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Must Do Problems Of Function for NDA

Must Do Problems For NDA

Must Do Problems For NDA
Q 2814401350

The function `f : R -> R` is defined by `f(x) = 3^(-x)`
I. `f` is one - one function.
II. `f` is onto function.
III. `f` is a decreasing function.
Which of the above statement(s) is/ are correct ?

(A)

I and II

(B)

II and III

(C)

I and III

(D)

All of these

Solution:

since `f : R -> R` such that ` f (x) = 3^(- x)`

Let `y_1` and `y_2` be two elements of ` f(x)`

such that ` y_1 = y_2`

` => 3^(-x_1) = 3^(-x_2) => x_1 = x_2`

Since, if two images are equal, then their

elements are equal, therefore it is

one-one function.

Since, `f(x)` is positive for every value of

`x`, therefore `f(x)` in into.

On differentiating w.r.t. x, we get

`(dy)/(dx) => - 3^(-x) log 3 < 0` , for every value of ` x` .

`:.` it is decreasing function.

`:.` Statements I and III are true .
Correct Answer is `=>` (C) I and III
Q 2813091840

T'he function `f : R -> R` defined by `f (x) = 4^x + 4^(|x|)` is

(A)

one - one and into

(B)

many-one and into

(C)

one-one and onto

(D)

many - one and onto

Solution:

Since, for different values of `x, 4^x`

and `4^(|x|) ` are different positive numbers.

`:. f` is one - one.

Also, `f` is not onto as its range is `(0 , oo)`

and it is subset of its codomain `R`.
Correct Answer is `=>` (A) one - one and into
Q 2884701657

Consider the function ` f(x) = sqrt( 3x^2 - 4x + 5)`.
lf `g(x) = log_e x^2` then range of the function `g[f(x)]` is

(A)

`( - oo , log_e (11)/3 ]`

(B)

`[ log_e (11)/3 , oo)`

(C)

`[ - log_e (11)/3 , log_e (11)/3]`

(D)

None of these

Solution:

`g [f(x)] = log_e ( sqrt( 3x^2 - 4x + 5)^2)`

`= log_e ( 3x^2 - 4x + 5)`

Let `y = log_e ( 3x^2 - 4x + 5)`

` => e^y = 3x^2 - 4x + 5`

` => 3x^2 - 4x + ( 5 - e^y) = 0`

For `x` to be real, discriminant `>= 0`

`:. 16 - 12 ( 5 - e^y) >= 0`

`=> 12 e^y >= 44`

`=> e^y >= (11)/3 => y >= log_e (11)/3`

`:.` Range of ` f = [ log_e (11)/3 , oo)`
Correct Answer is `=>` (B) `[ log_e (11)/3 , oo)`
Q 2713267140

Let `f(x) = x^2 + 4x - 1 ` and `g(x) =| x |`

If `h(x) = f(g(x))+ 10`. Find rangle of `h(x)`.

Solution:

`h(x) = f(g(x)) + 10 =| x|^2 +4| x |- 1 + 10`

`h(x) =| x |^2 + 4| x |+ 9 = (| x|+ 2)^2 + 5`

Hence range of `h(x)` is `[9, oo)`.
Q 2813080849

The range of the function `f(x) = 1/(2 - sin 3x)` is

(A)

`[ 1/3 , 1 [`

(B)

`[ 1/3 , 1 ]`

(C)

`[ 1/3 , 1 [`

(D)

`[ 1 , 1/3 ]`

Solution:

We have, `2y - y sin 3x = 1`

` => sin 3x = (2y - 1)/y`

Since, `- 1 <= sin 3x <= 1`

We have , ` - 1 <= ( 2y - 1)/y <= 1` ... (i)

Since, `y > 0` multiplying the inequality

Eq. (i) by `y`. we obtain

` - y <= 2y - 1 <= y` or `1 <= 3 y ` and `y <= 1`

`=> 1/3 <= y <= 1`
Correct Answer is `=>` (B) `[ 1/3 , 1 ]`
Q 2873080846

The domain of the function ` f(x) = 1/(log_(10) (1 - x)) + sqrt ( x + 2)` is

(A)

