Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.8
Question: (1) Verfy Rolle,s theorem for the function `f(x)=x^2 -2x-8, x in [-4,2]`
Solution:
Here `f(x)=x^2 -2x-8, x in [-4,2]`
(i)Since `f(x)`is a polynominal , so it is continuous in `[-4,2]`
(ii)Since `f(x)` is a polynomial, so it is differentiable in `(-4, 2)`
(iii)`f(-4)=(-4)^2 + 2xx -4 -8`
` = 16- 8=0`
`f(2)=(2)^2+2xx2 -8`
` = 4+4 -8=0`
`:. f(-4)=f(2)=0`
So there exists a real numer `C in (-4, 2)`such that `f(c)=0`
`:. f(x) = 2x +2`
`f(c)=2c +2=0`
`:. 2c + 2=0`
`=> c= -1 in (-4,2)`
Thus Rolle's theorem is verified.
Question: (2) Examine if Rolle's theorem is applicable to any of the following functions. can you say some thing about the converse of rolle's theorem from these examples?
(i)`f(x)= [x]`for `x in [5, 9]`
(ii)`f(x)= [x], x in [-2,2]`
(iii)` f(x)=x^2-1` for `x in [1, 2]`
Solution:
(i) Here `f(x)=[x], x in [5,9]`
At`x=7`
L.H.L. `=lim_(x->7^-)f(x)`
`= lim_(x->7^-)[x]`
put `x= 7 -h, as x ->, h-> 0`
`:. lim_(h->0)[7-h]`
` lim_(h->0)(6)=6`
R.H.L. = `lim_(x->7^+)f(x) `
`lim_(x->7^+)[x]`
put `x= 7+ h as x-> 7^+,h-> 0`
`:. lim_(h->0)[7+h]`
`= lim_(h->0)[7+h]`
`lim_(h->0)(7)=7`
`:.` L.H.L.`!=` R.H.L.
So `f(x)` is not continuous at `x=7`
`f(x)` is not continuous in `[5, 9]`
Thus Rolle's theorem is not applicable.
(ii) Here `f(x)=[x],x in [-2, 2]`
At `x=1`
L.H.L. `=lim_(x->1^-)f(x)`
` =lim_(x->1^-)[x]`
Put `x= 1-h, as x-> 1^-1, h-> 0`
`:. lim_(h->0)[1-h]`
`=lim_(h->0) (0)=0`
R.H.L.`= lim_(x->1^+)f (x) = lim_(x->1^+)[x]`
Put `x= 1+h, as x-> 1^+, h-> 0`
`:. lim_(h->0)[1+h]`
`= lim_(h->0)(1) =1`
`:.` L.H. L.`!=` R.H.L.
So `f(x)` is not continuous at `x=1`
`:. f(x)` is not continuous in `[-2, 2]`
Thus rolle's theorem is not applicable.
(iii)Here `f(x)= x^2-1, x in [1,2]`
(i)since `f(x)` is a polynomial,so it is continuous in `[1,2]`
(iii) `f(1)=(1)^2-1`
`1-1 =0`
`f(2)=(2)^2-1=4-1=3`
`:. f(1) != f(2)`
So all conditions of rolle's theorem are not satisfied.
Thus Rolle's theorem is not applicable.
Question: (3) If ` f:[-5, 5]-> R` is a differentiable function and if `f(x)` does not vanish anywhere, then prove that`f(-5)!= f(5)`.
Solution:
It is given that `f(x)=!0=> f(c)!= 0`
This means that Rolle's theorem is not applicable.
It is given that the function is continuous and differentiable. So two conditions of Rolle's theorem are satisfied. The third condition is not satisfied `=>f(-5!= f(5)`
Question: (4) Verify mean value theorem for if`f(x) =x^2 -4x-3 ` in the interval [a, b], where `a= 1` and `b=4 `
Solution:
Here `f(x)= x^2- 4x-3, x in [1, 4]`
(i) Since `f(x)` is a polynomial , so it is continuous in `[1, 4]`
(ii) Since `f(x)` is a polynomial , so it is differentiable in `(1, 4)`
So, there exists a real number c in `(1, 4)` such that
`f(c)=(f(4)-f(1))/(4-1)`
`:.f(x)=2x-4`
`=>f(c)=2c -4`
` f(4)=(4)^2 -4xx4-3`
`=16-16-3=-3`
` f(1)= (1)^2 -4 xx1-3`
` 1-4-3=-6`
` 2c-4= (-3-(-6)/(4-1)`
`=> 2c-4=3/3=> 2c=1+4`
` c= 5/ 2 in (1, 4)`
Thus mean value theorem is verified.
