`:. lim_(x->0)f(x)`

`=lim_(x->0)(5x-3)`

`= 5xx0-3`

`= -3`

`=`

`=`

`=`

`:. lim_(x->0)f(x)`

`=lim_(x->0)(5x-3)`

`= 5xx0-3`

`= -3`

`=`

`=`

`=`

अंशो का योग

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अंशो का योग

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`:. lim_(x->0)f(x)`

`=lim_(x->0)(5x-3)`

`= 5xx0-3`

`= -3`

`=`

`=`

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अंशो का योग

=

अंशो का योग

+

अंशो का योग

अंशो कक

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\[x + 3\mathop{\left){\vphantom{1\begin{array}{l}3{x^2} + 5x + 6\\\frac{{3{x^2} + 6}}{{3{x^2} + 6}}\end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^2} + 5x + 6\\\frac{{3{x^2} + 6}}{{3{x^2} + 6}}\end{array}}}} \limits^{\displaystyle\,\,\, {3x + 2}}\]

`:. lim_(x->0)f(x)`

`=lim_(x->0)(5x-3)`

`= 5xx0-3`

`= -3`

`=`

`=`

`=`

`:. lim_(x->0)f(x)=lim_(x->0)(5x-3)`

NCERT Exercise 5.1

Question: (1) Prove that the function `f(x)=5x-3` is continuous at `x=0`, at `x=-3` and at `x=5`.

Solution:

Here `f(x)=5x-3`

At `x=0`

`:. lim_(x->0)f(x)=lim_(x->0)(5x-3)`

`=5xx0-3=-3`

`f(0)=5xx0-3=-3`

`:.lim_(x->0)(x)=f(0)`

Thus `f(x)` is continuous at `x=0`

At `x=-3`

`:.lim_(x->-3) f(x)`

`lim_(x->-3)(5x-3)`

=`5xx(-3)-3`

=`-15-3=-18`

`f(-3)=5xx-3-3=-18`

`:. lim_(x->-3)f(x)=f(-3)`

Thus `f(x)` is continuous at `x=-3`

At `x=5`

`:. lim_(x->5)f(x)=lim_(x->5)(5x-3)`

`=5xx5-3=25-3=22`

`f(5)= 5xx5-3=25-3=22`

`:.lim_(x->5)f(x)=f(5)`

Thus `f(x)` is continuous at `x=5`

Question: (2) Examine the continuity of the functio `f(x)=2x^2-1` at `x-3`.

Solution:

Here `f(x)=2x-1`

`:.lim_(x->3)f(x)=lim_(x->3)xx2x^2-1`

`=2xx(3)^2-1=18-1=17`

`f(3)=2xx(3)^2-1=18-1=17`

`:. lim_(x->3)f(x)=f(3)`

Thus `f(x)` is continuous at `x=3`

Question: (3) Examine the following functions for continuity

(a)`f(x)=x-5`

Solution:

`f(x)=1/(x-5)` is a quotient function of two polynomial functions, so `f(x)` is continuous at. all values of `x` provided `x!=5`

(b)`f(x)=1/(x-5),x!=5`

Solution:

`f(x)=1/(x-5)` is a quotient function of two polynomial functions, so`f(x)` is continuous at all values of x provided`x!=5.`

(c) `f(x)= (x^2-25)/(x+5), x!=5`

Solution:

`f(x)= (x^2-25)/(x+5)`

`=((x+5)(x-5))/(x+5)=x-5`

`:.f(x)=x-5` is a polynomial function, so `f(x)` is continuous at all values of `x`

(d)`f(x)=|x-5|`

Solution:

`f(x)=|x-5|={(x-5, text(for), x>=5), (5-x, text(for), x<5):}`

For `x->5^+ lim_(x->5^+)f(x)`

`lim_(x->5^+)(x-5)=5-5=0`

For `x->5^, lim_(x->5^-)f(x)`

`=lim_(x->5^-)(5-x)=5-5=0`

`f(5)=5-5=0`

∵ L.H.L.= R.H.L.=`f(5)=>` The function is continuous at `x=5`

Question: (4) Prove that the function `f(x)=x^n` is continuous of `x=n`, where n is a positive integer.