`] - 3 , - 2 .5 [ cup ] - 2.5 - 2 [`

(B)

`[-2 , 0 [ cup] 0 , 1 [`

(C)

`] 0 ,1 [`

(D)

None of these

Solution:

For `f(x)` to be defined,

`x + 2 >= 0 => x >= -2`

and ` 1 - x > 0 ` and ` 1 - x != 1`

`=> 1 - x > 0` and `1 - x != 1`

`=> x < 1` and `x != 0`

`:. x in [- 2, 0 [ cup ] 0, 1 [`
Correct Answer is `=>` (B) `[-2 , 0 [ cup] 0 , 1 [`
Q 2723856741

Let f be a real valued function of real and positive argument such that `f(x) + 3x f (1/x) =2 (x+1) AA x > 0`,

find the value of `f( 1 0099)`.

Solution:

`f(x) + 3x f (1/x) =2 (x+1)` ...........(1)

replacing x by `1/x`

`f (1/x) + 3/x f(x) = 2 (1/x +1 )`

`f (1/x) + 3/x f(x) = 2 (1/x +1 )`

`x f (1/x) +3 f(x) =2 (1+x)` ........(2)

On solving(1) and (2)

`8 f (x) =4 (1+x) => f(x) = (x+1)/2`

`f (10099) = 10100/2 =5050`
Q 2824601551

Consider the function ` f(x) = sqrt( 3x^2 - 4x + 5)`.
The domain of function `f(x)` is

(A)

`R`

(B)

`( - oo , 1)`

(C)

`(1, oo)`

(D)

`(2/3 , oo)`

Solution:

`f(x)` is defined , if `3x^2 - 4x + 5 >= 0`

`=> 3 [ x^2 - 4/3 x + 5/3 ] >= 0`

` => 3 [ ( x - 2/3 )^2 + (11)/9 ] >= 0`

Which is true for all real `x`.

`:.` Domain `(f) = (- oo , oo) = R`
Correct Answer is `=>` (A) `R`
Q 2743656543

Find the period of `f(x)` satisfying the condition

(i) `f(x- 1) +f(x + 3)= f(x + 1) +f(x + 5)`

(ii) `f(x) = {x} + cos pi x`

where `{ * }` denotes fraction part.

Solution:

(i) `f(x - 1) + f(x + 3)= f(x + 1)+ f(x + 5)` ............(1)

Replacing `x` by `x + 2`

`f(x + 1) +f(x + 5) = f(x + 3) + f(x + 7)`

Adding (1) and (2), we get

`f(x - 1) = f(x +7)` i.e. `f(x) = f(x + 8)`


Hence, period is `8`.

(ii) ` f(x) = {x} + cos pi x`

Period of `{x} = 1`

Period of `cos pi x = (2 pi)/pi = 2`

Hence period of `f(x) = 2`.
Q 2814212159

Consider `f(x) = ([x])/x` and ` g(x) = | x | ` where `[·]` denotes the greatest integer function.
What is the value of `fof (-7 // 4) + gog (-1) `?

(A)

`0`

(B)

`-1`

(C)

`1/4`

(D)

`- 1/8`

Solution:

`fof (- 7/4) + gog (-1)`

`= f (f (-7 // 4)) + g (g(-1))`

`= f (( [-7// 4])/(-7//4) ) + g (-1 xx 1) `

` = f ( (-2)/(-7//4) ) + g ( -1) = f (8/7) + (-1)`

` = ([8//7])/(8//7) -1 = 7/8 - 1 = - 1/8`
Correct Answer is `=>` (D) `- 1/8`
Q 2783691547

Let `f(a) =(a-1)/(a+1)` Consider the following :

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

2. `f(1/a) =-f(a)`

Which of the above is /are correct?
NDA Paper 1 2017
(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

`f(a) =(a+1)/(a+1)`

1. `f(2a) =(2a-1)/(2a+1)`

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

`f(a) ne f(a) +1`

Here (1) is incorrect

2. `f(1/a) =(1-a)/(1+a)`

`-f(a) =(1-a)/(1+a)`

`f(1/a) =-f(a)`

Hence (2) is correct
Correct Answer is `=>` (B) 2 only
Q 1502556438

If `f(x) = In ((1+x)/(1-x)),` then

(A)

