KC Sinha Mathematics Solution Class 12 Chapter 10 अवकलनीयता (Differentiability) Exercise 10.1 (Q7-Q10)
Sol :
$\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$
$=\lim _{x \rightarrow 2} \frac{x \cdot f(2)-2 f(2)-2 f(x)+2f(2)}{x-2}$
$=\lim_{x\rightarrow 2} \frac{f(2)[x-2]-2[f(x)-f(1)]}{x-2}$
$=\lim_{x \rightarrow 2}\left\{\left(\frac{f(2) \cdot[x-2]}{x-2}-\frac{2[f(x)-f(2)]}{x-2}\right.\right\}$
$=\lim_{x \rightarrow 2} f(2)-2 \lim_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}$
$=f(2)-2 f^{\prime}(2)$
[x-2=h
when x🠖2 ,x🠖0
$\lim_{h\rightarrow0}\frac{(2+h)-f(2)}{h}$ ]
=4-2(1)
=4-2
=2
Question 8
यदि (If) $f(x)=\left\{\begin{array}{ll}-x, & x<0 \\ x^{2}, & 0 \leq x \leq 1 \\ x^{2}-x+1, & x>1\end{array}\right.$f(x) का x=0 तथा x=1 पर अवकलनीयता की जाँच करें।
[examine the differentiability of f(x) at x=0 and x=1]
Sol :
L.H.D
$f^{\prime}\left(0^{-}\right)=\lim_{h \rightarrow 0^-}\frac{f(0-h)-f(0)}{-h}$
$=\lim _{h \rightarrow 0} \frac{-(0-h)-(-0)}{-h}$
$=\lim_{h \rightarrow 0} \frac{h}{{h}}=-1$
R.H.D
$f^{\prime}\left(0^{+}\right)=\lim_{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}$
$=\lim_{h \rightarrow 0}=\frac{(0+h)^{2}-(0)^{2}}{h}$
$=\lim_{h \rightarrow 0} \frac{h^{2}}{h}=0$
$f^{\prime}\left(0^{-}\right) \neq f^{\prime}\left(0^{+}\right)$
[]
L.H.D
$f^{\prime}\left(1^{-}\right)=\lim_{h \rightarrow 0^{-}} \frac{f(1-h)-f(1)}{-h}$
$=\lim _{h \rightarrow 0} \frac{(1-h)^{2}-1^{2}}{-h}$
$=\lim_{h\rightarrow 0} \frac{1-2 h+h^{2}-1}{-h}$
$=\lim_{h \rightarrow 0}\frac{-h(2-h)}{-h}$
=2-0
=2
R.H.D
$f^{\prime}\left(1^{+}\right)=\lim_{h{\rightarrow 0}}\frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{(1+h)^{2}-(1+h)+1-\left(1^{2}-1+1\right)}{h}$
$=\lim_{h \rightarrow 0}\frac{1+2 h+h^{2}-1-h+1-1}{h}$
$=\lim _{h \rightarrow 0} \frac{h+h^{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{k(1+h)}{h}$
=1+0
=1
$f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)$
[]
Sol :
L.H.D
$f^{\prime}\left(0^{-}\right)=\lim_{h \rightarrow 0^-}\frac{f(0-h)-f(0)}{-h}$
$=\lim _{h \rightarrow 0} \frac{-(0-h)-(-0)}{-h}$
$=\lim_{h \rightarrow 0} \frac{h}{{h}}=-1$
R.H.D
$f^{\prime}\left(0^{+}\right)=\lim_{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}$
$=\lim_{h \rightarrow 0}=\frac{(0+h)^{2}-(0)^{2}}{h}$
$=\lim_{h \rightarrow 0} \frac{h^{2}}{h}=0$
$f^{\prime}\left(0^{-}\right) \neq f^{\prime}\left(0^{+}\right)$
[]
L.H.D
$f^{\prime}\left(1^{-}\right)=\lim_{h \rightarrow 0^{-}} \frac{f(1-h)-f(1)}{-h}$
$=\lim _{h \rightarrow 0} \frac{(1-h)^{2}-1^{2}}{-h}$
$=\lim_{h\rightarrow 0} \frac{1-2 h+h^{2}-1}{-h}$
$=\lim_{h \rightarrow 0}\frac{-h(2-h)}{-h}$
=2-0
=2
R.H.D
$f^{\prime}\left(1^{+}\right)=\lim_{h{\rightarrow 0}}\frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{(1+h)^{2}-(1+h)+1-\left(1^{2}-1+1\right)}{h}$
$=\lim_{h \rightarrow 0}\frac{1+2 h+h^{2}-1-h+1-1}{h}$
$=\lim _{h \rightarrow 0} \frac{h+h^{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{k(1+h)}{h}$
