Mathematics previous year question of AOD for NDA

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previous year question of AOD for NDA

Q 2703680548

What is the length of the longest interval in which the function

`f(x) = 3 sin x - 4 sin^3 x` is increasing?
NDA Paper 1 2017

(A)

`pi/3`

(B)

`pi/2`

(C)

`(3 pi)/2`

(D)

`pi`

Solution:

`f(x) = sin 3x`

Longest interval `=pi/6 + pi/6`

`=pi/3`

Correct Answer is `=>` (A) `pi/3`

Q 2743780643

What is the maximum value of the function `f(x) = 4 sin ^2 x + 1` ?
NDA Paper 1 2017

(A)

`5`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

`f(x) = 4 sin^2 x+1`

`-1 le sin x le 1`

`sin^2 x le 1`

`4 sin^2 x le 4`

`4 sin^2 x +1 le 5`

`max[(f(x)] =5`

Correct Answer is `=>` (A) `5`

Q 2713691549

What is the maximum area of a triangle that can be inscribed in a circle of radius `a` ?
NDA Paper 1 2017

(A)

`(3a^2)/4`

(B)

`a^2/2`

(C)

`(3 sqrt 3 a^2)/4`

(D)

`(sqrt 3 a^2)/4`

Solution:

the maximum area of a triangle that can be inscribed in a circle of radius `a`, triangle should be equilateral.

So area of equilateral triangle `A= (sqrt 3)/4 a^2`

Correct Answer is `=>` (D) `(sqrt 3 a^2)/4`

Q 2733791642

Let `f(x) = x+ 1/x` where `x in (0, 1)` . Then which one of the following is correct?
NDA Paper 1 2017

(A)

`f(x)` fluctuates in the interval

(B)

`f(x)` increases in the interval

(C)

`f(x) ` decreases in the interval

(D)

None of the above

Solution:

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

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

`x in (0,1) => x^2 in (0,1) => 1/(x^2) > 1`

` => f' (x) < 0 `

` => f (x)` is decreasing

Correct Answer is `=>` (C) `f(x) ` decreases in the interval

Q 2176201176

Consider the function
`f(x) = | x - 1| + x^(2)` where `x in R`.

Which one of the following statements is correct.
NDA Paper 1 2016

(A)

`f(x)` is increasing in `(- oo , 1/2)` and decreasing in `(1/2 ,oo) `

(B)

`f(x)` is decreasing in `(- oo , 1/2)` and increasing in `(1/2 ,oo) `

(C)

`f(x)` is increasing in `(- oo, 1/2)` and dicreasing in `(1, oo)`

(D)

`f(x)` is decreasing in `(- oo, 1)` and increasing in `(1, oo)`

Solution:

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

`:. f(x) = { tt (( x^2 + (x - 1) , text(for x > 1) ), ( x^2 - x + 1 ,text(for x < 1)) )`

From the graph between `(- oo, 1/2)`.

`f(x)` decreases when x increases.

`:. f(x)` is decreasing in `(- oo, 1/2)`

and from the graph between `(1/2 - oo),`

`f(x)` increases when x increases

`:. f(x)` is increasing in `(1/2 - oo),`

Correct Answer is `=>` (B) `f(x)` is decreasing in `(- oo , 1/2)` and increasing in `(1/2 ,oo) `

Q 2176401376

Consider the function
`f(x) = | x - 1| + x^(2)` where `x in R`.

