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Previous year question Of Function for NDA

Previous Year NDA Question Of Functions :

Set 1
Q 2723591441

The function `f: X → Y` defmed by `f(x) = cosx`, where `x in X`, is one-one and onto if `X` and `Y` are respectively equal to
NDA Paper 1 2017
(A)

`[0, pi]` and `[-1, 1]`

(B)

`[-pi/2 , pi/2]` and `[-1 , 1]`

(C)

`[0 , pi]` and `(-1 , 1)`

(D)

`[0, pi]` and `[0, 1]`

Solution:

Please see the graph of `cos x` between 0 and `pi`

No line parallel to x-axis cuts it twice so one-one

Range is [-1,1] which is equal to co-domain so onto

Graph will be one one onto `[0, pi]-> [-1 , 1]`
Correct Answer is `=>` (A) `[0, pi]` and `[-1, 1]`
Q 2753591444

If `f(x) = x/(x-1)` then what is `( f(a))/(f(a+1))` equal to?
NDA Paper 1 2017
(A)

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

(B)

`f(a^2)`

(C)

`f(1/a)`

(D)

`f(-a)`

Solution:

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

`(f(a))/(f(a+1)) =(a/(a-1))/((a+1)/a) = a^2/(a^2-1) = f(a^2)`
Correct Answer is `=>` (B) `f(a^2)`
Q 2763691545

Let `f(x) = px + q` and `g(x) = mx + n` Then `f {g(x)} = g(f(x))` is equivalent to
NDA Paper 1 2017
(A)

`(p) = g(m)`

(B)

`(q) = g(n)`

(C)

`f(n) = g(q)`

(D)

`f(m) = g(p)`

Solution:

`f(x) =px+q`

`g(x) =mx +n`

`f(g(x)) =g(f(x))`

`f(mx+n)+q =m(px+q)+n`

`pn+q =mq+n`

`f(n)=g(q)`
Correct Answer is `=>` (C) `f(n) = g(q)`
Q 2753680544

Let `f(x) : { tt (( x, x, is, rational),(0, x, is, irrational))`

and

`g(x) : { tt (( 0, x, is, rational), (x, x, is, irrational))`

If `f : R -> R` and `g: R -> R`, then `(f - g)` is
NDA Paper 1 2017
(A)

one-one and into

(B)

neither one-one nor onto

(C)

many-one and onto

(D)

one-one and onto

Solution:

`f-g(x) : { tt (( x, x, is, rational),(-x, x, is, irrational))`

`=> (f-g)(x_1)=(f-g)(x_2)`

`x_1 =x_2` { in both cases when `x`, is rational or irrational

Hence, `(f-g)(x)` is one-one.

As f and g are from R -> R so f-g will also be from R->R

now for for all y element of R there is an x in f-g which equal to y or -y depending upon whether y is rational or irrational.
Correct Answer is `=>` (D) one-one and onto
Q 2743591443

Which one of the following functions is neither even nor odd?
NDA Paper 1 2017
(A)

`x^2-1`

(B)

`x+ 3/x`

(C)

`|x|`

(D)

`x^2 (x-3)`

Solution:

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

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

`f (x ) = f (-x) ` Hence it is even function

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

`f (-x) = - ( x + 3/x ) = - f (x )`

`f (-x ) = - f (x ) `

Hence its a odd function

3) `f (x ) = | x |`

` f (-x ) = | x |`

`f (x ) = (-x)`

hence it a even function

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

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

`f (x ) != f (-x )`

`f(-x) != - f (x )`

Hence it is neither even nor odd function
Correct Answer is `=>` (D) `x^2 (x-3)`
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 2116578479

If `f(x_1 ) - f(x_2 ) = f ((x_1 - x_2)/(1 - x_1x_2) )` for `x_1 , x_2 in (-1, 1)`, then
what is `f(x)` equal to ?
NDA Paper 1 2016
(A)

ln `( (1 - x)/(1+x))`

(B)

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

(C)

`tan^(-1) ( (1 - x)/(1+x))`

(D)

`tan^(-1) ( (1 + x)/(1 - x))`

Solution:

Check by option

`f(x) = ln( (1 - x)/(1+x))`

Now, `f(x_1) - f(x_2 ) = ln ( (1 - x_1)/(1+x_1)) -ln ( (1 - x_2)/(1+x_2))`

`= ln ( ((1 - x_1) (1 + x_1))/ ((1 - x_1)(1 + x_1)) )`

`= ln ( (1 + x_2 - x_1 - x_1x_2 )/ (1 + x_1 - x_2 - x_1x_2 ) )` ........(i)


and `f ( (x_1 - x_2)/(1 - x_1x_2) ) = ln ( ( 1 - (x_1 - x_2)/(1 - x_1x_2) )/( 1 + (x_1 - x_2)/(1 - x_1x_2) ))`

