Math Twelve

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 `

So , the least value of `K` is `-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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