Which one of the following statements is correct?
NDA Paper 1 2016

(A)

`f(x)` has local minima at more than one point in `(-oo , oo)`

(B)

`f(x)` has local maxima at more than one point in `(-oo , oo)`

(C)

`f(x)` has local minima at one point only in`(-oo , oo)`

(D)

`f(x)` has neither maxima nor minima in `(-oo , oo)`

Solution:

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

`:. f(x) = { tt (( x^2 + (x - 1) , text(for x > 1) ), ( x^2 - x + 1 ,text(for x < 1)) )`

From the graph there is only one minimum point which

is at point `(1/2 - 3/4).`

`[∵` when `x =1/2` then `y =| 1/2 - 1| + 1/4 = 3/4 ]`

`:. f(x)` has local maxima at one point only in `(-oo , oo)`

Correct Answer is `=>` (C) `f(x)` has local minima at one point only in`(-oo , oo)`

Q 2116512470

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

What is the number of points of local minima of the
function `f( x )?`

NDA Paper 1 2016

(A)

None

(B)

One

(C)

Two

(D)

Three

Solution:

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

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

`f'(x) = (x- 1 )^(2) d/dx [(x + 1) (x- 2)^(3)]`

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

`= (x- 1)^(2) [(x + 1) 3(x- 2)^(2) + (x- 2)^(3)]`

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

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

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

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

`+ 2(x + 1) (x - 2)]`

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

`+ 2x^(2) -2x-4`

Put ` f'(x) = 0 => x = 2 , 1 , (5 - sqrt(25 +120) )/(12) , (5 + sqrt(25 +120) )/(12)`

`=> x = 2 , 1 , (5- sqrt(145) )/12 , (5+ sqrt(145) )/12`

Now, `f''(x) = 2(x- 2) (3x- 5) 5x^(2) - 5x -1)`

For `x = 2, f''(x) = 0` For `x = 1 , f'' (x) < 0`

Hence, at `x = 1` there is local maxima.

For `x = (5 pm sqrt(145))/12 , f''(x) < 0`

Hence, at `x = (5 pm sqrt(145))/12` , there is loca1 minima .

Correct Answer is `=>` (C) Two

Q 2146512473

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

What is the number of points of local maxima of the
function `f( x) ?`
NDA Paper 1 2016

(A)

None

(B)

One

(C)

Two

(D)

Three

Solution:

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

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

`f'(x) = (x- 1 )^(2) d/dx [(x + 1) (x- 2)^(3)]

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

`= (x- 1)^(2) [(x + 1) 3(x- 2)^(2) + (x- 2)^(3)]`

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

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

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

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

`+ 2(x + 1) (x - 2)]`

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

`+ 2x^(2) -2x-4`

Put ` f'(x) = 0 => x = 2 , 1 , (5 - sqrt(25 +120) )/(12) , (5 + sqrt(25 +120) )/(12)`

`=> x = 2 , 1 , (5- sqrt(145) )/12 , (5+ sqrt(145) )/12`

Now, `f''(x) = 2(x- 2) (3x- 5) 5x^(2) - 5x -1)`

For `x = 2, f''(x) = 0` For `x = 1 , f'' (x) < 0`

Hence, at `x = 1` there is local maxima.

For `x = (5 pm sqrt(145))/12 , f''(x) < 0`

Hence, at `x = (5 pm sqrt(145))/12` , there is loca1 minima.

Correct Answer is `=>` (B) One

Q 2146678573

Consider the function `f(theta) = 4(sin^(2) theta + cos^(4) theta)`

What is the maximum value of the function `f(theta)?`
NDA Paper 1 2016

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

`f(theta) = 4(sin^( 2) theta + cos^( 4) theta)`

Also, `1-2 sin ^(2) theta =cos( 2)theta`

` => sin^( 2) theta = (1- cos(2)theta)/2` .........(i)

and `2 cos^(2) theta -1 =cos( 2) theta`

` => cos^( 2) theta = (1+ cos(2)theta)/2`

` => cos^( 4) theta = ((1+ cos2theta)/2)^2` .........(ii)

` :. f(theta) = 4(sin^( 2) theta + cos^(4) theta)`

` = 4 { (1- cos2theta)/2 + ((1+ cos2theta)/2)^2}`

`= 4 { (1- cos2theta)/2 + (1+ cos^(2)2theta + 2cos2theta)/4}`

` = 4{ (2 - 2 cos 2theta + 1 + cos ^(2) 2 theta + 2cos 2 theta)/4}`

`= 4 {( 3 + cos^( 2) 2theta)/4} = 3 + cos ^(2) 2theta`

`:. f(theta)_(max) = 3 + 1 = 4 `

Correct Answer is `=>` (D) `4`

Q 2156678574

Consider the function `f(theta) = 4(sin^(2) theta + cos^(4) theta)`

What is the minimum value of the function `f(theta)?`
NDA Paper 1 2016

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Correct Answer is `=>` (D) `3`

Q 2781380227

Which one of the following statements is correct in respect of the function `f(x) = x^3 sinx ?`
NDA Paper 1 2016

(A)

It has local maximum at x = 0.

