Sol :
At $x=\frac{1}{2}$
L.H.L
$\lim _{x \rightarrow \frac{1}{2}^-}=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}-x\right)$
$=\frac{1}{2}-\frac{1}{2}=0$
R.H.L
$\lim _{x \rightarrow \frac{1}{2}^+} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{3}{2}-x\right)$
$=\frac{3}{2}-\frac{1}{2}=\frac{2}{2}$
=1
∴$\lim _{x \rightarrow \frac{1}{2}^{-}}f(x) \neq \lim _{x \rightarrow \frac{1}{2}+} f(x)$
f(x),$x=\frac{1}{2}$ पर असंतता है ।
Question 18
यदि (If )$\begin{aligned} f(x) &=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1 \left\{\begin{array}{l}x<1 \\ x>1\end{array}\right.\\ &=2, \quad x=1 \end{aligned}$
तो f(x) के सांतत्य की जाँच करें ।
[examine the continuity of f(x)]
Sol :
At x=1 ,
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$
$=\lim _{x \rightarrow 1}\left(\frac{x^{2}-3 x-2 x+3}{x^{2}-1^2}\right)$
$=\lim _{x \rightarrow 1}\left[\frac{x(x-3)-1(x-3)}{(x-1)(x+1)}\right]$
$=\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{(x-1)(x+1)}$
$=\frac{1-3}{1+1}$
$=\frac{-2}{2}$
=-1
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$
=-1
f(1)=2
f(x),x=1 पर असंतता है ।
Question 19
If $f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$
तो f(x) के सांतत्य की जाँच करें ।
examine the continuity of f(x)
Sol :
At x=1
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}\right)$
$=1^{2}$
=1
R.H.L
$\lim_{x\rightarrow1^+}f(x)=\lim _{x \rightarrow 1}(2 x-1)$
=2(1)-1
=1
f(1)=2(1)-1
=2-1
=1
∴$\lim _{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^+}f{(x)}$=f(1)
f(x),x=1 पर संतता है
Question 19
If $f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$
examine the continuity of f(x)
Sol :
At x=2,
L.H.L
$\lim_{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(2 x-1)$
=2(2)-1
=4-1
=3
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(x+3)$
=2+3
=5
∴$\lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$
f(x), x=2 पर असंतता है ।
Question 20
(यदि) If $f(x)=\frac{x^{2}-4}{x-2}, \quad 0<x<2$
=x+1 , 2≤x≤5
तो f(x) के सांतत्यता की जाँच करें
test the continuity of f(x)
Sol :
At x=2 ,
L.H.L
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$
$=\lim _{x \rightarrow 2}\left(\frac{x^{2}-2^{2}}{x-2}\right)$
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$
$=\lim _{x \rightarrow 2}\left(\frac{x^{2}-2^{2}}{x-2}\right)$
$=\lim _{x \rightarrow 2}\left(\frac{(x-2)(x+2)}{x-2}\right)$
=2+2
=4
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}$
=4
R.H.L
$\lim _{x \rightarrow 2}+f(x)=\lim _{x \rightarrow 2}(x+1)$
=2+1
=3
$\therefore \lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$
f(x), x=2 पर असंतता(discontinuous) है
Question 17
(यदि) If $\begin{aligned} f(x) &=\frac{1}{2}-x, 0 \leq x<\frac{1}{2} \\ &=\frac{1}{2}, x=\frac{1}{2} \\ &=\frac{3}{2}-x, \frac{1}{2}<x \leq 1 \end{aligned}$
तो f(x) के सांतत्य की जाँच करें ।
[examine the continuity of f(x)]
Sol :
At $x=\frac{1}{2}$
L.H.L
$\lim _{x \rightarrow \frac{1}{2}^-}=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{1}{2}-x\right)$