Solution:

Here`f(x)=x^n`

`:.lim_(x->n) f(x)= lim_(x->n)x^n=n^n`

`f(n)=n^n`

`[∵ f(x)=x^n]`

`:.lim_(x->n)f(x)=f(n)`

Thus `f(x)` is continuous at `x=n`, where n is a positive integer

Question:5 .

Is the function f defined by

`f(x)={(x, if x>=1), (5, if x>1):}` continuous at `x=0?` at `x=1?` at `x=2?`

Solution:

Here `f(x)={(x, if x>=1), (5, if x>1):}`

At `x=0`

L.H.L.`=lim_(x->0^-)f(x)`

`=lim_(x->0^-)(x)`

Put `x=0-h` as `x->0^-,h->0`

`lim_(h->0^-)(0-h)=0-0=0`

R.H.L.=`lim_(x->0^+)f(x)`

=`lim_(x->0^+)(x)`

Put `x=0+h` as `x->0^+,h->0`

`lim_(h->0)(0+h)=0+0=0`

`[∵ f(x)=x]`

`f(0)=0`

∵ L.H.L = R.H.L=`f(0)`

Thus `f(x)` is continuous at `x=0`

At `x=1`

L.H.L =`lim_(x->1^-)f(x)`

`lim_(x->1^-)(x)`

`x=1-hasx->1^-,h->0`

`lim_(h->0)(1-h)=1-0=1`

R.H.L.=`lim_(x->1^+)f(x)=5`

`:.L.H.L !=R.H.L.`

Thus `f(x)` is discontinuous at `x=1`

At `x=2`

`f(2)=5`

`:.L.H.L.=R.H.L.=f(2)`

Thus all points of discontinuity of f, where f is defined by

Question:6 .

`f(x)={:{(2x+3, if x<=2), (2x-3, ifx>2):}`

Solution:

Here `f(x)={:{(2x+3, if x<=2), (2x-3, ifx>2):}`

L.H.L=`lim_(x->2)f(x)`

=`lim_(x->2)(2x+3)`

Put `x=2-h\   as\   x->2,h->0`

`:.lim_(h->0)[2(2-h)+3]`

=`lim_(h->0)(7-2h)=7-2xx0=7`

R.H.L=`lim_(x->2^+)f(x)`

`lim_(x->2^+)(2x-3)`

Put `x=2-h`   as   `x->2^-`, `h->0`

Put x`-2+h\    as\    x->2^+,h->0`

`:. lim_(h->0)[2(2+h)-3]`

`=lim_(h->0)(1+2h)`

`=1+2xx0=1`

`:.`L.H.L `!=` R.H.L

Question: 7 .

`f(x)={:{(|x|+3, if x<= -3), (-2x, if -3<\x\<\3), (6x+2, if x>=3):}`

Solution:

Here `f(x)={:{(|x|+3, if x<= -3), (-2x, if -3<\x\<\3), (6x+2, if x>=3):}`

At `\ x =-3`

L.H.L.`lim_(x->-3^-)f(x)`

`=lim_(x->-3^-)\(|x|+3)`

Put `x=(-3-h)as x->-3-, h->0`

`:.lim_(h->0)[|-3-h|+3]`

=`lim_(h->0)(6+h)=6+0=6`

R.H.L=`lim_(x->3^+)f(x)`

=`lim_(x->3^+)(-2x)`

put `x=-3+hasx->3^+,h->0`

`lim_(h->0)-2(-3+h)`

`lim_(h->0)(6-2h)`

=`-2xx0=6`

`f(-3)=|-3|+3=3+3`

`=6[:.f(x)=|x|+3]`

`:.` L.H.L = R.H.L.= `f(-3)`

Thus `f(x)` is continuoud at `x=-3`

At `x=3`

L.H.L.=`lim_(x->-3^-)f(x)`

`=lim_(x->-3^-)(-2x)`

Put`x=3-h as x->3^-,h->0`

`:. =lim_(h->0)-2(3-h)`

=`lim_(h->0)(-6+2h)`

R.H.L=`=lim_(x->-3^+)f(x)`

=`lim_(x->-3^+)(6x+2)`

Put `x=3+h as x->3^+,h->0`

`:.lim_(h->0)[6(3+h)+2]`

=`lim_(h->0)(18+6h+2)`

=`lim_(h->0)(20+6h)`

`=20+6xx0+20`

Thus `f(x)` is discontinuous at `x=3`

Question:8 .

