Exercise 10.1
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?
KC Sinha Mathematics Solution Class 12 Chapter 10 अवकलनीयता (Differentiability) Exercise 10.1 (Q4-Q6)
=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)
[]
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)
[]
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)
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
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
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