$=\frac{1}{2}-\frac{1}{2}=0$
R.H.L
$\lim _{x \rightarrow \frac{1}{2}^+} f(x)=\lim _{x \rightarrow \frac{1}{2}}\left(\frac{3}{2}-x\right)$
$=\frac{3}{2}-\frac{1}{2}=\frac{2}{2}$
=1
∴$\lim _{x \rightarrow \frac{1}{2}^{-}}f(x) \neq \lim _{x \rightarrow \frac{1}{2}+} f(x)$
f(x),$x=\frac{1}{2}$ पर असंतता है ।
Question 18
यदि (If )$\begin{aligned} f(x) &=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1 \left\{\begin{array}{l}x<1 \\ x>1\end{array}\right.\\ &=2, \quad x=1 \end{aligned}$
तो f(x) के सांतत्य की जाँच करें ।
[examine the continuity of f(x)]
Sol :
At x=1 ,
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$
$=\lim _{x \rightarrow 1}\left(\frac{x^{2}-3 x-2 x+3}{x^{2}-1^2}\right)$
$=\lim _{x \rightarrow 1}\left[\frac{x(x-3)-1(x-3)}{(x-1)(x+1)}\right]$
$=\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{(x-1)(x+1)}$
$=\frac{1-3}{1+1}$
$=\frac{-2}{2}$
=-1
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(\frac{x^{2}-4 x+3}{x^{2}-1}\right)$
=-1
f(1)=2
f(x),x=1 पर असंतता है ।
Question 19
If $f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$
तो f(x) के सांतत्य की जाँच करें ।
examine the continuity of f(x)
Sol :
At x=1
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}\right)$
$=1^{2}$
=1
R.H.L
$\lim_{x\rightarrow1^+}f(x)=\lim _{x \rightarrow 1}(2 x-1)$
=2(1)-1
=1
f(1)=2(1)-1
=2-1
=1
∴$\lim _{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^+}f{(x)}$=f(1)
f(x),x=1 पर संतता है
Question 19
If $f(x)=\left\{\begin{array}{lr}x^2,0\leq x<1\\2x-1,1\leq x<2\\x+3,x\geq 2\end{array}\right.$
examine the continuity of f(x)
Sol :
At x=2,
L.H.L
$\lim_{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(2 x-1)$
=2(2)-1
=4-1
=3
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(x+3)$
=2+3
=5
∴$\lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$
f(x), x=2 पर असंतता है ।
Question 20
(यदि) If $f(x)=\frac{x^{2}-4}{x-2}, \quad 0<x<2$
=x+1 , 2≤x≤5
तो f(x) के सांतत्यता की जाँच करें
test the continuity of f(x)
Sol :
At x=2 ,
L.H.L
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$
$=\lim _{x \rightarrow 2}\left(\frac{x^{2}-2^{2}}{x-2}\right)$
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^{2}-4}{x-2}\right)$
$=\lim _{x \rightarrow 2}\left(\frac{x^{2}-2^{2}}{x-2}\right)$
$=\lim _{x \rightarrow 2}\left(\frac{(x-2)(x+2)}{x-2}\right)$
=2+2
=4
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}$
=4
R.H.L
$\lim _{x \rightarrow 2}+f(x)=\lim _{x \rightarrow 2}(x+1)$
=2+1
=3
$\therefore \lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$
f(x), x=2 पर असंतता(discontinuous) है
Question 21
निम्नलिखित फलनों के असांत्यता के बिन्दुओं को ज्ञात करें ।
[Find the points of discontinuity of the following functions]
(i) $f(x)=\left\{\begin{array}{l}x+1,(\text { if }) x \geq 1 \\ \left.x^{2}+1, \text {(if }\right) x<1\end{array}\right.$
Sol :
At x=1
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}+1\right)$
$=1^{2}+1$
=2