(B)

It has local minimum at x = 0.

(C)

It has neither maximum nor minimum at x= 0

(D)

It has maximum value 1.

Solution:

`f(x) = x^3 sinx`

`f'(x) = x^3 cosx +sinx (3x^2)`

For maxima & minima

`f'(x) = 0`

`x^2 (x cosx+3sinx) = 0`

` x = 0` or `x cosx +3sinx = 0`

`f''(x) = x^3 (- sinx) +cosx (3x^2)+ sinx 6x +3x^2 cosx`

For checking maxima & minima at `x = 0`

`f''(0) = 0`

so `f'''(0) = 0`

`f''''(x) > 0` on `( x = 0)`

So It has minimum at `x = 0`

Correct Answer is `=>` (B) It has local minimum at x = 0.

Q 2711380220

`f(x) = { tt ((3x^2+12x-1 , -1 le x le 2 ) , (37 -x , 2 < x le 3))`
Which of the following statements is / are correct ?

1. `f(x)` is increasing in the interval `[-1 , 2]`.

2. `f(x)` is decreasing in the interval `{ 2 , 3]`

Select the correct answer using the code given below:
NDA Paper 1 2016

(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given,
`f(x) = { tt ((3x^2+12x-1 , -1 le x le 2 ) , (37 -x , 2 < x le 3))`

1. `x in [-1 , 2]`

`f'(x) = 6x+12 > 0`

`f(-1) = 3-12-1 = -10`

`f(2) = 12+24-1 = 35`

`f(-1) > f(2)` means increasing in interval `[-1 , 2]`

2. For `2 < x < 3`

`f'(x) = -1 < 0`

`f(2) = 35`

`f(3) = 34`

`f(2) > f(3)`

Decreasing in `2 < x le 3`

Correct Answer is `=>` (C) Both 1 and 2

Q 2711380220

`f(x) = { tt ((3x^2+12x-1 , -1 le x le 2 ) , (37 -x , 2 < x le 3))`
Which of the following statements is / are correct ?

1. `f(x)` is continuous at x = 2.

2. `f(x)` attains greatest value at x = 2.

3. `f(x)` is differential at x = 2.
Select the correct answer using the code given below:
NDA Paper 1 2016

(A)

1 and 2 only

(B)

2 and 3 only

(C)

1 and 3 only

(D)

1 , 2 and 3

Solution:

1. Continuity at `x = 2`

`lim_(x&;0) f(2-h)`

` = lim_(h→0) 3(2-h)^2+12(2-h)-1`

` = 12+24-1 = 35`

`lim_(x→2^+) = lim_(h→0) f(2+h)`

` = lim_(h →0) [37-(2+h)]`

` = 35`

`lim_(x→2) f(x) = lim_(x→2) 3x^2+12x-1 = 35`

2. `f(x) given greatest value of `x` as

`f(x)` is increasing

`f(2) = 35`

3. RHD `lim_(h→0) (f(2+h)-f(2))/h = lim_(h→0) (37 - (2+h)-35)/h`

`lim_(h→0) (35-h-35)/h = -1`

LHD `lim_(h→0) (f(2-h)-f(2))/h = lim_(h→0) (3(2-h)^2+12(2-h)-1)-(3(2)^2+12*2-1))/h`

`RHD ne LHD`

`f(x)` is not differentiable at `x = 2`

Hence 3 is incorrect

Correct Answer is `=>` (A) 1 and 2 only

Q 2721280121

Let `f(x) = [ |x| - | x- 1 |]^2`
What is `f'(x)` equal to when `x > 1 ?`
NDA Paper 1 2016

(A)

`0`

(B)

`2x-1`

(C)

`4x-2`

(D)

`8x-4`

Solution:

When `x > 1`

`x - 1 > 0`

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

`f'(x) = 0`

Correct Answer is `=>` (A) `0`

Q 2721280121

Let `f(x) = [ |x| - | x- 1 |]^2`
Which of the following equations is /are correct ?