Question: (5) Veryfy mean value theorem for `f(x) = x^2 - 5x^2 - 3` in the interval `[a, b]`, where `a=1` and `b =3` find all `c` in `(1, 3)`for which `f(c) =0`
Solution:
Here `f(x) = x^3--5x^2-3, x in [1,3]`
(i) Since `f(x)` is a polynomial , so it is continuous in `[1, 3]`
(ii) Since `f(x)` is a polynomial , so it is differentiable in `(1, 3)`
So there exists a real number c in (1, 3) such that
`f(c)=(f(3)-f(1))/(3-1)`
`:.f(x)=3x^2-10x`
`=>f(c)=3c^2 -10c`
` f(c)= (3)^3-5(3)^2-3`
` =27 - 45 - 3= -21`
`f(1)=(1)^3-5 (1)^2-3`
` 1-5-3=-7`
`:. 3c^2-10c`
`= (-21-(-7))/(3-1)`
`=>3c^2-10c=-7`
`=>3c^2-10c +7=0`
`:. 3c^2-10c`
`=(-21- (-7))/(3-1)`
`=> 3c^2- 10c =-7`
`=> 3c^2- 10c+7=0`
`=> 3c^2-7c-3c+7=0`
`=> c(3c-7)(3c-7)=0`
`=> (3c-7)(c-1)=0`
`=>c= 7/3 or c=1`
`c= 1` is not possible`:. c= 7/3 in (1,3)`
`f(c)=0`
` =>3c^2-10=0`
` => c=10/3 not in (1, 3)`
Question: (6) Examine the applicablity of mean value theorem for all three functions given in the above exercise 2.
(i)`f(x)=[x, x in [5,9]`
(ii)` f(x) = [x, x in [-2, 2]`
(iii)`f(x)=x^2 -1`, for `x in [1, 2]`
Solution:
Here `f(x)=[x], x in [5,9]`
At `x =7`
L.H.L.`=lim_(x->7^-)f(x)`
`= lim_(x->7^-)[x]`
Put `x=7-h`, as `x->7^-,h->0`
`:. lim_(h-> 0)[7-h]`
` =lim_(h->0)(6)=6`
R.H.L. `= lim_(x->7^+)f(x)`
`= lim_(x->7^+)[x]`
put `x= 7+h`, as `x-> 7^+, h->0`
`:.lim_(h->0)[7+h]`
`= lim_(h->0)(7)=7`
L.H.L.`!=`R.H.L.
So `f(x)` is not continuous at `x=7`
`:. f(x)` is not continuous in `[5, 9]`
Thus mean value theorem is not applicable.
(ii) Here `f(x)=[x], x in [5, 9]`
At `x= 1`
L.H.L `= lim_(x->1^-)f(x)`
` lim_(x->1^-)[x]`
`Put x=1-h, as x-> 1^-, h->0`
`:. lim_(h->0)[1-h]`
` = lim_(h->0)(0)=0`
R.H.L. `= lim_(x->1^+)f(x)`
`= lim_(x->1^+)[x]`
put `x= 1+h`, as `x-> 1^+, h->0`
`:.lim_(h->0)[1+h]`
`= lim_(h->0)(1)=1`
L.H.L.`!=`R.H.L.
So `f(x)` is not continuous at `x=1`
`:. f(x)` is not continuous in `[-2, 2]`
Thus mean value theorem is not applicable.
(iii)Here `f(x) = x^2-1, x in [1,2]`
(i) Since `f(x)` is a polynomial , so it is continuous in `[1, 2]`
(ii) Since `f(x)` is a polynomial , so it is differentiable in `(1, 2)`
So there exists a real number c in (1, 2) such that
`f(c)=(f(2)-f(1))/(2-1)`
`:. f(x)= 2x => f(c)=2c`
`f(2)=(2)^2-1=4-1=3`
`f(1)= (1)^2-1=1-1=0`
`:. 2c = (3-0)/(2-1)`
`=> 2c =3`
`=> c= 3/2 in (1,2)`
Thus mean value theorem is verified.
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