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x+1)$
=1+1
=2
f(1)=1+1=2
∴$\lim _{x \rightarrow 1^{-}} f(x)=\lim_{x+1^+}f(x)=f(1)$
f(x) <to be added>
(ii) $f(x)= \begin{array}{ll}2 x+3, & \text {(if) } x \leq 2 \\ 2 x-3, & \text {(if) } x>2\end{array}$
Sol :
(iii) Find the points of discontinuity of the following functions:
$f(x)=\left\{\begin{array}{ll}x, 0 \leq x<\frac{1}{2} \\ 1, x=\frac{1}{2} \\ 1-x, \frac{1}{2}<x \leq 1\end{array}\right.$
Sol :
At $x=\frac{1}{2}$
L.H.L
$\lim _{x \rightarrow \frac{1}{2}^-} f(x)=\lim _{x \rightarrow \frac{1}{2}}(x)=\frac{1}{2}$
R.H.L
$\lim_{x \rightarrow \frac{1}{2}+} f(x)=\lim _{x \rightarrow \frac{\pi}{2}}(1-x)$
$=1-\frac{1}{2}=\frac{1}{2}$
$f\left(\frac{1}{2}\right)=1$
∴$\lim _{x \rightarrow \frac{1}{2}-}f(x)=\lim _{x \rightarrow \frac{1}{2}+}f( x)\neq f\left(\frac{1}{2}\right)$
f(x) , $x=\frac{1}{2}$ पर असंतता(discontinuous) है
(iv) $f(x)= \begin{array}{ll}x^{3}-3, & \text { (if) } x \leq 2 \\ x^{2}+1, & \text { (if) } x>2\end{array}$
Sol :
(v) $f(x)= \begin{array}{ll}x^{3}-3, & \text {(if) } x \leq 2 \\ x^{2}+1, & \text {(if) } x>2\end{cases}$
Sol :
(vi) $f(x)=\left\{\begin{array}{l}\frac{\sin x}{x},\text{if }x<0 \\ x+1,\text{if }x\geq 0\end{array}\right.$
Sol :
At x=0,
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin x}{x}$
=1
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(x+1)$
=0+1
=1
f(0)=0+1
=1
∴$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
f(x), <to be added>
(vii) $f(x)=\left\{\begin{array}{ll}2x+3, -3 \leq x<-2 \\ x+1,-2\leq x<0 \\ x+2, 0\leq x \leq 1\end{array}\right.$
Sol :
At x=-2
At x=0
Question 22
k का मान तिनकालें ताकि निम्नलिखित फलन दिए हुए बिन्दु पर संतत हों ।
[Find the values of k such that the following functions are continuous at the indicated point]
(i) $f(x)=\left\{\begin{array}{l}k x+1,\text{if }x\leq 5 \\ 3 x-5,\text{if }x>5\end{array}\right.$
Sol :
At x=5
L.H.L
$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 3}(k x+1)$
=k(5)+1
=5k+1
R.H.L
$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(3 x-5)$
=3(5)-5
=10
∵f(x),x=5 पर संतता है
5k+1=10
5k=9
$k=\frac{9}{5}$
(ii) $f(x)=\left\{\begin{array}{ll}k x^{2}, \text {(if ) } x \leq 2 \\ 3, \text {(if) } x>2\end{array}\right.$
Sol :
At x=2 ,
L.H.L
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(k x^{2}\right)$
$=k(2)^{2}$
=4k
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(3)=3$
∵f(x),x=2 पर संतता है ।
$\lim _{x \rightarrow 2^{-}} f(x)=\lim_{x\rightarrow 2^{+}} f (x)$
4k=3
$k=\frac{3}{4}$
(iii) $f(x)=\left\{\begin{array}{ll}k x+1, \text {(if ) } x \leq \pi \\\cos x, \text {(if) } x>\pi\end{array}\right.$
Sol :
At x=π
L.H.L
$\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi}(k x+1)=$
=kπ+1
R.H.S
$\lim_{x\rightarrow\pi^+}f(x)=\lim _{x \rightarrow \pi}(\cos x)$
=cosπ
=-1
∵f(x),x=π
$\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{+}} f(x)$
kπ+1=-1
kπ=-2
$k=-\frac{2}{\pi}$