`f(x)={:{(|x|/x, if x!=0), (0, if x=0):}`

Solution:

Here `f(x)={:{(|x|/x, if x!=0), (0, if x=0):}`

L.H.L=`lim_(x->0^-)f(x)`

`=lim_(x->0^-)|x|/x`

Put `x=0-h as x->0^-,h->0`

`:. lim_(h->0)(|0-h|)/(0-h)`

`lim_(h->0)h/-h =-1`

R.H.L.=`lim_(x->0^+)f(x)`

=`lim_(x->0^+)(|x|)/x`

Put `x= 0+h as x->0^+,h->0`

`:. lim_(h->0)(|0+h|)/(0+h)`

=`lim_(h->0)h/h=1`

`:.` L.H.L `!=` R.H.L.

Thus `f(x)`is discontinuous at `x=0`

Question:9 .

`f(x)={(x/|x|, if x<0), (-1, ifx>=0):}`

Solution:

Here `f(x)={(x/|x|, if x<0), (-1, ifx>=0):}`

L.H.L.`=lim_(x->0^-)f(x)`

`=lim_(x->0^-)x/|x|`

Put `x=0-h` as `x->0^-,h->0`

`:.=lim_(h->0)(0-h)/(|0-h|)`

`=lim_(h->0)(-h)/h=-1`

R.H.L.`=lim_(x->0^+)f(x)=-1`

`f(0)=-1`

`:.` L.H.L=R.H.L= `f(0)`

Thus `f(x)` is continuous at `x=0` ,There is no point of disconinuity for this function `f(x)`

Question:(10)

`f(x)={(x+1, if x>=1), (x^2+1, if x<1):}`

Solution:

Here `f(x)={(x+1, if x>=1), (x^2+1, if x<1):}`

L.H.L. =`=lim_(x->1^-)f(x)`

`=lim_(x->1^-)(x^2+1)`

Put` x=1+h` as `x->1^+,h->0`

`lim_(h->0)[(1-h)^2+1]`

`=lim_(h->0)[1+h^2-2h+1]`

`=lim_(h->0)[2+h^2-2h]`

`=2+0-0=2`

R.H.L.`=lim_(x->1^+)f(x)`

`lim_(x->1^+)(x+1)`

Put `x=1+h` as `x->1^+, h->0`

`:.lim_(h->0)(1+h+1)`

`lim_(h->0)(2+h)`

`=2+0=2`

`f(1)=1+1=2`

`[∵ f(x) =x+1]`

`:.` L.H.L.=R.H.L.=`f(1)`

Thus `f(x)` is continuous at `x=1`. There is no point of discontinuity for this function `f(x)` .

Question: 11 .

`f(x)={(x^3-3, if x>=2), (x^2+1, if x<2):}`

Solution:

Here `f(x)={(x^3-3, if x>=2), (x^2+1, if x<2):}`

L.H.L. =`lim_(x->2^-)f(x)`

`=lim_(x->2^-)f(x^3-3)`

Put `x=2-h` as `x->2^-, h->0`

`:.lim_(h->0)[(2-h)^3-3]`

`=lim_(h->0)[(8-h^3-12h+6h^2-3]`

`lim_(h->0)(5-h^3-12h+6h^2)=5`

R.H.L.`lim_(x->2^+)f(x)`

`=lim_(x->2^+)(x^2+1)`

Put `x=2+h` as `x->2^+,h->0`

`:. lim_(h->0)[(2+h)^2+1]`

`=lim_(h->0)(4+h^2+4h+1]`

` lim_(h->0)(5+h^2+4h]=5`

`f(2)=(2)^3-3`

`=8-3=5`

`[∵ f(x)=x^3-3]`

`:.` L.H.L.=R.H.L. `=f(2)`

Thus `f(x)` is continuous at `x=2` There is no point of discontinuity for this function `f(x)`.

