Contents
Question 1
यदि (If) $f(x)=\frac{x-2}{x^{2}-3 x+2}$ , when x≠2
=1 , when x=2
तो (then find) $f^{\prime}(2)$
Sol :
When x≠2 , $f(x)=\frac{x-2}{x^{2}-3 x+2}$
$=\frac{x -2}{(x-1)(x-2)}=\frac{1}{x-1}$
$f(2)=\left\{\begin{array}{cl}\frac{1}{x-1} & \text { when } x\neq 2 \\ 1 & , \text{when }x=2\end{array}\right.$
$f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2+h-1}-1}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{1+h}-1}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1-(1+h)}{1+h}}{h}$
$=\lim_{h \rightarrow 0} \frac{1-1-h}{h(1+h)}$
$=\lim_{h \rightarrow 0} \frac{-h}{h(1+h)}$
$=\frac{-1}{1+0}=-1$
Question 2
यदि (If) $f(x)=x+x^{3} \sin \frac{1}{x}$ , when x≠0
=0 , when x=0
तो x=0 पर f(x) का अवकलन निकालें (then find the derivative of f(x) at x=0)
Sol :
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{h+h^{3} \cdot \sin \frac{1}{h}-0}{h}$
$=\lim _{h \rightarrow 0} \frac{h\left(1+h^{2} \sin \frac{1}{h}\right)}{h}$
$=\lim _{h \rightarrow 0}\left[1+h \cdot \frac{\sin \frac{1}{h}}{\frac{1}{h}}\right]$
$=\lim_{h \rightarrow 0}(1)+\lim_{h \rightarrow 0} h . \lim_{h \rightarrow 0} \frac{\sin \frac{1}{h}}{\frac{1}{h}}$
=1+0×1=1
Question 3
क्या (Is) |x+1| पर अवकलनी differentiable at x=-1 ?
Sol :
A+x=-1
L.H.D
$f^{\prime}\left(-1^{-}\right)=\lim_{h\rightarrow 0^-} \frac{f(-1-h)-f(-1)}{-h}$
$=\lim _{h \rightarrow 0} \frac{|-1-h+1|-|-1+1|}{-h}$
$=\lim_{h\rightarrow 0} \frac{|-h|-0}{-h}$
$=\lim _{h \rightarrow 0} \frac{h}{-h}=-1$
R.H.D
$f^{\prime}\left(-1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(-1+4)-f(-1)}{h}$
$=\lim_{h \rightarrow 0} \frac{|-1+h+1|-|-1+1|}{h}$
$=\lim_{h \rightarrow 0} \frac{|h|-0}{h}$
$=\lim _{h \rightarrow 0} \frac{h}{h}=1$
$f^{\prime}\left(-1^{-}\right) \neq f\left(-1^{+}\right)$
Question 4
यदि (If) f(x)=x2 , when x≥0
=x , when x<0
तो x=0 पर f(x) का बाएँ और दाएँ से अवकलन निकालें । क्या f(x) , x=0 पर अवकलनीय है?
(then find the L.H. derivative R.H. derivative of f(x) at x=0. Is f(x) differentiable at x=0)
Sol :
L.H.D
$f^{\prime}\left(0^{-}\right)=\lim_{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$ or $\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{(0-h)-0}{-h}$ or $\lim _{h \rightarrow 0} \frac{\left(0+h\right)-0}{h}$
$=\lim _{h \rightarrow 0} \frac{-h}{-h}$ or $=\lim _{h \rightarrow 0} \frac{h}{h}$
=1
R.H.D
$f^{\prime}\left(0^{+}\right)=\lim_{h\rightarrow{0}^+} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{(0+h)^{2}-0^{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{h^{2}}{h}=0$
$f^{\prime}\left(0^{+}\right)=0$
$f^{\prime}\left(0^{-}\right) \neq f^{\prime}\left(0^{+}\right)$
[]
Question 5
यदि दिखाएँ कि x=0 पर f अवकलनीय है।