(iv) $f(x)=\left\{\begin{array}{ll}\frac{1-\cos 4 x}{8 x^{2}}, & x \neq 0\left\{\begin{array}{ll}x< 0\\x>0\end{array}\right. \\ k ;x=0\end{array}\right.$
Sol :
At x=0
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos {4} x}{8 x^{2}}\right)$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{{8 x^{2}}}$
$=\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)^{2}$
$=(1)^{2}=1$
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos 4 x}{8 x^{2}}\right)$
=1
f(0)=k
∵f(x),x=0 पर संतता है
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
1=1=k
∴k=1
(v) $f(x)=\left\{\begin{array}{cl}\frac{2^{x+2}-16}{4^{x}-16} & x \neq 2\left\{\begin{array}{l}x<2 \\ x>2\end{array}\right. \\ k, x=2\end{array}\right.$
Sol :
At x=2 ,
L.H.L
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}$
$=\lim_{x{\rightarrow 2}} \frac{2^{x} \cdot 2^{2}-16}{4^{x}-16}$
$=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}\right)^{2}-4^{2}}$
$=\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}$
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=\frac{1}{2}$
f(2)=k
∵ f(x), x=2 पर संतता है ।
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$
$\frac{1}{2}=\frac{1}{2}=k$
∴$k=\frac{1}{2}$
Question 23
(यदि) If $f(x)=\left\{\begin{array}{ll}1, \text {(if ) } x<5 \\ a x+b, \text { (if ) } 3<x<5 \\ 7, \text { (if ) } x \geq 5\end{array}\right.$
a और b का मान ज्ञात करें जिसके लिए f एक संतत फलन है।
find the value of a and be for which f(x) si continuous function
Sol :
At x=3
L.H.L
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}(1)=1$
R.H.L
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}(a x+b)$
=a(3)+b
=3a+b
∵f(x),x=3 पर संतता है ।
$\lim _{x \rightarrow 3}-f(x)=\lim _{x \rightarrow 3^{+}} f(x)$
1=3a+b
3a+b=1..(i)
$f(x)=\left\{\begin{array}{ll}1, \text {(if ) } {x \leq 3} \\ a x+b, \text { (if }) 3<x<5 \\ 7, \text {(if ) } x \geq 5\end{array}\right.$
At x=5,
L.H.L
$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(a x+b)$
=5a+b
R.H.L
$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5}(7)=7$
∵f(x),x=5 पर संतता है
$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)$
5a+b=7..(ii)
(i) तथा (ii) को जोड़ने पर
$\begin{array}3 a+b=1\\5 a+b=7 \\\hline -2a=-6 \end{array}$
a=3
a का मान (i) मे रखने पर,
3a+b=1
3(3)+b=1
b=1-9
b=-8
∴a=3 , b=-8
Question 24
(यदि) If $\begin{aligned} f(x) &=a x^{2}-b, 0 \leq x<1 \\ &=2, x=1 \\ &=x+1, 1<x \leq 2 \end{aligned}$
x=1 पर संतत है तो a और b में सम्बन्ध ज्ञात करें ।
[is continuous at x=1. find the relation between a and b]
Sol :
At x=1
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(a x^{2}-b\right)$
$=a(1)^{2}-b$
=a-b
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(x+1)$
=1+1
=2
∵f(x),x=1 पर संतत है।
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
a-b=2 or a=2+b
Question 25
माना कि (Let) $\begin{aligned} f(x) &=x+1,(\text { if }) x \leq 1 \\ &=3-a x^{2},(\text { if }) x>1 \end{aligned}$
a के किस मान के लिए फलन f(x) संतत है।
[for what value of a will the function f(x) be continuous?]