Question: 12 .

`f(x)={:{(x^10-1, if x<=1), (x^2, if x>1):}`

Solution:

Here`f(x)={(x^10-1, if x<=1), (x^2, if x>1):}`

L.H.L`=lim_(x->1^-)f(x)`

`=lim_(x->2^-)(x^10-1)`

Put `x=1-h` as `x->1^-,h->0`

`:. lim_(h->0)[(1-h)^10-1]`

`=(1-0)^10-1`

`=1-1=0`

R.H.L.=` lim_(x->1^+)f(x)`

`= lim_(x->1^+)(x^2)`

Put `x=1+h` as `x->1^+,h->0`

`lim_(h->0)(1+h)^2`

`lim_(h->0)1+h^2+2h=1`

`:.L.H.L. `!=` R.H.L.

Thus `f(x)` is not continuous at `x=1`

`[[a,b],[c,d]]`

`(i) (-3, 0) = {x : x ∈ R, -3 < x < 0}`

`sum_(i=1)^n i^3=((n(n+1))/2)^2`

\[B(m,n)=\int\limits_{0}^{1}{{{x}^{m-1}}{{\left( 1-x \right)}^{n-1}}dx}\]

`(i) (-3, 0) = {x : x ∈ R, -3 < x < 0}` \(e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right ) \\ = e^{u} \left ( \frac{sin ^{2} \;\frac{u}{2} + cos ^{2} \;\frac{u}{2} + 2 sin \;\frac{u}{2} \;cos\; \frac{u}{2}}{2\; cos ^{2} \;\frac{u}{2}} \right ) \\ = \frac{e^{u} \left ( sin \;\frac{u}{2} + cos \;\frac{u}{2} \right ) ^{2}}{2 cos ^{2} \;\frac{u}{2}} \\ = \frac{1}{2} e^{u} \left ( \frac{sin \;\frac{u}{2} + cos \;\frac{u}{2}}{cos \;\frac{u}{2}} \right ) ^{2} \\ = \frac{1}{2} e^{u} \left [ tan \;\frac{u}{2} + 1 \right ] ^{2} \\\) \(= \frac{1}{2} e^{u} \left [1 + tan \;\frac{u}{2} \right ] ^{2} \\ = \frac{1}{2} e^{u} \left [ 1 + tan ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ = \frac{1}{2} e^{u} \left [ sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = \left [ \frac{1}{2} sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \; \; \; ….. (1) \\ Suppose,\; tan\; \frac{u}{2} = f (u) \; so \; f’ (u) = \frac{1}{2} sec ^{2} \;\frac{u}{2}\) As we know, \(\int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C\) Considering equation (1), we get, \[\int \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = e^{u} tan\; \frac{u}{2} + C\] \begin{align} I&=\int \frac{1}{1 + cot\; \theta} d\theta \\ &= \int \large\frac{1}{1 + \frac{cos\; \theta}{sin\; \theta}} d\theta \\ &= \int \frac{sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ &= \frac{1}{2} \int \frac{2 sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ &= \frac{1}{2} \int \frac{(sin\; \theta + cos\; \theta) + (sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ &= \frac{1}{2} \int 1 d\theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ &= \frac{1}{2} \theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ \end{align} \begin{align} \int_0^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx&=\int_0^{\Large\frac\pi2}\frac1{(\sqrt{\cos x})^4\ (1+\sqrt{\tan x})^4}\ dx\\ &=\int_0^{\Large\frac\pi2}\frac{\sec^2x}{(1+\sqrt{\tan x})^4}\ dx\\ &\stackrel{\color{red}{[1]}}=\int_0^\infty\frac{du}{(1+\sqrt{u})^4}\\ &\stackrel{\color{red}{[2]}}=2\int_0^\infty\frac{t}{(1+t)^4}\ dt\\ &\stackrel{\color{red}{[3]}}=2\cdot\text{B}(2,2)\\ &=\large\color{blue}{\frac13}. \end{align} \begin{align} \int cos3x.sin2x \, dx &= \int \cos(2x+x) \sin2x \,\frac{a}{b} dx \\ &= \int [\cos2x\cos x - \sin2x\sin x ]\sin2x \, dx \\ &= \int \cos2x \cos x \sin2x - \sin^2 2x \sin x \, dx \\ &= \int (\cos^2 x - \sin^2 x) \cos x (2 \sin x \cos x) - (2 \sin x \cos x)^2 \sin x \, dx \\ &= \int 2 \cos^4 x \sin x - 2 \cos^2 x \sin^3 x - 4 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x \, dx - \int 6 \sin^3 x \cos^2 x \, dx \\ \end{align} \begin{align} \int cos3x.sin2x \, dx &= \int \cos(2x+x) \sin2x \,\frac{a}{b} dx \\ &= \int [\cos2x\cos x - \sin2x\sin x ]\sin2x \, dx \\ &= \int \cos2x \cos x \sin2x - \sin^2 2x \sin x \, dx \\ &= \int (\cos^2 x - \sin^2 x) \cos x (2 \sin x \cos x) - (2 \sin x \cos x)^2 \sin x \, dx \\ &= \int 2 \cos^4 x \sin x - 2 \cos^2 x \sin^3 x - 4 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x \, dx - \int 6 \sin^3 x \cos^2 x \, dx \\ \end{align}