=1+0
=1
$f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)$
[]
Question 9
a और b का मान निकालें ताकि(Find the value of a and b so that)
$f(x)=\left\{\begin{array}{ll}x^{2}+3 x+a, & x \leq 1 \\ b x+2, & x>1\end{array}\right.$
x=1 पर अवकलनीय है।
$f(x)=\left\{\begin{array}{ll}x^{2}+3 x+a, & x \leq 1 \\ b x+2, & x>1\end{array}\right.$
x=1 पर अवकलनीय है।
(is differentiable at x=1)
Sol :
A+x=1
L.H.D
$f^{\prime}\left(1^{-}\right)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$=\lim _{h \rightarrow 0} \frac{(1-h)^{2}+3(1-h)+a-\left(1^{2}+3(1)+a\right)}{-h}$
$=\lim _{h \rightarrow 0} \frac{1-2 h+h^{2}+3-3 h+a-1-3-a}{-h}$
$=\lim_{h{\rightarrow 0}} \frac{-5 h+h^{2}}{-h}$
$=\lim_{h \rightarrow 0} \frac{-h(5-h)}{-h}$
=5-0
=5
R.H.D
$f^{\prime}\left(1^{+}\right)=\lim_{h\rightarrow0^{+}} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{b(1+h)+2-(b \cdot 1+2)}{h}$
$=\lim_{h \rightarrow 0} \frac{b+b h+2-b-2}{h}$
$=\lim_{h \rightarrow 0} \frac{b h}{h}$
=b
[]
$f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)$
5=b
∵[]
[]
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}+3 x+a\right)$
$=1^{2}+3(1)+a$
=1+3+a
=4+a
R.H.L
$\lim_{x \rightarrow 1^+}f(x)=\lim_{x \rightarrow 1}(b z+2)$
=b(1)+2
=b+2
[]
$\lim_{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^{+}} f(x)$
4+a=b+2
∵ b=5
∴ 4+a=5+2
4+a=7
a=3 , b=5
$f(x)=\left\{\begin{array}{ll}2 x-2, & x<-1 \\ A x+B, & -1 \leq x \leq 1 \\ 5 x+7, & x>1\end{array}\right.$
A और B ज्ञात करें कि ताकि f सभी x के लिए संतत है। क्या f सभी x के लिए अवकलनीय है।
Sol :
A+x=1
L.H.D
$f^{\prime}\left(1^{-}\right)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$=\lim _{h \rightarrow 0} \frac{(1-h)^{2}+3(1-h)+a-\left(1^{2}+3(1)+a\right)}{-h}$
$=\lim _{h \rightarrow 0} \frac{1-2 h+h^{2}+3-3 h+a-1-3-a}{-h}$
$=\lim_{h{\rightarrow 0}} \frac{-5 h+h^{2}}{-h}$
$=\lim_{h \rightarrow 0} \frac{-h(5-h)}{-h}$
=5-0
=5
R.H.D
$f^{\prime}\left(1^{+}\right)=\lim_{h\rightarrow0^{+}} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{b(1+h)+2-(b \cdot 1+2)}{h}$
$=\lim_{h \rightarrow 0} \frac{b+b h+2-b-2}{h}$
$=\lim_{h \rightarrow 0} \frac{b h}{h}$
=b
[]
$f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)$
5=b
∵[]
[]
L.H.L
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}+3 x+a\right)$
$=1^{2}+3(1)+a$
=1+3+a
=4+a
R.H.L
$\lim_{x \rightarrow 1^+}f(x)=\lim_{x \rightarrow 1}(b z+2)$
=b(1)+2
=b+2
[]
$\lim_{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^{+}} f(x)$
4+a=b+2
∵ b=5
∴ 4+a=5+2
4+a=7
a=3 , b=5
Question 10
माना कि f : R→R निम्न प्रकार परिभाषित है।
(Let f :R→R be defined by)$f(x)=\left\{\begin{array}{ll}2 x-2, & x<-1 \\ A x+B, & -1 \leq x \leq 1 \\ 5 x+7, & x>1\end{array}\right.$
A और B ज्ञात करें कि ताकि f सभी x के लिए संतत है। क्या f सभी x के लिए अवकलनीय है।
[Determine A and B so that f is continuous for all x . Is f differentiable for all x ?]