Sol :
At x=1,
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(x+1)$
=1+1
=2
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(3-a x^{2}\right)$
$=3-a(1)^{2}$
=3-a
∵f(x),x=1 पर संतता है
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
2=3-a
a=3-2
a=1
Question 26
a और b का निर्धारण करें ताकि फलन f जो निम्न प्रकार प्रदत्त है,
[Determine a and be so that the function f given by]
$\begin{aligned} f(x) &=\frac{1-\sin ^{2} x}{3 \cos ^{2} x}, x<\frac{\pi}{2} \\ &=a, \quad x=\frac{\pi}{2} \\ &=\frac{b(1-\sin x)}{(\pi-2 x)^{2}}, x>\frac{\pi}{2} \end{aligned}$
$x=\frac{\pi}{2}$ पर संतत है
[is continuous at $x=\frac{\pi}{2}$]
Sol :
At $x=\frac{\pi}{2}$
L.H.L
$\lim _{x \rightarrow \frac{\pi}{2}-} f(x)=\lim _{x \rightarrow \frac{\pi}{2}^-}\frac{1-sin^2x}{3\cos^2 x}$
$=\lim _{h \rightarrow 0} \frac{1-\sin ^{2}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}$
$=\lim _{h \rightarrow 0} \frac{l-\cos ^{2} h}{3 \sin ^{2} h}$
$=\lim _{h \rightarrow 0} \frac{1-\cos ^{2} h}{3 \sin ^{2} h}$
$=\lim _{h \rightarrow 0} \frac{\sin ^{2} h}{3 \sin^ 2 h}$
$=\frac{1}{3}$
R.H.L
$\lim _{x \rightarrow \frac{\pi}{2}^+}f(x)=\lim _{x \rightarrow \frac{\pi}{2}^+} \frac{b(1-\sin x)}{(\pi-2 x)^{2}}$
$=\lim _{h \rightarrow 0} \frac{b\left[1-\sin \left(\frac{\pi}{2}+h\right)\right]}{\left[\pi-2\left(\frac{\pi}{2}+h\right)\right]^{2}}$
$=\lim _{h \rightarrow 0} \frac{b[1-\cos h]}{[ \pi-\pi-2 h]^{2}}$
$=\lim _{h \rightarrow 0} \frac{b \cdot 2 \sin^2 \frac{h}{2}}{(-2 h)^{2}}$
$=\lim _{h \rightarrow 0} \frac{2 b \sin^2 \frac{h}{2}}{2^{4} h^{2}}$
$=\frac{b}{2} \lim _{h \rightarrow 0} \frac{\sin 2^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{1}{4}$
$=\frac{b}{8} \lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2}$
$=\frac{b}{8}(1)^{2}=\frac{b}{8}$
$f\left(\frac{\pi}{2}\right)=a$
∵f(x),$x=\frac{\pi}{2}$ पर <to be added>
$\lim _{x \rightarrow \frac{\pi}{2}^-}f(x)=\lim _{2 \rightarrow \frac{\pi}{2}+} f(x)=f\left(\frac{\pi}{2}\right)$
$\frac{1}{3}=\frac{b}{8}=a$
$\begin{array}{r|l}\frac{1}{3}=9 &\frac{b}{8}=\frac{1}{3} \\ \therefore a=1&b=\frac{8}{3}\end{array}$
Question 27
यदि (If) $f(x)=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x},-1 \leq x<0$
$\quad=\frac{2 x+1}{x-1}, 0 \leq x<1$
अन्तराल [-1,1] में संतत है, तो p निकालें ।
[is continuous in the interval [-1,1], find p.]