\begin{align}I&=\int sin^3x.cos^3x \ dx \\ &=\int sin x.sin^2x.cos^3x\,dx\end{align}

`=>I=int\ sin x.sin^2x.cos^3x\dx `

`=>I=int\ sin x.(1-cos^2x).cos^3x\dx `

`I=int\ sin x.(cos^3x-cos^5x)\dx `

माना `cosx=t `

`:.\-sinx\dx=dt `

`I=-int (t^3-t^5)dt `

`=- t^4/4+t^6/6+c`

`= 1/6cost^6x-1/4cos^4x\+c`

`sum_(i=1)^n i^3=((n(n+1))/2)^2`

\[B(m,n)=\int\limits_{0}^{1}{{{x}^{m-1}}{{\left( 1-x \right)}^{n-1}}dx}\]

(Theorem: Euclid’s Division Lemma)

यूक्लिड विभाजन प्रमेयिका के अनुसार दो धनात्मक पूर्णांक `a` तथा `b`, दिये रहने पर, ऐसी अद्वितीय पूर्ण संख्याएँ `q` तथा `r` विद्यमान हैं कि `a=bq+r, 0\<=\r\<\b`

यहाँ `q` = भागफल (Quotient) तथा `r` = शेष (Remainder) है।

यूक्लिड विभाजन एल्गोरिथ्म (कलन विधि) इसी प्रमेयिका (Lemma) पर आधारित है। यूक्लिड विभाजन एल्गोरिथ्म (कलन विधि) दिये गये दो धनात्मक पूर्ण संख्याओं का HCF (महत्तम समापवर्तक) निकालने की एक विधि है।

दो धनात्मक पूर्ण संख्याओं, यथा `c` तथा `d`, जहाँ `c\>\d`, का HCF (महत्तम समापवर्तक) निम्नांकित steps (चरणों) के अनुसरण द्वारा निकाला जा सकता है :

Step: 1. `c` और `d` के लिए यूक्लिड विभाजन प्रमेयिका का प्रयोग कीजिए। इसलिए, हम ऐसे `q` और `r` ज्ञात करते हैं कि `c=dq+r, 0\<=r\<\d`.

Step: 2. यदि `r=0` है, तो `d` पूर्णांकों `c` और `d` का HCF है। यदि `r!=0` है, तो `d` और `r` के लिये यूक्लिड विभाजन प्रमेयिका का प्रयोग कीजिए।

Step: 3. इस प्रक्रिया को तबतक जारी रखा जाता है, जबतक कि शेषफल 0 न प्राप्त हो जाए। इसी स्थिति में, प्राप्त भाजक ही वांछित HCF है।

NCERT अभ्यास प्रश्नावली 1.1

`sum_(i=1)^n i^3=((n(n+1))/2)^2`