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.
KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q29-Q33)
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
a और b का मान ज्ञात करें ताकि निम्न प्रकार परिभाषित फलन f
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}
[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
[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 ।
फलन (The function) f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}, 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)
(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।
\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}
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}
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