Sol :
At x=0,
L.H.L
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{1+p{x}}-\sqrt{1-p{x}}}{x} \times \frac{\sqrt{1+px}+\sqrt{1-px}}{\sqrt{1+p{x}}+\sqrt{1-p{x}}}$
$=\lim _{x \rightarrow 0} \frac{(\sqrt{1+px})^{2}-(\sqrt{1-px})^{2}}{x[\sqrt{1+px}+\sqrt{1-px}]}$
$\left.=\lim _{x \rightarrow 0} \frac{1+p x-1+p x}{x[\sqrt{1+p x}+\sqrt{1-p x}]}\right.$
$=\lim _{x \rightarrow 0} \frac{2 P x}{x[\sqrt{1+p x}+\sqrt{1-p x}]}$
$=\frac{2 p}{\sqrt{1+p(0)}+\sqrt{1-p(0)}}$
$=\frac{2 p}{1+1}=\frac{2f}{2}=p$
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{2 x+1}{x-1}$
$=\frac{2(0)+1}{0-1}$
$=\frac{1}{-1}=-1$
∵f(x),x=0 [-1,1] पर संतता है ।
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
p=-1
Question 28
माना (Let) $\begin{aligned}f(x)&=-2\sin x,\text{(if) }x \leq-\frac{\pi}{2}\\&=A \sin x+B,\text{(if) }-\frac{\pi}{2}<x<\frac{\pi}{2}\\&=\cos x,\text{(if) }x \geq \frac{\pi}{2}\end{aligned}$
A और B ज्ञात करें जो f(x) को $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ में संतत बना दें ।
(find A and B so as to make f(x) continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )
Sol :
[a,b]
$\lim _{x \rightarrow a^{+}} f(x)=f(a)$
$\lim_{x\rightarrow b^-}f(x)=f{(b)}$
At $x=-\frac{\pi}{2}$
R.H.L
$\lim _{x \rightarrow-\frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow-\frac{\pi}{2}}(A \sin x+B)$
$=A \sin \left(-\frac{\pi}{2}\right)+B$
=-A+B
$f\left(-\frac{\pi}{2}\right)=-2 \sin \left(-\frac{\pi}{2}\right)$
$=2 \sin \frac{\pi}{2}$
=2(1)=2
∵f(x), $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ पर संतता है ।
$\lim _{x \rightarrow-\frac{\pi}{2}^+}f(x)=f\left(-\frac{\pi}{2}\right)$
-A+B=2..(i)
At $x=\frac{\pi}{2}$ ,
L.H.L
$\lim_{x\rightarrow \frac{\pi}{2}^-} f(x)=\lim _{x \rightarrow \frac{\pi}{2}}(A \sin x+B)$
$=A\sin\frac{\pi}{2}+B$
=A+B
$f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0$
∵ $f(x),\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ पर संतता है ।
$\lim _{x \rightarrow \frac{\pi}{2}-} f(x)=f\left(\frac{\pi}{2}\right)$
$\lim _{x \rightarrow \frac{\pi}{2}^-} f(x)=f\left(\frac{\pi}{2}\right)$
A+B=0..(ii)
(i) तथा (ii) को जोड़ने पर,
$\begin{aligned}-A+B &=2 \\ A+B &=0 \\\hline 2 B &=2 \end{aligned}$
B=1
B का मान (iii) मे रखने पर ,
A+B=0
A+1=0
A=-1
∴A=-1 , B=1
Question 29
फलन $\int$ निम्नप्रकार परिर्भाषित है
[The function f is defined as]
$f(x)=\left\{\begin{array}{ll}x^{2}+a x+b, & 0 \leq x<2 \\ 3 x+2, &{2\leq x \leq 4} \\ 2 a x+5 b, & 4<x \leq 8\end{array}\right.$
यदि f(x),[0,8] में संतत है तो a और b का मान निकालें ।
[If f(x) is continuous on [0,8], find the value of a and b]
Sol :
At x=2,
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(x^{2}+ax+b\right)$
$=2^{2}+a(2)+b$
=2a+b+4
R.H.L
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(3 x+2)$
=3(2)+2
=8
∵f(x),x=2 पर संतता है ।
$\lim _{x \rightarrow 2^-}f(x)=\lim _{x \rightarrow 2+} f(x)$