`f(x_1 ) · f(x_2 ) = f(x_1 + x_2)`

(B)

`f(x + 2)- 2f(x + 1) + f(x) = 0`

(C)

`f(x) + f(x + 1) = f(x2 + x)`

(D)

`f(x_1) + f(x_2) = f ((x_1+x_2)/(1+x_1x_2))`

Solution:

`f(x_1)+ f(x_2) = In ((1+x_1)/(1-x_1)) + In ((1+x_2)/(1-x_2))`

` = In {((1+x_1)(1+x_2))/((1-x_1)(1-x_2))}`

` = In {(1+(x_1+x_2)+x_1x_2)/(1-(x_1+x_2)+x_1x_2)}`

` In {(1+((x_1+x_2)/(1+x_1x_2)))/(1-((x_1+x_2)/(1+x_1x_2)))} = f ((x_1+x_2)/(1+x_1x_2))`
Correct Answer is `=>` (D) `f(x_1) + f(x_2) = f ((x_1+x_2)/(1+x_1x_2))`
Q 2468878705

The set of all `x` satisfying the inequality `(4x-1)/(3x+1) ge 1`
BCECE Stage 1 2016
(A)

`(-oo,(-1)/3) cup (1/4,oo)`

(B)

`(-oo,(02)/3) cup [5/4,oo)`

(C)

`(-oo,(-1/3)) cup [2,oo)`

(D)

`(-oo, -2/3] cup[4,oo)`

Solution:

Given, `(4x-1)/(3x+1) ge 0`

`=> (x-2)/(3x+1) ge 0`

`x - 2 ge 0` and `3x + 1 > 0`

or `x - 2 le 0` and `3x + 1 < 0`

`x ge 2` and `x < -1/3`

orb `x le 2` and `x < -1/3 => x in (-oo,-1/3) cup [2, oo)`
Correct Answer is `=>` (C) `(-oo,(-1/3)) cup [2,oo)`
Q 2883091847

If `f (x)` satisfies the relation `2 f ( x) + f (1 - x) = x^2` for all real `x`, then `f(x)` is

(A)

` (x^2 + 2x - 1)/6`

(B)

` (x^2 + 2x - 1)/3`

(C)

` (x^2 + 4x - 1)/3`

(D)

` (x^2 + 4x - 1)/6`

Solution:

Given, `2f(x )+ f ( 1 - x)` ... (i)

Replacing `x` by `( 1 - x )`, we get

`2f ( 1 - x) + f (x) = ( 1 - x)^2`

`=> 2 f (1 - x) + f(x) = 1 + x^2 - 2x` .... (ii)

Multiplying Eq. (i) by `2` and subtracting

Eq. (ii) from Eq. (i), we get

`3 f(x) = x^2 + 2x -1`

` => f(x) = ( x^2 + 2x - 1)/3`
Correct Answer is `=>` (B) ` (x^2 + 2x - 1)/3`
Q 2560545415

Let `f: R -> R` be defined as `f(x) = (x^2 -x +4)/(x^2 +x +4)`

Then, range of the function `f(x)` is
WBJEE 2015
(A)

`[3/5,5/3]`

(B)

`(3/5,5/3)`

(C)

`(-oo, 3/5) cup (5/3 , oo )`

(D)

`[-5/3 , -3/5]`

Solution:

Let `y = (x^2 -x +4)/(x^2 +x+4)`

`=> x^2 y + xy + 4y = x^2 -x +4`

`=> (y-1)x^2 +(y+1)x+4y-4 = 0`

For `x` to be real, discriminant of the above
quadratic equation should be greater than or
equal to `0`.

