KC Sinha Solution Class 11 Chapter 26 अवकलज (Derivatives) Exercise 26.1

 Exercise 26.1

Question 1

यदि(If) f x=x², find f '(2) निकाले ।

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$=\lim _{h \rightarrow 0} \frac{(2+h)^{2}-2^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{2^{2}+2 \cdot 2 \cdot h+h^{2}-2^{2}}{h}$

$=\lim _{h\rightarrow 0} \frac{h(4+h)}{h}$

=4+0

=4

Question 2

यदि (If) f x=x³+1, find f '(3) निकाले ।

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{(3+h)-f(3)}{h}$

$=\lim _{h \rightarrow 0} \frac{(3+h)^{3}+1-\left(3^{3}+1\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{3^{3}+h^{3}+3 \cdot 3^{2} \cdot h+3 \cdot 3 \cdot h^{2}+1-3^{3}-1}{4}$

$=\lim _{h \rightarrow 0} \frac{ h^{3}+27 h+9 h^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{h\left(h^{2}+27+9h\right)}{h}$

f '(3)=27

Question 3

यदि (If)  f x=x²+2x+7, (find)  f '(3) निकाले ।

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$

$=\lim _{h \rightarrow 0}\frac{(3+h)^{2}+2(3+h)+7-\left(3^{2}+2(3)+7\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{3^{3}+2 \cdot 3 \cdot h+h^{2}+6+2 h+7-9-6-7}{h}$

$=\lim _{h \rightarrow 0} \frac{6+h^{2}+2 b}{4}$

$=\lim_{h\rightarrow 0}\frac{8 h+h^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{\operatorname{h}(8+4)}{h}$

=8+0

=8

Question 4

यदि (If)  f x=mx+c, (find)  f '(0) निकाले ।

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{m(0+h)+c-(m \cdot 0+c)}{h}$

$=\lim _{h \rightarrow 0} \frac{m h+c-c}{h}$

$=\lim _{h \rightarrow 0} \frac{m h}{h}$

=m

Question 5

यदि (If) f x=3t²+1 ,(find) f '(1) निकाले ।

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$=\lim _{h \rightarrow 0} \frac{3(1+h)^{2}+1-\left(3(1)^{2}+1\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{3\left(1+2 h+1^{2}\right)+1-(3+1)}{h}$

$=h_{\rightarrow 0} \frac{3+6h+3 h^{2}+1-3-1}{h}$

$=\lim _{h \rightarrow 0} \frac{h(6+3 h)}{h}$

=6+3(0)

=6

Question 6

x=1 पर x का अवकलज निकाले।

[Find the derivative of x at x=1]

Sol :

Let f x=x

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$= \lim_{h \rightarrow 0} \frac{1+h-1}{h}$

$=\lim _{h \rightarrow 0} \frac{h}{h}=1$

Question 8

x=100 पर 99x का अवकलज निकाले।

[Find the derivative of 99x at x=100]

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(100)=\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h}$

$=\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h}$

$=\lim _{h \rightarrow 0} \frac{9900+99 h-9900}{h}$

$=\lim _{h \rightarrow 0} \frac{99k}{h}$

=99

Question 9

x=100 पर x²-2 का अवकलज निकाले।

[Find the derivative of x²-2 at x=10]

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

माना f(x)=x²-2

$f^{\prime}(10)=\lim _{h \rightarrow 0} \frac{f(10+4)-f(10)}{h}$

$=\lim _{h \rightarrow 0}\frac{(10+h)^{2}-2-\left(10^{2}-2\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{10^{2}+2.10.h+h^2-2-10^2+2}{h}$

$=\lim_{h \rightarrow 0} \frac{20 h+h^{2}}{h}$

$=\lim _{h \rightarrow 0} h\left(\frac{20+h}{h}\right)$

=20+0

=20

Question 10

x=0 तथा x=3 पर f(x)=3 का अवकलन निकाले।

[Find the derivative of f (x)=3 at x=0 and x=3]

Sol :

$\left[f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]$

$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+4)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{3-3}{h}$

$=\lim _{h \rightarrow 0} \frac{0}{h}$

$=\lim _{h \rightarrow 0} 0$

=0

$f^{\prime}(3)=\lim_{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$

$=\lim_{h\rightarrow 0}\frac{3-3}{h}$

=0

Question 11

अचर फलन f(x)=a का अवकलन किसी निश्चित वास्तविक संख्या पर निकालें

[Find the derivative of the constant function f(x)=a for a fixed real number]

Sol :