Sol :
A+x=1
L.H.L
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1}(2 x-2)$
=2(-1)-2
=-2-2
=-4
R.H.L
$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1}(A x+B)$
=A(-1)+B
=-A+B
∵ []
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)$
-4=-A+B
A-B=4..(i)
A+x=1
L.H.L
$\lim_{x{\rightarrow 1^-}}{f(x)}=\lim _{x \rightarrow 1}(A x+B)$
=A(1)+B
=A+B
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(5 x+7)$
=5(1)+7
=12
[]
$\lim_{x+1^-}f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
A+B=12..(ii)
From equation (i) and (ii)
$\begin{aligned}A-B=4\\A+B=12\\ \hline2 A=16 \end{aligned}$
A=8 , B=4
$f(x)=\left\{\begin{array}{ll}2 x-2, & x<-1 \\ 8 x+4, & -1 \leq x \leq 1 \\ 5 x+7, & x>1\end{array}\right.$
A+x=1
L.H.D
$f^{\prime}(-1)=\lim_{h \rightarrow 0^1} \frac{f(-1-h)-f(-1)}{-h}$
$=\lim_{h \rightarrow 0} \frac{2(-1-h)-2-\{2(-1)-2\}}{-h}$
$=\lim_{h\rightarrow0} \frac{-2-2h-2+2+2}{-h}$
$=\lim_{h\rightarrow 0}-\frac{2h}{-h}=2$
R.H.D
$f^{\prime}\left(-1^{+}\right)=\lim_{h \rightarrow 0^{+}} \frac{f(-1+h)-f(-1)}{h}$
$=\lim_{h \rightarrow 0^+}\frac{8(-1+h)+4-\{8(-1)+4\}}{h}$
$=\lim _{h \rightarrow 0} \frac{-8+8h+4+8-4}{h}$
$=\lim _{h \rightarrow 0} \frac{8 h}{h}=8$
$f^{\prime}\left(-1^{-}\right) \neq f^{\prime}\left(-1^{+}\right)$
[]
$f(x)=\left\{\begin{array}{ll}2 x-2, & x<-1 \\ 8 x+4, & -1 \leq x \leq 1 \\ 5 x+7, & x>1\end{array}\right.$
A+x=1
L.H.D
$f^{\prime}(1^-)=\lim_{h \rightarrow 0^-}\frac{f(1-h)-f(1)}{-h}$
$=\lim_{h\rightarrow0} \frac{8(1-h)+4-(8 \cdot 1+4)}{-h}$
$=\lim _{h \rightarrow 0} \frac{8-8 h+4-8-4}{-h}$
$=\lim_{h\rightarrow 0}-\frac{8 h}{-h}=8$
R.H.D
$f^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f^{\prime}(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{5(1+h)+7-(5\times 1+7)}{h}$
$=\lim _{h \rightarrow 0} \frac{5+5 h+7-5-7}{h}$
$=\lim_{h\rightarrow 0} \frac{5 h}{h}=5$
$f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)$
[]
Sol :
A+x=1
L.H.L
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1}(2 x-2)$
=2(-1)-2
=-2-2
=-4
R.H.L
$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1}(A x+B)$
=A(-1)+B
=-A+B
∵ []
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)$
-4=-A+B
A-B=4..(i)
A+x=1
L.H.L
$\lim_{x{\rightarrow 1^-}}{f(x)}=\lim _{x \rightarrow 1}(A x+B)$
=A(1)+B
=A+B
R.H.L
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(5 x+7)$
=5(1)+7
=12
[]
$\lim_{x+1^-}f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
A+B=12..(ii)
From equation (i) and (ii)
$\begin{aligned}A-B=4\\A+B=12\\ \hline2 A=16 \end{aligned}$
A=8 , B=4
$f(x)=\left\{\begin{array}{ll}2 x-2, & x<-1 \\ 8 x+4, & -1 \leq x \leq 1 \\ 5 x+7, & x>1\end{array}\right.$
A+x=1
L.H.D
$f^{\prime}(-1)=\lim_{h \rightarrow 0^1} \frac{f(-1-h)-f(-1)}{-h}$
$=\lim_{h \rightarrow 0} \frac{2(-1-h)-2-\{2(-1)-2\}}{-h}$
$=\lim_{h\rightarrow0} \frac{-2-2h-2+2+2}{-h}$
$=\lim_{h\rightarrow 0}-\frac{2h}{-h}=2$
R.H.D
$f^{\prime}\left(-1^{+}\right)=\lim_{h \rightarrow 0^{+}} \frac{f(-1+h)-f(-1)}{h}$
$=\lim_{h \rightarrow 0^+}\frac{8(-1+h)+4-\{8(-1)+4\}}{h}$
$=\lim _{h \rightarrow 0} \frac{-8+8h+4+8-4}{h}$
$=\lim _{h \rightarrow 0} \frac{8 h}{h}=8$
$f^{\prime}\left(-1^{-}\right) \neq f^{\prime}\left(-1^{+}\right)$
[]
$f(x)=\left\{\begin{array}{ll}2 x-2, & x<-1 \\ 8 x+4, & -1 \leq x \leq 1 \\ 5 x+7, & x>1\end{array}\right.$
A+x=1
L.H.D
$f^{\prime}(1^-)=\lim_{h \rightarrow 0^-}\frac{f(1-h)-f(1)}{-h}$
$=\lim_{h\rightarrow0} \frac{8(1-h)+4-(8 \cdot 1+4)}{-h}$
$=\lim _{h \rightarrow 0} \frac{8-8 h+4-8-4}{-h}$
$=\lim_{h\rightarrow 0}-\frac{8 h}{-h}=8$
R.H.D
$f^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f^{\prime}(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{5(1+h)+7-(5\times 1+7)}{h}$
$=\lim _{h \rightarrow 0} \frac{5+5 h+7-5-7}{h}$
$=\lim_{h\rightarrow 0} \frac{5 h}{h}=5$
$f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)$
[]
No comments:
Post a Comment