Application of Derivatives NCERT Solutions
NCERT Exercise 6.2 Q:10-19
Question:(10) Prove that the logarthmic function is strictly increasing on `(0, oo)`.
Solution:
Let `f(x)=log x`
`f'(x) =1/x > 0 when x> 0`
`:. f(x) ` is increasing in ` (0, oo)`.
Question: (11) Prove that the function `f ` given by ` f(x)=x^2 -x+ 1` is neither increasing nor decreasing strictly on `(-1,1)`
Solution:
We have `f(x)= x^2 -x+1`
`f(x) = 2x-1`
Now `f'(x=0`
`=> 2x -1 =0 => x= 1/2`
` Case 1: When `-1, In this case `f' (x) < 0` `:. f(x) ` is strictly decreasing in `(-1 1/2)` Case 11: When ` 1/2 < x< 1` In this cas `f(x)> 0` `:. f(x)` is strictly increasing in `(1/2, 1)` Hence `f(x)` is neither strictly increasing nor decreasing in `(-1,1)` Question:(12) Which of the functions are strictly decreasing on `(0, pi/2)` (A) `cos x` (B) `cos2x` (C) `cos 3x` (D) `tan x` Solution: (A) We have `f(x) = cos x` `:. f'(x)=-sin x` `f'(x) =0` `=>-sin x= 0 => sin x =0` `=> x = 0` When `0 < x < pi/2` In this case `f'(x)< 0` `:. f(x)` is strictly decreasing on `(0, pi/2)` (B) We have `f(x) = cos 2x` `:. f'(x) =-2sin 2x` `f'(x)=0` `=> -2 sin2x ` `=0=> sin 2x =0` `=> 2x =0` `x=0` When `0< x < pi/2 or 0< 2x < pi` `:. sin 2x > 0` In this case `f'(x)< 0` `:. f(x)` is strictly decreasing on `(0, pi/2)` (C) In case of `cos 3x` the ` f'(x) < 0` When `0 < x < pi/2`, Therefore `f(x)` is strictly decreasing in `( 0, pi/2)` (D) When `f(x)= tan x` `:. f'(x) =sec^2 x` clearly `f'(x) > 0` `:. f(x)` is strictly increasing on `(0, pi/2)` Hence , function `cos x, cos 2x` and `cos 3x` //* are strictly decreasing. Question: (13) On which of the following intervals is the function `f` given by `f(x)=x^(100+ sin x-1` strictly decreasing? (A)`(0, 1)` (B)`(pi/2 , pi)` (c)`(0, pi/ 2)` (D)None of these. Solution: `f(x) =x^(100) + sin x-1` `f'(x) = 100x^(99) + cos x` (A) `x in (0,1)=> x^(99) > 0 ` `=> 100 x^(99) > 0` `=> x` lies between 0 and 1 radian `=> x` lies between `0^0` and ` 57^0 [1= 57^0]` `=>x` lies in first quadrant `=> cos x> 0` `:. x in (0,1)=> 100x^(99)+ cos x > 0` `=> f (x)> 0` Thus `f(x)` is increasing on `(0,1)`, (B)`x in (pi/2, pi)` `pi/2 < x < pi` `[pi/2 < x < pi => 22/14 < x < 22/7]` `=> x^(99) > 1` ` 100 x^(99) >100` Also `x in (pi/2, pi)` `=> -1< cos x < 0` `=> 0 > cos x > -1` ---(ii) From (i) and (ii), we obtain that ` x in ( pi/2 , pi)` ` => 100x^(99)+ cos x > 100 -1 = 99 > 0` ` => 100x(99) + cos x > 0` `=> f(x)> 0 ` Thus `f(x) ` is increasing on `( pi/2, pi)` (C) `x in ( 0, pi/2)` ` 0 < x < pi/2 ` `=> x^(99)> 0 ` `100x ^(99)+ cos x > 0 ` => f'(x) > 0 ` Thus `f(x)` is increasing on ` ( 0, pi/2)` Thus option (D). None of these is correct . Question: (14) Find the least value of a such that the function `f given by `f(x) =x^2+ ax +1` is strictly increasing on `(1, 2)` Solution: We have `f(x)=x^2 +ax +1` `f'(x) = 2x +a` Now `1 < x < 2` `=> 2 < 2x < 4 ` `=> 2+ a < 2x + a < 4 + a ` now, ` f(x)` is strictly increasing when ` f'(x)> 0 ` So, `0 < 2 + a ` Now `1 < x < 2 ` `=> 2 < 2x < 4 ` ` => 2+ a < 2x + a < 4 +a` Now , `f(x)` is strictly increasing when `f'(x) > 0` So, `0 < 2 +a` `:. 0 < 2+a` `=> -2< a ` `=> a > -2 `
