KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q29-Q33)

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KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q29-Q33)

Exercise 9.1

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

$\lim_{x \rightarrow \frac{\pi}{4}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)$

$=f\left(\frac{\pi}{4}\right)$

$\frac{1}{2} \quad=\frac{1}{2}=f\left(\frac{\pi}{4}\right)$

∴

$f\left(\frac{\pi}{4}\right)=\frac{1}{2}$

KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q29-Q33)

Exercise 9.1

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