`= ln ( (1 - x_1x_2 - x_1 + x_2) /(1 - x_1x_2 + x_1 - x_2))` ..........(ii)

`:.` From Eqs. (i) and (ii), we get

` f(x_1) - f(x_2) = f ( (x_1 - x_2)/(1 - x_1x_2) )`
Correct Answer is `=>` (A) ln `( (1 - x)/(1+x))`
Q 2116678570

What is the range of the function

` y = x^(2)/(1+(x)^(2))` where `x in R?`

NDA Paper 1 2016
(A)

`[0, 1)`

(B)

`[0, 1]`

(C)

`(0, 1)`

(D)

`(0, 1]`

Solution:

Let ` f(x) = x^(2)/(1+(x)^(2))`

Clearly, domain `(f) = R`

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

`=> y + x^(2)y = x^(2)`

`=> x^(2) - x^(2)y = y`

`=> x^(2)( 1-y) = y`

`=> x = pm sqrt(y/(1-y))`

Clearly, `x` will take real values, if `y/(1-y) >= 0`

` => (y-0)/(y-1) <= 0`

` => 0 <= y <1`

` => y in [0, 1)`

Hence, range `f(x)` is `[0, 1)`.
Correct Answer is `=>` (A) `[0, 1)`
Q 2156778674

Consider the curves `f(x) = x | x | - 1` and `g(x) = { tt (((3x)/2 ,x > 0 ) , (2x, x <= 0))`

Where do the curves intersect?
NDA Paper 1 2016
(A)

At `(2, 3)` only

(B)

At `(-1,- 2)` only

(C)

At `(2, 3)` and `(-1,- 2)`

(D)

Neither at `(2, 3)` nor at `(-1,- 2)`

Solution:

Given curves are

`f(x) =x | x |- 1` and `g(x) = { tt (((3x)/2 ,x>0) , (2x, x <= 0))`

When `x > 0, f(x) = x^(2) - 1` and `g(x) = (3x)/2`

Both curves having common points, if they intersect
each other.

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

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

` => 2(x^(2) -1) =3x`

`=> 2x^(2) -2 -3x = 0`

`=> 2x^(2) -3x -2=0`

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

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

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

`=> 2x + 1 = 0` or `x - 2 = 0`

` => x = -1/2 ` or `x = 2`

`:. x = 2` and `f(x) = g(x) =3`

When `x <= 0, f(x) = -x^(2) -1` and `g(x) = 2x`

Both curves having common points, if they intersect

each other.

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

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

`=> -x^(2) - 2x - 1 = 0`

` => x^(2) + 2x + 1 =0`

`=> (x + 1)^(2) = 0`

`=> (x+ 1)(x+ 1)=0`

` => x= -1`.

since `x <= 0`

`:. x = -1` and `f(x) = g(x) = -2`

So, both curves intersect at `(2, 3)` and `(-1, -2)`.

Hence, correct option is (c).
Correct Answer is `=>` (C) At `(2, 3)` and `(-1,- 2)`
Q 2136078872

Consider the function. `f( x) = (27(x^(2//3) - x))/4` .

How many solutions does the function `f(x) = -1` have?

NDA Paper 1 2016
(A)

One

(B)

Two

(C)

Three

(D)

Four

Solution:

Given, `f( x) = (27(x^(2//3) - x))/4`

We have, `f( x) =-1`

`=> (27(x^(2//3) - x))/4 = -1`

` => x^(2//3) - x = - 4/(27)`

Let `x^(1//3) = alpha => x = alpha^(3)`

Then `alpha^(2) -alpha^(3) = - 4/27`

`=> alpha^( 3) -alpha^(2) = 4/ 27 `

` => alpha^(2) (alpha -1) = 4/27 `

` => alpha.alpha .(alpha - 1) = 2/3 xx 2/3 xx 1/3 `

` => alpha = 2/3 ` then ` alpha - 1 = - 1/3 `which represents a

contradiction.

Thus, no solution exists.

Hence, no option is correct.
Q 2781480327

What is the domain of the function `f(x) = 1/sqrt(|x| - x)` ?
NDA Paper 1 2016
(A)

`(-oo , 0)`

(B)

`(0 , oo)`

(C)

`0 < x < 1`

(D)

`x > 1`

Solution:

`f(x) = 1/sqrt(|x| - x)`

For Domain

`|x| - x > 0`

`x - x > 0` and `-x -x > 0`

`-2x > 0`

`x < 0`
Correct Answer is `=>` (A) `(-oo , 0)`
Q 2157023884

Let `f : R -> R` be a function such that
` f(x) = x^(3) + x^(2)f' (1) + xf'' (2) + f'" (3)`
for `x in R`.