Question: (15) Let 1 be any interval disjoint from `(-1, 1, 1)` Prove that the function `f ` given by
`f(x) = x+ 1/x` is strictly increeasing on `1`
Solution:
we have `f(x)= x+ 1/x`
`f'(x) =1 - 1/x^2 => (x^2 - 1)/ x^2 `
` => (x^2 -1)/ x^2 `
Now `x in 1 => x !in (-1, 1)`
` x - <= -1 ` or `x >= 1`
`=> x^2 >= 1 => x^2 - 1 >= 0`
`=> (x^2-1)/x^2 >= 0`
`[∵ x^2 .= 0]`
f' (x) >= 0`
Thus `f(x) >= 0 ` for all `x in 1`
Hence `f(x)` increasing on 1
Question: (16) Prove that the function `f ` given by `f(x) = log sin x` is strictly ingrsing on `(0, pi/2)` and strictly decreasing on (pi/2, pi)`
Solution:
We have ` f(x)= log sin x`
`:. f' (x) =1/(sin x) . cos x = cot x`
`:. f'(x) -0 => cot x=0`
`=> cot x = 0 => x = pi/2 `
Thus , the critical point is `( x= pi/2)`
Case 1: when 0 < x < pi/ 2`
in this case, `cose , cot x < 0` and therefore ` f '(x) < 0`
Hence `f(x)` is strictly decreasing in `(pi/2, pi)`
Question: (17)Prove that the function `f ` givev by `f(x)=log cos x` is strictly decreasing on `(o, pi/2)` and strictly increasing on `( pi/2, pi)`
Solution:
`f(x) = log cos x`
` f(x) = 1/(cosx) (- sin x)=- tan x`
Case 1: `x in ( 0, pi/2)`
`tan x > 0`
`- tan x < 0 `
`f' (x) < 0 `
Hence `f(x) ` is decreasing function on ` (o, pi/2)`
Case 11: `x in ( pi/2, pi)`
`tan x< 0 `
`-tan x > 0`
`f(x) > 0`
Hence `f(x) ` is an increasing function on `( pi/2, pi)`
Question: (18) Prove that the function given by `f(x)=x^2 -3x^2 +3x -100` is increasing in `R`
Solution:
We have `f(x)= x^3-3x^2 + 3x- 100`
`f'(x) = 3x^2-6x +3`
`=3(x^2-2x+1)`
`=3(x-1)^2`
Now,` f'(x) =0`
`=> 3(x-1)^2`
` = 0=> (x-1)^2 =2`
`=> X-1 = 0 => x=1`
When `x < 1 `
Then f' (x) = 3(+)= +ve`
`:. f' (x) >0 `
`:. f(x)` is increasing for `x < 1 `
when ` x > 1`
` f(x) = 3 (+)= +ve`
`:. f' (x) > 0`
`:. f(x)` is increasing for ` x>1`
Hence `f(x)` is increasing in `(-oo,1) uu (1, oo)` i.e. in R.
Question:(19) The interval in which ` y= x^2e^(-x) ` is increasing is :
(A)`(-oo, oo)`
(B)`(-2,0)`
(C)`(2, oo)`
(D)`(0, 2)`
Solution:
`y = x^2 e^(-x)`
`(dy)/(dx) = x^2 e^(-x)(-1)+e^(-x)(2x)`
`(dy)/(dx)= xe^(-x)[ -x + 2]`
For the increasing and decreasing nature `(dy)/(dx) = 0`
`xe^(-x)(-x +2)`
`x =0 v e^(-x) =0 `
`x=0 v e^(-x) = 0 ` not possible `v x=2`
`()
for `x=0, 2`,We get the intervals as `(-oo, 0),(0,2), (2,oo)`
for `x in (-oo, 0)`
`f'(x)` =(+1)(+1)(+1)(+1)= +ve`
For `x in ( 2, oo) `
`f'(x) = (+2)(+)(-1)=-ve`
Hence option (D), `(0,2)` is correct.
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