2a+b+4=8
2a+b=4..(i)
At x=4
L.H.L
$\lim_{x \rightarrow 4^{-}}f(x)=\lim _{x \rightarrow 4}(3 x+2)$
R.H.L
$\lim _{x \rightarrow 4^+}f(x)=\lim _{x \rightarrow 4}(2 a x+5 b)$
=2a(4)+5b
=8a+5b
∵f(x),x=4 पर संतता है ।
$\lim_{x\rightarrow 4^+} f(x)=\lim _{x \rightarrow 4^-}f(x)$
8a+5b=14..(ii)
(i) तथा (ii) को जोड़ने पर
$\begin{aligned}10 a+5 b=20\\8 a+5 b=14\\\hline 2a=6\end{aligned}$
a=3
a का मान (i) मे रखने पर ,
2a+b=4
2(3)+b=4
6+b=4
b=-2
∴a=3 , b=-2
Question 30
a और b का मान ज्ञात करें ताकि निम्न प्रकार परिभाषित फलन f
[Find the value of a and be such that the functions f defined by]
$f(x)=\left\{\begin{array}{ll}x+a \sqrt{2} \sin x, & 0 \leq x<\frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2}<x \leq \pi\end{array}\right.$
$0 \leq x \leq \pi$ में सभी x के लिए संतत है ।
[is continuous for all values of x in 0≤x≤π]
Sol :
At $x=\frac{\pi}{4}$,
L.H.L
$\lim _{x \rightarrow \frac{\pi}{4}-f} f(x)=\lim _{x \rightarrow \frac{\pi}{4}}(x+a \sqrt{2} \sin x)$
$=\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1\pi}{2}$
$=\frac{\pi}{4}+a \sqrt{2} \times \frac{1}{\sqrt{2}}$
$=a+\frac{\pi}{4}$
R.H.L
$\lim _{x \rightarrow \frac{\pi}{4}^+} f(x)=\lim _{x \rightarrow \frac{\pi}{4}}(2 x \cot x+b)$
$=\frac{2 \pi}{4} \cot\frac{\pi}{4}+b$
$=\frac{x}{2}(1)+b$
$=b+\frac{\pi}{2}$
∵f(x),$x=\frac{\pi}{4}$ पर संतता है ।
$\lim _{x \rightarrow \frac{\pi}{4}^-} f(x)=\lim _{x \rightarrow \frac{\pi}{4}^+} f(x)$
$a+\frac{\pi}{4}=b+\frac{\pi}{2}$
$a=b+\frac{\pi}{4}$..(i)
At $x=\frac{\pi}{2}$
L.H.L
$\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}}(2x \cot x+b)$
$=2 \frac{\pi}{2} \cdot \operatorname{cot} \frac{\pi}{2}+b$
=π.0+b=b
R.H.L
$\lim _{x \rightarrow \frac{\pi}{2}+} f(x)=\lim _{x \rightarrow \frac{\pi}{2}}(a \cos 2 x-b \sin x)$
$=a \cos 2\left(\frac{\pi}{2}\right)-b \sin \frac{\pi}{2}$
=a(-1)-b(1)
=-a-b
∵ f(x),$ x=\frac{\pi}{2}$ पर संतता है ।
$\lim _{x \rightarrow \frac{\pi}{2}^-} f(x)=\lim _{x \rightarrow \frac{\pi}{2}+} f(x)$
b=-a-b
a=-b-b
a=-2b..(ii)
समीकरण(equation) (i) तथा (ii) से ,
$b+\frac{\pi}{4}=-2 b$
$b+2 b=-\frac{\pi}{4}$
$3 b=-\frac{\pi}{4}$
$b=-\frac{\pi}{12}$
b का मान समीकरण(equation) (i) मे रखने पर,
$a=b+\frac{\pi}{4}$
$a=\frac{-\pi}{12}+\frac{\pi}{4}$
$a=\frac{-x+3 \pi}{12}$
$=\frac{2 \pi}{12}=\frac{\pi}{6}$
∴ $a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
Question 31
एक फलन f निम्नप्रकार परिभाषित है,
[A function f(x) is defined as follows]
$\begin{array}{l}f(x)=\frac{\sin x}{x}, (\text { when }) x \neq 0,\left\{\begin{array}{l}x<a \\ x>0\end{array}\right.\\=2,\text{when }x=0\end{array}$
क्या f(x), x=0 पर संतत है ? यदि नहीं तो x=0 पर f(x) का मान क्या होना चाहिए ताकि f(x), x=0 पर संतत हो जाय ।
[Is f(x) continuous at x=0 ? If not what should be the value of f(x) at x=0 so that f(x) becomes continuous at x=0 ?]