What is `f(1)` equal to?
NDA Paper 1 2016
(A)

`-2`

(B)

`-1`

(C)

`0`

(D)

`4`

Solution:

`f(x) = x^( 3) + x ^(2) f'(1) + xf''(2) + f'''(3) ... (i)`

On differentiating Eqs. (i) w.r.t.x, thricely we get

`f'(x)=3x ^(2) + 2xf'(1)+0+ f''(2)+0+0 ... (ii)`

`f''(x) = 6x + 2f'(1) ... (iii)`

and `f'''(x) = 6 ... (iv)`

On putting `x = 1, 2, 3` in Eqs. (ii), (iii) and (iv), we get

`f'(1) = 3 + 2f'(1) + f''(2)`

`f''(2) = 12 + 2f'(1)`

`f'''(3) = 6`

On solving these equations, we get

`f'(1) = -5, f''(2) = 2, f'''(3) = 6`

`:. f(x) = x^( 3) + x^(2) (-5) + x(2) + 6`

`=> f(x) = x^( 3) - 5x^(2) + 2x + 6`

`f(1) = (1)^(3) - 5(1)^(2) + 2(1) + 6`

`= 1- 5+ 2 + 6 = 4`
Correct Answer is `=>` (D) `4`
Q 2137078882

Let `f(x)` be the greatest integer function and `g(x)` be the modulus
function.

What is `( gof) (- 5/3) - ( fog ) (- 5/3)` equal to?
NDA Paper 1 2016
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`( gof) (- 5/3) - ( fog ) (- 5/3)`

`=g(f(-5/3))-f(g(-5/3))`

`= g ([ -5/3])- f ( |- 5/3 | )`

`= g(-2) - f (5/3)`

`= |-2| - [5/3]= 2 - 1 = 1`
Correct Answer is `=>` (C) `1`
Q 2157078884

Let `f(x)` be the greatest integer function and `g(x)` be the modulus
function.

What is `( fof) (- 9/5) + (gog)( -2)` equal to?
NDA Paper 1 2016
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`( fof) (- 9/5) + (gog)( -2)`

`= f (f ( -9/5)) + g(g(-2))`

`= f [( - 9/5)] + g ( |-2|)= f(-2) + g(2)= -2 + 2 = 0`
Correct Answer is `=>` (B) `0`
Q 2271378226

The domain of the function `f( x) = 1/(sqrt(|x| -x)` is
NDA Paper 1 2015
(A)

`[0 ,oo)`

(B)

`(-oo, 0)`

(C)

`[1,oo)`

(D)

`(- oo, 0]`

Solution:

Given , `f( x) = 1/(sqrt(|x| -x)`

Here, `f(x)` is defined only when

`| x | - x > 0` or `| x | > x`

which is possible only for negative values of `x`.

`:.` Domain of `f` is `(- oo, 0)`.
Correct Answer is `=>` (B) `(-oo, 0)`
Q 2251201124

If `g( x) = 1/(f(x))` and `f( x) = x, x != 0`, then which one of the
following is correct ?
NDA Paper 1 2015
(A)

`f(f(f(g(g(f(x)))))) = g(g(f(g(f(x)))))`

(B)

`f(g(g(g(f(x))))) = g(g(f(g(f(x)))))`

(C)

`f(g(f(g(g(f(g(x))))))) = g(g(f(g(f(x)))))`

(D)

`f(f(f(g(g(f(x)))))) = f(f(f(g(f(x)))))`

Solution:

Given, `g(x) = 1/(f(x))` and `f(x) = x`

`:. g(x) = 1/x , x != 0`

Clearly, `g(f(x)) = 1/(f(x)) = 1/x ; g(g(f(x))) = 1/(1/x) = x`

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

`f(g(g(g(f(x))))) = 1/x` and `g(f(x)) = 1/x`

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

`g(f(g(f(x)))) = 1/(1/x) = x ; g(g(f(g(f(x))))) = 1/x`

From option (b), `LHS = RHS`
Correct Answer is `=>` (B) `f(g(g(g(f(x))))) = g(g(f(g(f(x)))))`
Q 1688680507

For each non-zero real number `x`, let `f(x) =x/(|x|)` ,The
range of `f` is
NDA Paper 1 2015
(A)

a null set

(B)

a set consisting of only one element

(C)

a set consisting of two elements

(D)

a set consisting of infinitely many elements

Solution:

We have,

`f(x) =x/(|x|)` for ` x != 0`

i.e., `f(x) = tt{(x/x)`, if ` x >0 (x/(-x))` , if ` x < 0`

` = { (1),` if ` x >0 , (-1)` if ` x < 0`

Thus, range set of `f = {1, -1}`.
Correct Answer is `=>` (C) a set consisting of two elements
Q 1618180909

If `f(x) = log_e ((1+x)/(1-x)) , g(x) = (3x + x^3)/(1 + 3x^2)` and
`gof (t) = g (f(t))`, then what is `gof ( (e -1)/(e +1))` equal to?
NDA Paper 1 2015
(A)

`2`

(B)

`1`

(C)

`0`

(D)

`1/2`

Solution:

We have, `f(x) = log_e ((1+x)/(1-x)) ,` and ` g(x) = (3x + x^3)/(1 + 3x^2)`

To find `gof ( (e -1)/(e +1)) `

Clearly, `gof ( (e -1)/(e +1)) = g [f ( (e -1)/(e +1))]` ...........(1)

Let us first find ` f ( (e -1)/(e +1)) ` which is given by

` f ( (e -1)/(e +1)) = (log_e) ( 1 + (e -1)/(e +1))/( 1 - (e -1)/(e +1)) = ( log_e) ( (e + 1 + e - 1)/(e + 1 - e + 1))`

` = log_e ( (2e)/2) = log_e (e) =1`

Now, from Eq. (i), we have

go ` f ( (e -1)/(e +1)) = g (1) = ( 3(1) + (1)^3 )/( 1+ 3(1)^2) = 4/4 = 1`
Correct Answer is `=>` (B) `1`
Q 1608291108

Consider the following functions

I. `f(x) = x^ 3, x in R`
II. `f(x) = sinx, 0 < x < 2 pi`
III.`f(x) = e^x, x in R`

Which of the above functions have inverse defined
on their ranges?
NDA Paper 1 2015
(A)

I and II

(B)

II and III

(C)

I and III

(D)

I, II and III

Solution:

I. We have, `f(x) = x^ 3, x in R`

`=> f'(x) = 3x^2 >= 0`

`=> f` is increasing function

So, f is one-one .

Hence, `f` is invertible on their range.

II. We have,` f(x) = sin x, 0 < x < 2 pi`

Clearly `f (pi/3) = sin pi/3 = sqrt(3)/2`

and `f ((2pi)/3) = sin( pi - pi/3 ) = sin(pi/3 )= sqrt(3)/2`

So, `f` is not one-one.

Hence, `f` is not invertible on their range.

3. We have, `f(x) =e^x. x in IR`

`=>f'(x ) = e^x > 0`

So, `f` is increasing function. `=> f` is one-one

Hence, `f` is invertible on their range.
Correct Answer is `=>` (C) I and III
Q 2211878720

If `f : R -> R, g : R -> R` are two functions given by
`f(x) = 2x- 3` and `g(x) = x^ 3 + 5`, then `(fog)^(-1) (x)` is equal
to
NDA Paper 1 2015
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Given, `f(x) = 2x - 3` and `g(x) = x^3 + 5`

Now, `(fog ) (x) = f (g(x)) = 2(x^3 + 5) - 3`

`= 2x^3 + 10 - 3 = 2x^3 + 7`

Let `(fog (x) = y)`, then

`y =2x^3 + 7`

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

`:. fog^(-1) (x) = ( (x-7)/2 ) ^(1//3)`
Correct Answer is `=>` (B) `( (x-7)/2 ) ^(1//3)`
Q 2201280128

Consider the following statements
Statement I The function `f : R -> R` such that `f( x) = x^ 3`
for all `x in R` is one-one.
Statement II `f(a) = f(b) => a= b` for all `a, b in R`, if the
function `f ` is one-one.

Which one of the following is correct in respect of the above
statements?
NDA Paper 1 2015
(A)

Both the statements are true and Statement II is the correct explanation of Statement I

(B)

Both the statements are true and Statement II is not the correct explanation of Statement I

(C)

Statement I is true but Statement II is false.

(D)

Statement I is false but Statement II is true.

Solution:

Statement I is correct, since `f(x) = x^3` is one-one,

`AA x in R`.