1. `f(-2) = f(5)`

2. `f''(-2)+f''(0.5)+f''(3) = 4`

Select the correct answer using the code given below .

NDA Paper 1 2016

(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Correct Answer is `=>` (D) Neither 1 nor 2

Q 2721280121

Let `f(x) = [ |x| - | x- 1 |]^2`
What is `f'(x)` equal to when `0< x < 1 ?`
NDA Paper 1 2016

(A)

`0`

(B)

`2x-1`

(C)

`4x-2`

(D)

`8x-4`

Solution:

`0 < x < 1`

`x > 0`

`x-1 < 0`

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

`f'(x) = 2(2x-1) * 2 = 8x-4`

Correct Answer is `=>` (D) `8x-4`

Q 2721067821

Let `f(x) = { tt (((e^x-1)/x , x > 0 ) , ( 0 , x = 0))` be a real valued function
Which one of the following statements is correct ?
NDA Paper 1 2016

(A)

`f(x)` is a strictly decreasing function in ( 0 , x)

(B)

`f(x)` is a strictly increasing function in ( 0 , x)

(C)

`f(x)` is neither increasing nor decreasing function in ( 0 , x)

(D)

`f(x)` is not decreasing in ( 0 , x)

Solution:

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

`f'(x) = { tt (((x* e^x+ e^x -1)/(x^2) , x > 0 ) , ( 0 , x = 0))`

Here, `f'(x)>0` for `(x > 0)`

So we can say f(x) is a strictly increasing function in `(0,x)`

Correct Answer is `=>` (B) `f(x)` is a strictly increasing function in ( 0 , x)

Q 2721067821

Let `f(x) = { tt (((e^x-1)/x , x > 0 ) , ( 0 , x = 0))` be a real valued function
Which one of the following statements is/are correct ?

1. `f(x)` is right continuous at x= 0

2. `f(x)` is discontinuous at x= 1.

NDA Paper 1 2016

(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Correct Answer is `=>` (B) 2 only

Q 1628680501

Consider the following statements

I. `y = (e^x +e^(-x))/2 `is an increasing function on `[0, oo )`.

II. `y = (e^x - e^(-x))/2 ` is an increasing function on `( -oo, oo )`.

Which of the above statements is/are correct?
NDA Paper 1 2015

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Let `f(x) =(e^x +e^(-x))/2 => f'(x) = (e^x - e^(-x))/2 `

` = 1/2 ( e^x - 1/e^x ) = 1/2 ( (e^(2x) -1) /e^x) ` ......(1)

Now, for` x >= 0`, we have `2x >= 0`

`= e^(2x) >= e^0 (:.e^x` is an increasing function) = `e^(2x) >= 1`

also for `x >= 0 => e^x >= 1`

:. From Eq. (i), we have `f'(x) =1/2( (e^(2x) -1) /e^x) >= 0`

So, `f(x)` is increasing function on `(0, oo)`.

II. Let `g(x) = (e^(x) - e^(-x) )/2`

`=> g'(x) = (e^(x) - e^(-x) )/2 > 0`

[:.`e^x` and `e^(-x)` both are greater than zero in `(-oo, oo)]`

So,` g(x)` is an increasing function on `(-oo, oo)`.

Hence, both the statements are true.

Correct Answer is `=>` (C) Both I and II

Q 1679601516

Consider the function `f(x) = (x^2 -1)/(x^2 +1)` where `x in R`.