(If f(x)=x|x| ,show that f(x) is differentiable at x=0)
Sol :
L.H.D
$f^{\prime}\left(0^{-}\right)=\lim_{h\rightarrow{0}^-} \frac{f(0-h)-f(0)}{-h}$
$=\lim_{h \rightarrow 0} \frac{(0-h)|0-h|-0|0|}{-h}$
$=\lim_{h \rightarrow 0} \frac{(-h)|-h|-0}{-h}$
$=\lim_{h \rightarrow 0}\frac{-h \times h}{-h}$
=0
R.H.D
$f^{\prime}\left(0^{+}\right)=\lim_{h \rightarrow 0^+}\frac{f(0+h)-f(0)}{h}$
$=\lim_{h \rightarrow 0} \frac{(0+h)|0+h|-0|0|}{h}$
$=\lim_{h-0^{-}} \frac{h|h|-0}{h}$
$=\lim _{h \rightarrow 0} \frac{h \cdot h}{h}=0$
$f^{\prime}\left(0^{-}\right)=f^{\prime}\left(0^{+}\right)$
[]
Question 6
यदि [x], x के पूर्णांक भाग को सूचित करता है तो x=f(x)=[x] sin 𝜋x का बाएँ और दाएँ तरफ से अवकलन निकालें , जहाँ k एक पूर्णांक है।
(If [x] denotes the integral part of x. find the left hand derivative of f(x)=[x]sin 𝜋x at x=k, where k is an integer)
Sol :
L.H.D
$f^{\prime}\left(k^{-}\right)=\lim _{h \rightarrow 0^{-}} \frac{f(k-h)-f / k}{-h}$
$=\lim _{h \rightarrow 0} \frac{[k-h] \sin \pi(k-h)-[k] \cdot \sin \pi k}{-h}$
[sin.n.𝜋=0 , n𝜖z]
$=\lim_{h \rightarrow 0} \frac{(k-1) \cdot \sin \left(k{\pi}-\pi h\right)-k \cdot 0}{-h}$
$=\lim _{h \rightarrow 0} \frac{(k-1) \cdot \sin (k \pi-\pi h)}{-h}$
$=-(k-1) \lim_{h \rightarrow 0} \frac{\sin (k \pi-\pi h)}{h}$
Value of sin(k𝜋-𝜋h)
sin(𝜋-𝜋h)=sin𝜋h
sin(2𝜋-𝜋h)=-sin𝜋h
sin(3𝜋-𝜋h)=-sin𝜋h
∴ sin(k𝜋-𝜋h)=(-1)k-1sin𝜋h
L.H.D
$=-(k-1) \lim _{h \rightarrow 0} \frac{\ln (k \pi-\pi h)}{h}$
$=-(1 .-1) \lim_{h \rightarrow 0}(-1)^{k-1} \frac{\sin \pi h}{\pi^{n}} \times \pi$
$=-1(k-1) \cdot(-1)^{k-1} \pi \lim _{h \rightarrow 0} \frac{\sin \pi h}{\pi h}$
$=(-1)^{k} \cdot \pi(k-1) \times 1$
$=(-1)^{k} \pi(k-1)$
R.H.D
$f^{\prime}(k^+)=\lim_{h\rightarrow{0^{+}}} \frac{f(k+h)-f(k)}{h}$
$=\lim_{h \rightarrow 0^{+}} \frac{[k+h] \cdot \sin \pi(k+h)-[k] \cdot \sin \pi k}{h}$
[sin.n.𝜋=0 , n𝜖z]
$=\lim _{h \rightarrow 0} \frac{k \cdot \sin (k \pi+\pi h)-0}{h}$
$=\lim_{h \rightarrow 0}\frac{\sin \left(k\pi+\pi h\right)}{h}$
Value of sin(k𝜋-𝜋h)
sin(2𝜋+𝜋h)=sin𝜋h
sin(3𝜋+𝜋h)=-sin𝜋h
sin(k𝜋+𝜋h)=(-1)ksin𝜋h
R.H.D
$=k \cdot \lim _{h \rightarrow 0} \frac{\sin\left(k{h}+\pi{h}\right)}{h}$
$=k \lim _{h \rightarrow 0} \frac{(-1)^{k} \sin \pi{h}}{h}$
$=k \cdot(-1)^{k} \lim _{h \rightarrow 0} \frac{\sin \pi h}{\pi h} \times \pi$
$=k \cdot(-1)^{k} \pi$
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 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
$f^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f^{\prime}(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{5(1+h)+7-(5\times 1+7)}{h}$
$=\lim _{h \rightarrow 0} \frac{5+5 h+7-5-7}{h}$
$=\lim_{h\rightarrow 0} \frac{5 h}{h}=5$
$f^{\prime}\left(1^{-}\right) \neq f^{\prime}\left(1^{+}\right)$