माना c एक निश्चित वास्तवीक संख्या है।

$f^{\prime}(c)=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$

$=\lim_{h \rightarrow 0} \frac{a-a}{h}$

$=\lim _{h \rightarrow 0} 0$

=0

Question 12

x=0 पर sinx का अवकलन निकाले ।

[Find the derivative of sin x at x=0]

Sol :

माना f(x)=sin x

$f^{\prime}(0)=\lim _{n \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{\sin (0+h)-\sin 0}{h}$

$=\lim _{h \rightarrow 0} \frac{\sinh -0}{h}$

$=\lim _{h \rightarrow 0} \frac{\sin h}{h}$

=1

Question 13

यदि (If) f(x)=x³+7x²+8x-9, (find) f '(4) निकाले ।

Sol :

$f^{\prime}(4)=\lim _{h \rightarrow 0} \frac{f(4+h)}{h}-f(4)$

$=\lim _{h \rightarrow 0}\dfrac{\left[(4+h)^{3}+7(4+h)^{2}+8(4+h)-9\right]-[4^3+7(4)^2+8(4)-9]}{h}$

$=\lim _{h \rightarrow 0} \dfrac{4^3+h^3+3.4^2.h+3.4.h^2+7(4)^2+7.8h+7h^2+32+8h-9-4^3-7(4)^2-32+9}{h}$

$=\lim _{h \rightarrow 0} \frac{h^{3}+48 h+56 h+84+12 h^{2}+7 h^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{h^{3}+112 h+19 h^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{h \left(h^{2}+112+19 h\right)}{h}$

=0²+112+19(0)

=112

Question 14

यदि (If) f(x)=x³-2x+1, दिखलाएँ कि(show that) f '(2)=10 f '(1)

Sol :

L.H.S

$f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$=\lim _{h \rightarrow 0}\dfrac{\left[(2-h)^{3}-2(2+h)+1\right]-\left[2^{3}-2(2)+1\right]}{h}$

$=\lim_{h\rightarrow 0}\dfrac{2^3+h^3.2^2.h+3.2.h^2-4-2h+1-2^3+4-1}{h}$

$=\lim _{h \rightarrow 0} \frac{h^{3}+10 h+6 h^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{h\left(h^{2}+10+6 h\right)}{h}$

=0²+10+6(0)

=10

R.H.S

$10 f^{\prime}(1)=10 \cdot \lim _{h \rightarrow 0} \frac{f(1+4)-f(1)}{h}$

$=10 \cdot \lim _{h \rightarrow 0}\dfrac{\left[(1+4)^{3}-2(1+4)+1\right] -\left(1^{3}-2(1)+1\right]}{h}$

$=10 \lim_{h\rightarrow 0}\dfrac{1^3+h^3+3.1^2.h+3.1.h^2-2-2h+1-1+2-1}{h}$

$=10 \cdot \lim _{h \rightarrow 0} \frac{h^{3}+h+3 h^{2}}{h}$

$=10 \lim_{h\rightarrow 0} \dfrac{h\left(\mathrm{h}^{2}+1+3h\right)}{h}$

=10(0²+1+3(0))

=10×1

=10

$f^{\prime}(2)=10 \cdot f^{\prime}(1)$

Question 16

x=-1 पर f(x)=2x²+3x-5 का अवकलन निकाले । साथ ही साबित करे कि  f '(0)+3 f '(-1)=0

Sol :

$f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}$

$=\lim _{h \rightarrow 0}\dfrac{\left[2(-1+h)^{2}+3(-1+h)-5\right]-\left[2(-1)^{2}+3(-1]-5\right]}{h}$

$=\lim _{h \rightarrow 0}\dfrac{\left[2(1-2h+h^2)-3+3h-5\right]-\left[2-3-5\right]}{h}$

$=\lim _{h \rightarrow 0} \dfrac{2-4 h+2 h^{2}-3+3 h-5-2+3+5}{h}$

$=\lim _{h \rightarrow 0} \frac{-h+2 h^{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{h(-1+2h)}{h}$

=-1+2(0)

=-1

Question 16

x=-1 पर f(x)=2x²+3x-5 का अवकलन निकाले । साथ ही साबित करे कि  f '(0)+3 f '(-1)=0

Sol :

$f^{\prime}(0)=\lim _{h \rightarrow 0} \dfrac{f(0+h)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}$