`=> (y+1)^2 -4(y-1)(4y-4) ge 0`

`=> (y+ 1)^2 -16(y-1)^2 ge 0`

`=> (y+1+4y-4)(y+1-4y+4) ge 0`

`=> (5y-3)(5-3y) ge 0`

`:. y in [3/5 ,5/3]`
Correct Answer is `=>` (A) `[3/5,5/3]`
Q 2416334270

Inverse of function `f(x)=(10^x-10^(-x))/(10^x+10^(-x))` i s
UPSEE 2010
(A)

`log_10 (2-x)`

(B)

`1/2 log_10 ((1+x)/(1-x))`

(C)

`1/2 log_10 (2x-1)`

(D)

`1/4 log_10 (2x)/(2-x)`

Solution:

Let `f(x) = y`, then

`(10^x-10^(-x))/(10^x +10^(-x))=y`

`=> (10^(2x) -1)/(10^(2x) +1) =y`

`=> 10^(2x) =(1+y)/(1-y)`

`=> x=1/2 log_10 ((1+y)/(1-y))`

`=> f^(-1) (y)=1/2 log_10 ((1+x)/(1-x))`

`=> f^(-1) (x)=1/2 log _10 ((1+x)/(1-x))`
Correct Answer is `=>` (B) `1/2 log_10 ((1+x)/(1-x))`
Q 2305078868

If `f(x + 2y, x - 2y) = xy`, then `f(x, y)` equals
BITSAT Mock
(A)

`(x^2 - y^2)/8`

(B)

`(x^2 - y^2)/4`

(C)

`(x^2 + y^2)/4`

(D)

`(x^2 -y^2)/2`

Solution:

Putting `u = x + 2y, v = x - 2y`,

`x= (u+v)/2 , y = (u-v)/4`

`:. f(u,v) = ( (u+v)/2) ((u-v)/4)`

`= (u^2 -v^2)/8`

`=> f(x,y) = (x^2 - y^2)/8`
Correct Answer is `=>` (A) `(x^2 - y^2)/8`
Q 2670323216

If `f(x) = 1/(1 - x)`, then the derivative of the composite function `f [f {f(x)}]` is equal to
BCECE Mains 2015
(A)

`0`

(B)

`1//2`

(C)

`1`

(D)

`2`

Solution:

Clearly, `f( x)` is not defined at `x = 1`.

For `n in R -{1}`, we have

`g(x) = f { f{f(x)}]`

` = f [ f(1/(1 - x)) ] = f ( 1/(1 - 1/(1 - x)))`

` => g(x) = f ( (x - 1)/x) = 1/( 1 - (x - 1)/x) = x`

`:. g'(x) = 1` for all `x in R - {1}`
Q 2823691541

If `f(x) = 3x + 10` and `g(x) = x^2 - 1` then `(fog)^(- 1)` is equal to

(A)

` ( (x - 7)/3)^(1//2)`

(B)

` ( (x + 7)/3)^(1//2)`

(C)

` ( (x - 3)/7)^(1//2)`

(D)

` ( (x + 3)/7)^(1//2)`

Solution:

`f(x) = 3x - 10` and `g (x) = x^2 - 1`

`:. fog = f[ g(x )]= 3 [g(x) ] + 10`

`= 3 (x^2 - 1) + 10 = 3x^2 + 7`

Let `3x^2 + 7 = y => x^2 = (y - 7)/3`

`=> x = ((y-7)/3)^(1//2)`

So, `(fog)^(-1) = ( (x - 7)/3 )^(1//2)`
Correct Answer is `=>` (A) ` ( (x - 7)/3)^(1//2)`
Q 2823891741

Let `f(x) = (- 1)^([x]) ` (where [·] denotes the greatest integer function), then



(A)

range of `f` is `{ -1 ,1 }`

(B)

`f` is an even function

(C)

`f` is an odd function

(D)

`f` is one - one function

Solution:

`f(x) = (-1)^([x]) = { -1, 1 }` , since ` [ x] in Z`
Correct Answer is `=>` (A) range of `f` is `{ -1 ,1 }`
Q 2853180944

The function `f(x) = log ( x + sqrt(x^2 + 1) )` is

(A)

an even function

(B)

an odd function

(C)

periodic function

(D)

None of these

Solution:

`f (-x) = log [ -x + sqrt(1 + x^2 ) ]`

`f(x) + f(-x) = log [ x + sqrt(1 + x^2 ) ]`

`+ log [-x + sqrt(1 + x^2 ) ]`

`= log [1 + x^2 - x^2 ] = log 1= 0`

` :. f(-x) = - f(x)`

So, `f( x)` is an odd function of `x`.
Correct Answer is `=>` (B) an odd function
 
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