At what value of `x` does `f(x)` attain minimum
value?
NDA Paper 1 2015

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

We have, `f(x) = (x^2 -1)/(x^2 +1)`

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

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

For critical points, put `f'(x) = 0`

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

`:. x = 0`

Correct Answer is `=>` (B) `0`

Q 1629701611

Consider the function `f(x) = (x^2 -1)/(x^2 +1)` where `x in R`.

What is the minimum value of `f(x)?`
NDA Paper 1 2015

(A)

`0`

(B)

`1/2`

(C)

`-1`

(D)

`2`

Solution:

Thus, `x = 0` is the only critical point which could

possible be the point of minima.

Note that for values close to `0` and to the right of `0`,

`f'(x) > 0` and for values close to 0 and to the left of `0`,

`f'(x) < 0.`

Therefore, by first derivative test, `x = 1` is a point of

minima and the minimum value of `f(x )` is given by

`f(0) = (0 - 1)/(0 + 1) = - 1`

Correct Answer is `=>` (C) `-1`

Q 2211378229

Consider the following statements
1. The function `f( x) = x ^2 + 2 cos x` is increasing in the
interval `(0, pi)`.
2. The function `f( x) =` In `( sqrt(1+x^2) - x)` is decreasing in
the interval `(- oo, oo )`.

Which of the above statements is/are correct?
NDA Paper 1 2015

(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and ` 2`

(D)

Neither `1` nor `2`

Solution:

1. We have, `f(x) = x^2 + 2 cos x`

On differentiating, we get `f' (x) = 2x - 2 sin x`

Here, `f'(x) > 0` in the interval `(0, pi)`.

`:. f(x)` is increasing function.

2. We have, `f(x) = ln (sqrt(1+x^2)-x)`

On differentiating, we get

`f'(x) = 1/(sqrt(1+x^2)-x) xx ( (x)/(sqrt(1+x^2)) - 1)`

Here, `f' (x) < 0` in the interval `(- oo , oo )`.

`:. f(x)` is decreasing function.

Correct Answer is `=>` (C) Both `1` and ` 2`

Q 2241678523

The function `f(x) = x^2/e^x` is monotonically increasing, if
NDA Paper 1 2015

(A)

`x < 0` only

(B)

`x > 2` only

(C)

`0 < x < 2`

(D)

`x in (- oo,0) cup (2,oo)`

Solution:

We have, `f(x) = x ^2 e^(-x)`

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

`=e^(-x) x(-x^2 + 2x) = e^(-x) (2-x)x`

Since, `f(x)` is monotonically increasing.

`:. f'(x) > 0`

`=> e^(-x) x(2- x) > 0` or `x(x - 2) < 0`

Hence, `0 < x < 2`

Correct Answer is `=>` (C) `0 < x < 2`

Q 1609001818

Consider the function
`f(x) = 0.75x^4 - x^3 - 9x^2 + 7`

What is the maximum value of the function?
NDA Paper 1 2015

(A)

`1`

(B)

`3`

(C)

`7`

(D)

`9`

Solution:

`f(x) = 0.75x^4 - x^3- 9x^2 + 7`

`=> f'(x) = 3x^3 - 3x^2 - 18x`

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

For critical points, put `f'(x) = 0`

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

`=> x = 0, x = 3` or `x = - 2`

Thus, there are only three critical points which could

possible be the point of local maxima or local minima.

Now, the sign of `f'(x)` is given by

Clearly, `-2` and `3` are the point of local minima and `0`

is the point of local maxima.

`:.` The maximum value of the function is
given by

`f(0) = 7`

Correct Answer is `=>` (C) `7`

Q 1619101910

Consider the function
`f(x) = 0.75x^4 - x^3 - 9x^2 + 7`

Consider the following statements
I. The function attains local minima at `x = -2` and
`x =3`.
II. The function increases in the interval `( -2, 0)`.