`∵ f(a) = f(b) => a^3 = b^3 => a = b`

Statement II is also correct and is the correct

explanation of Statement I.
Correct Answer is `=>` (A) Both the statements are true and Statement II is the correct explanation of Statement I
Q 2281191927

`f(xy) = f(x) + f(y)` is true for all
NDA Paper 1 2015
(A)

polynomial functions f

(B)

trigonometric functions f

(C)

exponential functions f

(D)

logarithmic functions f

Solution:

Let `f(x) = log x`

`:. f(xy) = log (xy)` and `f(x) = log x, f(y)= log y`

Now, `f(x) + f(y) =log x + logy = log (xy)= f(xy)`

Hence, `f(x) + f(y) = f(xy)`
Correct Answer is `=>` (D) logarithmic functions f
Q 1649591413

The function `f: N -> N`, where `N` being the set of natural
numbers, defined by `f(x) = 2x + 3` is
NDA Paper 1 2014
(A)

injective and surjective

(B)

injective but not surjective

(C)

not injective but surjective

(D)

neither injective nor surjective

Solution:

We have, `f : N -> N`

`:. f(x) = 2x + 3 => f'(x) = 2 > 0`

So, `f(x)` is increasing, ` AAx in N`

Hence, `f(x)` is injective.

Let `f(x) = y`

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

Let `y = 4`

`:. x = 1/2 `

and `y in N` but `x != N`

Hence,` f(x)` is not surjective.
Correct Answer is `=>` (B) injective but not surjective
Q 1619591419

If `f(x) = ax + b` and `g(x) = cx + d` such that
`f [g(x)] = g[f(x)]`, then which one of the following is
correct?
NDA Paper 1 2014
(A)

`f(c) = g(a)`

(B)

`f(a) = g(c)`

(C)

`f(c) = g(d)`

(D)

`f(d) = g(b)`

Solution:

We have, `f(x) = ax + b` and `g(x) =cx+ d`

`f [(g(x)] = a(cx +d) + b`

` = acx + ad + b`

and `g [f(x)] = c(ax + b)+d`

` = acx + dc + d`

`:. f [(gx)] = g [f(x)]`

`=> ad + b = bc +d`

`=> f(d) = g(b)`
Correct Answer is `=>` (D) `f(d) = g(b)`

Set 2

Q 1722380231

Let `N` denote the set of aU non-negative integers and `Z`
denote the set of all integers. The function `f : Z -> N`
given by `f(x) =| x|` is
NDA Paper 1 2014
(A)

one-one but not onto

(B)

onto but not one-one

(C)

Both one-one and onto

(D)

Neither one-one nor onto

Solution:

Given that. `f : Z -> N` and `f(x) = | x |`

Graph `h` of `f(x) = | x |`

We see that, if we draw a parallel line along `X` -axis. It cuts

the curve into more than one point. So, function `f(x) = | x |` is

not one-one.

Here, codomain of `f = N` (natural number)

and Range of `f = z^+` (positive integers)

Here, codomain of `f != ` Range of `f`

So, function is not onto.
Correct Answer is `=>` (D) Neither one-one nor onto
Q 1649691513

Consider the function `f(x) = (x - 1)/(x + 1)`

What is `(f(x) + 1)/(f(x) - 1) + x` equal to?
NDA Paper 1 2014
(A)

`0`

(B)

`1`

(C)

`2x`

(D)

`4x`

Solution:

We have , `f(x) = (x - 1)/(x + 1)`

Applying componendo and dividendo, we get

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

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

Now, ` (f(x) + 1)/(f(x) - 1) + x = -x + x = 0`
Correct Answer is `=>` (A) `0`
Q 1609691518

Consider the function `f(x) = (x - 1)/(x + 1)`

What is `f(2x)` equal to?


NDA Paper 1 2014
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

We have, `f(x) = (x - 1)/(x + 1)`

` => f(2x)= (2x - 1)/ (2x + 1) => f(2x) = ([ 2 ([f(x) + 1 ])/(1 - f(x))]-1)/ ([ 2 ([f(x) + 1 ])/(1 - f(x))]+1)`

` => f(2x)= (3 f (x) + 1)/ (f(x) + 3)`
Correct Answer is `=>` (C) `(3f(x) + 1)/(f(x) + 3)`
Q 1639791612

Consider the function `f(x) = (x - 1)/(x + 1)`

What is `f[f(x)]` equal to?
NDA Paper 1 2014
(A)

`x`

(B)

`- x`

(C)

` - 1/x`

(D)

None of these

Solution:

We have, `f(x) = (x - 1)/(x + 1)`

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

` => f [f(x)] = - 1/x quad [ ∵ x = - { ( f(x) + 1)/(-f(x) - 1)}]`
Correct Answer is `=>` (C) ` - 1/x`
Q 2318723600

What is the range of the function

`f(x)=(|x|)/x,x ne 0` ?
NDA Paper 1 2013
(A)

Set of all real numbers

(B)

Set of all integers

(C)

`{-1,1}`

(D)

