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)
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