Which of the above statements is/are correct?
NDA Paper 1 2015

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Correct Answer is `=>` (C) Both I and II

Q 2211180020

Consider the following statements
1. `f ( x) = ln x` is an increasing function on `( 0, oo )`.
2. `f(x) = e^x - x(ln x)` is an increasing function on `(1, oo)`.

Which of the above statements is/are correct?
NDA Paper 1 2015

(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

1. We have, `f(x) = ln x`

` :. f'(x) = 1/x`

Here, `f' (x) > 0` in the interval `(0, oo )`.

`:. f(x)` is increasing function in `(0, oo )`.

2. We have, `f(x) = e^x - x(ln x)`

`:. f'(x) = e^x - ln x- x . 1/x = e^x -ln x -1`

Here, `f' (x) > 0` in the interval `(1, oo )`.

`:. f(x)` is increasing function in `(1, oo )`.

Correct Answer is `=>` (C) Both `1` and `2`

Q 2261680525

Consider the function `f(x) = (1/x)^(2x^2)` where `x > 0`.

At what value of `x` does the function attain maximum value?
NDA Paper 1 2015

(A)

`e`

(B)

`sqrt(e)`

(C)

`1/sqrt(e)`

(D)

`1/e`

Solution:

We have,`f(x) = (1/x)^(2x^2) = x^ (-2x^2) = e ^ (-2x^2) log x`

` :. f' (x) = e^ (-2x^2) log x [ -2x^2 . 1/x + log x (-4x)]`

` => f'(x) = f(x)(-2x-4x log x)`

For maxima or minima, put `f'(x) = 0`

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

` => x = 0 ` or ` log x = -1/2 => x = e ^(-1/2) = 1/ sqrt(e)`

Now, `y '' (x) = f' '(x)=(-2x-4x logx)`

`f''(x)=( -2-4x 1/x -4logx)`

`= - 6 - 4 log x`

`:. y '' (x)` is negative at `x = 1/sqrt(e)`

Hence, `x = 1/sqrt(e)` is the point of maxima.

Correct Answer is `=>` (C) `1/sqrt(e)`

Q 2231880722

Consider the function `f(x) = (1/x)^(2x^2)`, where ` x > 0`.

The maximum value of the function is

NDA Paper 1 2015

(A)

`e`

(B)

`e ^(2/e)`

(C)

`e ^(1/e)`

(D)

`1/e`

Solution:

Given, `y = (1/x)^(2x^2) , x > 0`

Taking log on both sides, we get

`log y = 2x^2 log(1/x) , x > 0`

`=> log y = - 2x^2 log x, x > 0`

`=>( y')/y = - ( 4x log x + (2x^2)/x)`

`=> (y')/y =- (4x log x+ 2x)`

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

`=> y' = - xy (2log x + 1)`

For maximum or minimum, put `y' = 0`

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

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

`=> 2log x + 1 = 0`

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

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

Now, `y ' = - [ xy (2/x) + (2log x + 1) (y + xy')]`

`=> y'' = - [2y + (2log x + 1) (y + xy')]`

Now, `(y'') _(x = e ^(-1//2)) = -2y = -2 ( 1/e^(-1//2))^(2//e)`

`= - 2 ( e ^(1//2) )^(2//e) < 0`

`:. f(x)` is maximum when `x = e^(- 1// 2)`

Maximum value of the function

` = ( 1/ ( 1 //sqrt(e)))^(2)(1/sqrt(e))^2`

` = (sqrt(e))^(2//e) = [(sqrt(e))^2] ^(1//e) = (e)^(1//e)`

Correct Answer is `=>` (C) `e ^(1/e)`

set 2

Q 2221191021

Consider the function
`f(x) = - 2x^3 - 9x^2 - 12x + 1`

The function `f(x)` is an increasing function in the interval
NDA Paper 1 2015

(A)

`(- 2, -1)`

(B)

`(-oo,- 2)`

(C)

`(- 1, 2)`

(D)

`(-1, oo)`

Solution:

Given, `f(x) = - 2x^3- 9x^2- 12x + 1`

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

`f'(x) =- 6x^2 -18x -12`

For `f(x)` to be increasing function, ` f'(x) > 0`

`:. - 6x^2 - 18x - 12 > 0`

`=> x^2 +3x + 2 < 0 => (x+2) (x+1) < 0`

`:. - 2 < x < - 1 `

Correct Answer is `=>` (A) `(- 2, -1)`

Q 2211191029

Consider the function
`f(x) = - 2x^3 - 9x^2 - 12x + 1`

The function `f( x)` is a decreasing function in the interval
NDA Paper 1 2015

(A)

`(- 2,- 1)`

(B)

`(- oo,- 2)`

(C)

`(-1,oo)`

(D)

`(- oo,- 2) cup (- 1, oo)`

Solution:

Given, `f(x) = -2x^3 - 9x^2 -12x + 1`

`=> f'(x) =- 6x^2 -18x -12`

For `f(x)` to be decreasing function,

` f'(x) < 0`

`:. - 6x^2 - 18x - 12 < 0`

`=> x^2 +3x +2 > 0`

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

`:. x in (- oo,- 2) cup (- 1, oo)`

Correct Answer is `=>` (D) `(- oo,- 2) cup (- 1, oo)`

Q 1762191035

Consider the function
` f(x) = (x^2 - x + 1)/(x^2 + x + 1)`

What is the maximum value of the function?

NDA Paper 1 2014

(A)

`1/2`

(B)

`1/3`

(C)

`2`

(D)

`3`

Solution:

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

Now,` f' (x) = ([ (x^2 + x + 1) d/(dx) (x^2 - x + 1) - (x^2 - x + 1) d/(dx) (x^2 + x + 1) ])/(x^2 + x + 1)^2`

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

` => f' (x) = [ (2x^3 + 2x^2 + 2x - x^2 - x - 1 - 2x^3 + 2x^2 - 2x - x^2 + x -1])/(x^2 + x + 1)^2`

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

For maximum or minimum value of `f(x)`,

put `f' (x) = 0`

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

`:. x = pm 1`

Again ,

` f '' (x) = ((x^2 + x + 1)(4x)- (2x^2 - 2) 2(x^2 + x + 1)(2x + 1))/(x^2 + x + 1)^4`

` = ( 4(x^2 + x + 1)[x- (x^2 - 1) (2x + 1)])/(x^2 + x + 1)^4`

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

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

At ` x = 1`

` f '' (x) = ( 4(-2 -1 + 3+ 1))/(1 + 1 + 1)^3 = 4/(27) > (min)`

So, function `f(x)` is minimum at `x = 1`.

At ` x = -1`

` f' (-1) = ( 4(2- 1- 3 + 1))/((1 - 1 + 1)3)`

` = ( 4 xx (-1))/(1)^3 = - 4 < 0 (max)`

So, function `f(x)` is maximum at `x = -1`.

Now, maximum value of the function.

At `x = -1, f(-1) = ( 1 + 1 + 1)/(1 - 1 + 1) = 3`

Correct Answer is `=>` (D) `3`

Q 1702291138

Consider the function
` f(x) = (x^2 - x + 1)/(x^2 + x + 1)`

What is the minimum value of the function?

NDA Paper 1 2014

(A)

`1//2`

(B)

`1//3`

(C)

`2`

(D)

`3`

Solution:

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

Now,` f' (x) = ([ (x^2 + x + 1) d/(dx) (x^2 - x + 1) - (x^2 - x + 1) d/(dx) (x^2 + x + 1) ])/(x^2 + x + 1)^2`

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

` => f' (x) = [ (2x^3 + 2x^2 + 2x - x^2 - x - 1 - 2x^3 + 2x^2 - 2x - x^2 + x -1])/(x^2 + x + 1)^2`

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

For maximum or minimum value of `f(x)`,

put `f' (x) = 0`

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

`:. x = pm 1`

Again ,

` f '' (x) = ((x^2 + x + 1)(4x)- (2x^2 - 2) 2(x^2 + x + 1)(2x + 1))/(x^2 + x + 1)^4`

` = ( 4(x^2 + x + 1)[x- (x^2 - 1) (2x + 1)])/(x^2 + x + 1)^4`

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

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

At ` x = 1`

` f '' (x) = ( 4(-2 -1 + 3+ 1))/(1 + 1 + 1)^3 = 4/(27) > (min)`

So, function `f(x)` is minimum at `x = 1`.