`{ -1 ' 0, 1}`

Solution:

Given function, `f(x)=(|x|)/x,x ne 0`

Redefine the given function,

`f(x)= {tt((x/x, x >0), ((-x)/x, x <0))` `= {tt((1, x >0), (-1, x <0))`

`:.` Range of `f(x)= { -1, 1}`
Correct Answer is `=>` (C) `{-1,1}`
Q 2355001864

If `f` be a function from the set of natural numbers
to the set of even natural numbers given by
`f(x) = 2x`. Then, `f` is
NDA Paper 1 2013
(A)

one to one but not onto

(B)

onto but not one-one

(C)

Both one-one and onto

(D)

Neither one-one nor onto

Solution:

Given, f : N -> N even and `f(x) = 2x`

For one-one Let `x_1, x_2 in R`

Considering `f(x_1)` and `2 (x_2 )`

Such that `f(x_1) = f(x_2 )`

`=> 2x_1 = 2 x_2`

`=> x_1 = x_2`

So, `f` is one-one.

For onto Let `y = f(x)`

Then, `y = 2x => x = y/2` (expressing x in terms of `y`)

`:. x inN, y in N` (even)

(every element of codomain has pre-image in domain)

So, `f` is onto.

Hence, `f` is both one-one and onto.
Correct Answer is `=>` (C) Both one-one and onto
Q 2325323261

If `N` be the set of natural numbers and `f : N -> N`
be a function given by `f (x) = x + 1` for `x in N`, then
which one of the following is correct?
NDA Paper 1 2013
(A)

`f` is one-one and onto

(B)

`f` is one-one but not onto

(C)

`f` is only onto

(D)

`f` is neither one-one nor onto

Solution:

Given that `f : N -> N` and `f(x) = x + 1`, for `x in N`, if

`x_1 , x_2 in N`, then `f(x_1) = f(x_2 )`

`=> x_1 + 1 = x_2 + 1 => x_1 = x _2`

i.e., `f(x)` is one-one.

Range of `f(x) in N-{1}`

`:.` Range `subseteq` Codomain

So, `f(x)` is into function.

Hence, `f` is one-one but not onto.
Correct Answer is `=>` (B) `f` is one-one but not onto
Q 2355723664

If `f(xy) = f(x)f(y)`, then `f(t)` may be of the form

where, k is constants.
NDA Paper 1 2013
(A)

t + k

(B)

`ct + k`

(C)

`t^k + c`

(D)

`t^k`

Solution:

Given that,

`f(zy) = f(x) f(y)`

From option, we take `f (t) = t^k`

Then, `f(xy) = (xy)^k`

`= (x^k) (y)^k`

`= f(x) · f(y)`
Correct Answer is `=>` (D) `t^k`
Q 2308723608

If `f: R -> R` be a function whose inverse is `(x+5)/3`

Then, what is the value of `f(x)`?
NDA Paper 1 2012
(A)

`f(x) = 3x + 5`

(B)

`f(x) = 3x - 5`

(C)

`f(x) = 5x- 3`

(D)

`f(x)` does not exist

Solution:

Given, `f^(- 1) (x)=(x+5)/3=> f^(-1)(y)=(y+5)/3`..............(i)

`y=f(x)=> x=f^(-1)(y)`

`=>x=(y+5)/3=>3x=y+5`

`=>y=3x-5=>f(x)=3x-5`
Correct Answer is `=>` (B) `f(x) = 3x - 5`
Q 2375134966

What is the range of `f(x) =cos2x -sin2x`?
NDA Paper 1 2011
(A)

`[2, 4]`

(B)

`[- 1, 1]`

(C)

`[- sqrt(2), sqrt(2)]`

(D)

`(- sqrt(2), 2)`

Solution:

`∵ f(x) =cos 2x - sin 2x`

`[If f(x) = a cos x + b sin x \ \ \ \ [- sqrt(a^2 + b^2) <= f(x) <= sqrt(a^2 + b^2) ]`

`=>- sqrt(1+1) <= cos 2x - sin 2x <= sqrt(1+1)`

` = -sqrt(2) <= cos 2x - sin 2x <= sqrt(2)`

So, the range of `f(x)` is `[- sqrt(2), sqrt(2)]`.
Correct Answer is `=>` (C) `[- sqrt(2), sqrt(2)]`
Q 2378234106

If `f(x) = 2x + 7` and `g(x) = x^ 2 + 7 , x in R`, then
which value of `x` will satisfy `fog (x) = 25` ?
NDA Paper 1 2010
(A)

`-1` ana `1`

(B)

`-2` and `2`

(C)