Sol :
At x=0,
L.H.L
$\lim_{x\rightarrow 0^-}f(x)=\lim_{x \rightarrow 0}\frac{\sin x}{x} =1$
R.H.L
$\lim_{x\rightarrow 0^+}f(x)=\lim_{x \rightarrow 0}\frac{\sin x}{x} =1$
f(0)=2
∴$\lim _{x \rightarrow 0^{-}}f(x)=\lim_{x\rightarrow0^+}f(x)\neq f(0)$
f(x),x=0 पर संतता नही है ।
यदि f(x)=1, जब x=0 हो , तो f(x) संतता होगा ।
Question 32
फलन (The function) $f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}, x=0$
पर परिभाषित नहीं है । f(0) का मान निकलें ताकि f(x), x=0 पर संतत है ।
[is not defined at x=0.Find the value of f(0) so that f(x) is continuous at x=0]Sol :
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{2}$
$=\lim _{x \rightarrow 0}\left(\frac{\log \left(1+a{x}\right)}{x}-\frac{\log\left(1-bx\right)}{x}\right)$
[$\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$]
$=\lim _{x \rightarrow 0} \frac{\log (1+a x)}{x}-\lim _{x \rightarrow 0} \frac{\log (1-b x)}{x}$
$=\lim _{x \rightarrow 0} \frac{\log (1+a x)}{a x} \times a-\lim _{x \rightarrow 0} \frac{\log [1+(-b x)]}{-bx}\times (-b)$
=1×a-1×(-b)
=a+b
∵f(x),x=0 पर संतता(continuous) है
$\lim _{x \rightarrow 0} f(x)=f(0)$
a+b=f(0)
Question 33
(यदि) If $f(x)=\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}, x \neq \frac{\pi}{4}$ तो $f\left(\frac{\pi}{4}\right)$ निकालें यदि $f(x), x=\frac{\pi}{4}$ पर सतत है।
(Then find $f\left(\frac{\pi}{4}\right)$ so that f(x) is continuous at $x=\frac{\pi}{4}$ ).
Sol :
At $x=\frac{\pi}{4}$
L.H.L
$\lim _{x \rightarrow \frac{\pi}{4}^-} f(x)=\lim _{x \rightarrow \frac{\pi}{4}^-} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}$
$=\lim _{h \rightarrow 0} \frac{\tan \left[\frac{\pi}{4}-\left(\frac{\pi}{4}-h\right)\right]}{\cot 2\left(\frac{\pi}{4}-h\right)}$
$=\lim _{h \rightarrow 0} \frac{\tan \left[\frac{\pi}{4}-\frac{\pi}{4}+h\right]}{\cot \left(\frac{\pi}{2}-2 h\right)}$
$=\lim _{h \rightarrow 0} \frac{\tanh }{\tan 2 h}$
<to be added>
$=\lim _{h \rightarrow 0} \frac{\frac{\tan h}{h}}{\frac{\tan 2 h}{h}}$
$=\frac{\lim _{h \rightarrow 0} \frac{\tanh }{4}}{\lim _{h \rightarrow 0} \frac{\tan 2 h}{2h} \times 2}$
$=\frac{1}{2} \times \frac{1}{1}=\frac{1}{2}$
R.H.L
$\lim_{x\rightarrow \frac{\pi}{4}^{+}}=\lim _{x \rightarrow \frac{\pi}{4}+} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}$
$=\lim_{h \rightarrow 0} \frac{\tan \left[\frac{\pi}{4}-\left(\frac{\pi}{4}+h\right)\right]}{\operatorname{cot} 2\left(\frac{\pi}{4}+h\right)}$
$=\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}-h\right)}{\cot \left(\frac{\pi}{2}+2 h\right)}$
$=\lim _{h \rightarrow 0} \frac{-\tan h}{-\tan 2h}$
$=\frac{1}{2}$
∵f(x),$x=\frac{\pi}{4}$ पर संतता(continuous) है
$\lim_{x \rightarrow \frac{\pi}{4}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)$
$=f\left(\frac{\pi}{4}\right)$
$\frac{1}{2} \quad=\frac{1}{2}=f\left(\frac{\pi}{4}\right)$
∴$f\left(\frac{\pi}{4}\right)=\frac{1}{2}$