$=\lim _{h \rightarrow 0} \frac{2 h^{2}+3 h-5-2(0)^{2}-3(0)+5}{h}$

$=\lim _{h \rightarrow 0} \frac{2 h^{2}+3 h}{h}$

$=\lim _{h \rightarrow 0} \frac{h(2 h+3)}{h}$

=2(0)+3

=3

L.H.S

$f^{\prime}(0)+3 f^{\prime}(-1)$

=3+3(-1)

=3-3

=0

Question 17

फलन f के लिए जो f(x)=kx²+7x-4 से परिभाषित है f '(5)=97 तो k निकालेय़

[For the function f given by f(x)=kx²+7x-4 ,f '(5)=97 then find k]

Sol :

f '(5)=97

$\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}=97$

$\lim _{h \rightarrow 0}\frac{\left[k(5+h)^{2}+7(5+h)-4\right]-\left[k(5)^{2}+7(5)-4\right]}{4}=9$

$\lim _{h \rightarrow 0}\dfrac{\left[k\left(25+10 h+h^{2}\right)+35+7 h-4\right]

-\left[25 h^{2}+35-4\right]}{h}=57$

$\lim_{h\rightarrow 0}\dfrac{25 k+10 k h+k h^{2}+35+7 h-4 -25k-35+4}{h}=97$

$\lim _{h \rightarrow 0} \frac{10 k h+7 h+k h^{2}}{h}=97$

$\lim _{h \rightarrow 0} \frac{h (10 \mathrm{k}+7+\mathrm{kh})}{h}=97$

10k+7+k(0)=97

10k=90

k=9

Question 18

फलन f के लिए जो f (x)=x²+2ax+5 से परिभाषित है, f '(1)=10 तो a निकाले ।

[For the function f given by  f (x)=x²+2ax+5, f '(1)=10 find a ]

Question 19

यदि (If) $f(x)=\frac{x-2}{x^{2}-3 x+2}, x \neq 2$

=1, x=2

तो f '(2) निकाले (then find f '(2) ]

Sol :

L.H.D

$f^{\prime}\left(2^{-}\right)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$

$=\lim _{h \rightarrow 0} \dfrac{\frac{1}{{2-1-1}}-\frac{1}{2-1}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{1-h}-\frac{1}{1}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1-(1-h)}{1-h}}{-h}$

$=\lim_{h \rightarrow 0} \frac{\frac{1-1+h}{1-h}}{-h}$

$=\lim _{x \rightarrow 0} \frac{\frac{h}{1-h}}{-h}$

$=\lim_{h\rightarrow 0}\frac{-1}{1-h}=\frac{-1}{1-0}=-1$

R.H.D

$f^{\prime}\left(2\right)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{2+h-1}-\frac{1}{2-1}}{h}$

$=\lim _{h \rightarrow 0}\dfrac{ \frac{1}{1+h}- \frac{1}{1}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\frac{1-(1+h)}{1+h}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\frac{1-1-h}{1+h}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\frac{-h}{1+h}}{h}$

$=\frac{-1}{1+0}=-1$

$f^{\prime}(2)=-1$

$f^{\prime}\left(2^{-}\right)=f^{\prime}(2^+)$

Question 20

समय t=0 से शूरू कर एक कण एक सरल रेखा मे इस तरह से चसता है ताकि इसका स्थिति t सेकेण्ड के बाद s(t)=t²-6t+8 मीटर है। इसका चाल t=3 सेकेण्ड पर निकाले ।

Sol :

चाल 

$s^{\prime} (3)=\lim _{h \rightarrow 0} \frac{s(3+4)-s(3)}{h}$

$=\lim _{h \rightarrow 0}\dfrac{\left[(3+h)^{2}-6(3+h)+8\right]-\left[3^{2}-6(3)+8\right]}{h}$

$=\lim _{h \rightarrow 0} \dfrac{9+6 h+h^{2}-18-6 h+8-9+18-8}{h}$

$=\lim_{h\rightarrow 0} \frac{h^{2}}{h}$

=0

Question 21

एक कण एक सरल रेखा मे इस तरह से चलता है ताकि इसका स्थिति t सेकेण्ड के बाद 

s(t)=6t-t² है। इसका प्रारंभिक वेग क्या है।

[A particle moves along a line so that at time t, its position is s(t)=6t-t² .What is its initial velocity?]

Question 22

एक कण एक सरल रेखा मे इस तरह से चलता है ताकि t समय पर इसकी स्थिति $s(t)=\frac{t^{2}+2}{t+1}$ इकाई है। इसका वेग t=3 पर निकाले ।

[A particle moves along a line so that its position at time t is $s(t)=\frac{t^{2}+2}{t+1}$ units.Find its velocity at time t=3 ]