`-sqrt2` and `sqrt 2`

(D)

None of these

Solution:

`f(x) = 2x + 7` and `g(x) = x^2 + 7`

`:. fog(x) = f{g(x)} = f(x^ 2 + 7)`

`= 2 (x^2 + 7) + 7 = 2x ^2 + 14 + 7 = 2x ^2 + 21`

But `fog(x) = 25` (given)

`=> 2x 2 + 21 = 25 => x^ 2 = 2`

`x=±sqrt 2`
Correct Answer is `=>` (C) `-sqrt2` and `sqrt 2`
Q 2318534409

If `f(x) = 2/3 x+3/2, x in R` then what is `f^(-1)(x)` equal to ?
NDA Paper 1 2010
(A)

`3/2 x+2/3`

(B)

`3/2 x-9/2`

(C)

`2/3 x-4/9`

(D)

`2/3 x -2/3`

Solution:

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

`=> 4x+9=6y`

`=> x=(6y-9)/4=f^(-1)(y)`

`:. f^(-1)(x)=(6x-9)/4=(3x)/2-9/4`
Correct Answer is `=>` (B) `3/2 x-9/2`
Q 2388634507

If `f : R -> R, g : R -> R` and `g(x) = x + 3` and
`(fog) (x) = (x + 3)^2` , then what is the value of `f(-3)` ?
NDA Paper 1 2010
(A)

`-9`

(B)

`0`

(C)

`9`

(D)

`3`

Solution:

`fog(x) = (x + 3)^2 = f(g(x))`

and `g(x) = x + 3`

`=> f(x + 3) = (x + 3)^2`

`=> f(x) = x^ 2`

`=> f(-3) = (-3)^2 = 9`
Correct Answer is `=>` (C) `9`
Q 2328734601

Consider the following statements
I. Every function has a primitive.
II. A primitive of a function is unique.
Which of the statements given above is/are connect?
NDA Paper 1 2010
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Every function doesn't have a primitive and primitive
of a function is unique, by definition of function.

Primitive of a function is anti derivative of the function.
Correct Answer is `=>` (D) Neither I nor II
Q 2345556463

Consider the function `f : R -> {0, 1}` such that
`f(x) = { tt ((1, text( if x is rational)),(0, text(if x is irrational)))`

Which one of the following is correct?
NDA Paper 1 2010
(A)

The function is one-one into

(B)

The function is many-one into

(C)

The function is one-one onto

(D)

The function is many-one onto

Solution:

Since, on taking a straight line parallel to x-axis, the

group of given function intersect it at many points.

`:. f(x)` is many one.

and as range of `f(x)` = Co-domain

So, `f(x)` is onto.

Hence, `f(x)` is many one onto.
Correct Answer is `=>` (C) The function is one-one onto
Q 2385367267

The function `f(x) = e^x, x in R` is
NDA Paper 1 2010
(A)

onto but not one-one

(B)

one-one onto

(C)

one-one but not onto

(D)

Neither one-one nor onto

Solution:

It is clear from the graph that `f(x) = e^x, AA x in R` is

one-one but not onto. Since range `!=` Codomain, so `f(x)` is into.
Correct Answer is `=>` (C) one-one but not onto
Q 2355067864

The function `f(x) = x/(x^2 + 1)` from `R` to `R` is
NDA Paper 1 2010
(A)

one-one as well as onto

(B)

onto but not one-one

(C)

Neither one-one nor onto

(D)

one-one but not onto

Solution:

Given function, `f(x) = x/(x^2 + 1)`

For the function `f(x)` is one-one

`f(x_1) = f(x_2)`

`=> x_1/(x_1^2 + 1) = x_2/(x_2^2 + 1)`

`=> x_1x_2^2 + x_1 = x_1^2 x_2 + x_2`

`=> x_1x_2 (x_1 - x_2) - (x_1 - x_2 ) = 0`

`=> (x_1 - x_2 )(x_1x_2 - 1) = 0`

`=> x_1 = x_2` and `x_1x_2 != 1`

So, the function is one-one.

Let `y = x/(x^2 + 1) => yx^2 + y = x`

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

`=> x = (1 ± sqrt(1 - 4y^2))/(2y) 1 - 4y^2 >= 0`

` (1 - 2y) (1 + 2y) >= 0`

`(2y - i) (2y + i) <= 0`

`y in [- 1/2 , 1/2] ~ { 0}`

Here, the range of `f(x) = [- 1/2 , 1/2 ] ~ {0}` and codomain `= R`

=> Range `!=` Codomain

So, `f(x)` is not onto.

