Exercise 9.1
Question 25
KC Sinha Mathematics Solution Class 12 Chapter 9 संतता या सांतत्य (continuity) Exercise 9.1 (Q25-Q28)
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
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]
$\begin{aligned} f(x) &=\frac{1-\sin ^{2} x}{3 \cos ^{2} x}, x<\frac{\pi}{2} \\ &=a, \quad x=\frac{\pi}{2} \\ &=\frac{b(1-\sin x)}{(\pi-2 x)^{2}}, x>\frac{\pi}{2} \end{aligned}$
[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
[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>
$\lim _{x \rightarrow \frac{\pi}{2}^-}f(x)=\lim _{2 \rightarrow \frac{\pi}{2}+} f(x)=f\left(\frac{\pi}{2}\right)$
$\frac{1}{3}=\frac{b}{8}=a$
$\begin{array}{r|l}\frac{1}{3}=9 &\frac{b}{8}=\frac{1}{3} \\ \therefore a=1&b=\frac{8}{3}\end{array}$
यदि (If) $f(x)=\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x},-1 \leq x<0$
$\quad=\frac{2 x+1}{x-1}, 0 \leq x<1$
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
$\quad=\frac{2 x+1}{x-1}, 0 \leq x<1$
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
माना (Let) $\begin{aligned}f(x)&=-2\sin x,\text{(if) }x \leq-\frac{\pi}{2}\\&=A \sin x+B,\text{(if) }-\frac{\pi}{2}<x<\frac{\pi}{2}\\&=\cos x,\text{(if) }x \geq \frac{\pi}{2}\end{aligned}$
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 $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )
(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
[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
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