Exercise 9.1
(Q1-Q4) (Q5-Q8) (Q9-Q12) (Q13-Q16) (Q17-Q20) (Q21-Q24) (Q25-Q28) (Q29-Q33)
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
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
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