Hence, `f(x)` is one-one but not onto.
Correct Answer is `=>` (D) one-one but not onto
Q 2358734604

Let `f: R -> R` be defined by `f(x) =| x |//x, x ne 0,`
`f(0) = 2`, then what is range of `f` ?
NDA Paper 1 2009
(A)

`{1,2}`

(B)

`{1,-1}`

(C)

`{-1,1,2}`

(D)

`{1}`

Solution:

Given, `f(x) = {tt(((|x|)/x, x ne 0), (2, x=0))`

Now, redefine the given function

`={tt((1, x > 0),(2, x=0),(-1, x <0))`

So, the range of `f(x)` is `{ -1, 1, 2}`.
Correct Answer is `=>` (C) `{-1,1,2}`
Q 2325778661

A mapping `f : R -> R` which is defined as
`f(x) =cos x ; x in R` is
NDA Paper 1 2009
(A)

one-one only

(B)

onto only

(C)

one-one onto

(D)

Neither one-one nor onto

Solution:

Given, `f(x) = cos x`

It is clear from the figure that `f(x)` is neither one-one nor a onto

function.

Since, whenever we drawn a line parallel to `X`-axis, then it

intersects at infinite points to the curve. So, f(x) is not one-one.

and range of `f(x) = [-1, 1]`,

Codomain of `f(x) = R`

Range of `f(x) !=` Codomain of `f(x)`

So, `f(x)` is not onto.
Correct Answer is `=>` (D) Neither one-one nor onto
Q 2355280164

Which one of the following functions `f : R -> R` is
injective?
NDA Paper 1 2009
(A)

`f(x) = | x | , AA x in R`

(B)

`f(x) = x^2, AA x in R`

(C)

`f(x) = 11, AA x in R`

(D)

`f(x) = - x , AA x in R`

Solution:

An injective function means one-one.

In option `(d), f(x) = - x`.

For every values of `x`, we get a different value of `f `.

Hence, it is injective.
Correct Answer is `=>` (D) `f(x) = - x , AA x in R`
Q 2335880762

Let `f : R -> R` be a function defined as
`f(x) = x | x | `; for each `x in R, R` being the set of real
numbers. Which one of the following is correct?
NDA Paper 1 2009
(A)

f is one-one but not onto

(B)

f is onto but not one-one

(C)

f is both one-one and onto

(D)

f is neither one-one nor onto

Solution:

` ∵ f(x) = x | x |`

If `f(x_1) = f(x_2)`

`=> x_1 | x_1 | = x_2 | x_2|`

`=> x_1 = x_2`

So, `f(x)` is one-one.

Also, range of `f(x)` = codomain of `f(x)`

So, `f(x)` is onto.

Hence, `f(x)` is both one-one and onto.
Correct Answer is `=>` (C) f is both one-one and onto
Q 2378734606

Function 'j' defined by `f(x) = x +1/x` , then consider
the following statements,

I. `{f(x)}^2=f(x^2)+2`
II. `{f(x)}^3=f(x^3)+3f(x)`

Which of the correct in above statements?
NDA Paper 1 2008
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

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

`=(x+1/x)^2={f(x)}^2`

and `f(x^ 3) + 3f(x) = x^3+1/x^3+3(x+1/x)`

`=(x+1/x)^2={f(x)}^3`
Correct Answer is `=>` (C) Both I and II
Q 2358145004

If `f(x) = x` and `g(x) =|x |` then what is `(f + g) (x)`
equal to?
NDA Paper 1 2008
(A)

`0 , AA x in R`

(B)

`2x, AA x in R`

(C)

`{tt((2x, for x ge 0),(0, for x < 0))`

(D)

`{tt((0, for x ge 0),(2x, for x < 0))`

Solution:

`f(x)=x` and `g(x)=|x|`

`:. (f+g((x)=f(x)+g(x)`

`=x+|x|`

`={tt((x+x, fo r x ge 0),(x-x, fo r x < 0))`

(by redefining the function)

`= {tt((2x, for x ge 0),(0, for x < 0))`
Correct Answer is `=>` (C) `{tt((2x, for x ge 0),(0, for x < 0))`
Q 2318245100

If `g(x) =sin x, x in R` and `f(x)=1/(sin x), x in (0, pi/2)`

then what is `(gof)` `(x)` equal to?
NDA Paper 1 2008
(A)

`1`

(B)

`1/(sin(sin x))`

(C)

`1/(sin^2 x)`

(D)

`sin(1/(sin x))`

Solution:

`g(x)=sin x` and `f(x)=1/(sin x)`

`(gof)(x) = g[f(x)]= sin f(x)= sin(1/(sin x))`
Correct Answer is `=>` (D) `sin(1/(sin x))`
 
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