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KC Sinha Mathematics Solution Class 12 Chapter 10 अवकलनीयता (Differentiability) Exercise 10.1 (Q7-Q10)




Exercise 10.1

Question 7

यदि (If) f(2)=4 , f^{\prime}(2)=1 तो  निकालें। (then find) \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}

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)

[]


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।
(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


Question 10

माना कि f : R→R निम्न प्रकार परिभाषित है।
(Let :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)

[]

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