At ` x = -1`

` f' (-1) = ( 4(2- 1- 3 + 1))/((1 - 1 + 1)3)`

` = ( 4 xx (-1))/(1)^3 = - 4 < 0 (max)`

So, function `f(x)` is maximum at `x = -1`.

Now, minimum 'value of the function

At `x = 1, f(1) = ( 1 - 1 + 1)/(1 + 1 + 1) = 1/3`

Correct Answer is `=>` (B) `1//3`

Q 1713812740

What is the slope of the tangent to the curve
`y = sin^(-1) (sin^2 x)` at `x = 0`?
NDA Paper 1 2014

(A)

`0`

(B)

`1`

(C)

`2`

(D)

None of these

Solution:

Given curve,

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

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

` (dy)/(dx) = 1/sqrt(1-(sin^2 x)^2) . d/(dx) (sin^2x)`

` => (dy)/(dx) = (2 sin x .cos x)/sqrt(1 - sin^4 x)`

`=> (dy)/(dx) = (sin 2x)/sqrt(1-sin^4 x)`

` ((dy)/(dx))_(x=0) = (sin 0)/sqrt(1 - sin 0)`

` = 0/sqrt(1-0) = 0`

`:.` Slope of the curve `= 0`

Hence, slope of the tangent to the given curve = Slope of

that curve `= 0` .

Correct Answer is `=>` (A) `0`

Q 1713012849

Consider the curve `y = e^(2x)`.

What is the slope of the tangent to the curve at `(0, 1)`?
NDA Paper 1 2014

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`4`

Solution:

Given curve `y = e^(2x)`

We know that,

Slope of the tangent to the given curve = Slope of the given

curve .......... (i)

Given curve, `y = e^(2x)`

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

` (dy)/(dx) = 2e^(2e)`

` => ((dy)/(dx))_(0,1) = 2.e^0 =2 xx 1`

` = 2`

Hence, slope of the given curve `= 2`

which also the slope of the tangent to the given curve at

`(0, 1)`.

Correct Answer is `=>` (C) `2`

Q 1773112946

Consider the curve `y = e^(2x).`

Where does the tangent to the curve at `(0, 1)` meet the
`X` -axis?
NDA Paper 1 2014

(A)

`(1, 0)`

(B)

`(2, 0)`

(C)

`(-1//2, 0)`

(D)

`(1//2, 0)`

Solution:

Given curve, `y = e^( 2x)`

Now, equation of tangent to the curve `y = e^(2x)` at `(0, 1)` is

` (y-1) = ((dy)/(dx))_(0,1) (x-0)`

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

`=> y -1 = 2x`

` :. y = 2x + 1`

Since, the tangent meet X-axis.

So, put `y = 0` in equation of tangent, we get

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

Hence, `A (- 1/2 , 0)` the tangent to the curve at `(0, 1)` meet

the X-axis.

Correct Answer is `=>` (C) `(-1//2, 0)`

Q 1741501423

A cylinder is inscribed in a sphere of radius `r`.

What is the height of the cylinder of maximum volume?
NDA Paper 1 2014

(A)

` (2r)/sqrt(3)`

(B)

` 1/sqrt(3)`

(C)

`2r`

(D)

` sqrt(3)r`

Solution:

Let `h` be tile height, `R` be the radius and `V` be

the volume of cylinder.

In `Delta OAB`, we have

`r^2 = R^2 + (h/ 2)^2 ..... (i)`

`(∵ OA = h/2 ` as ` Delta OAB = Delta OCD)`

Clearly, ` V = piR^2h`

`=> V(h) =pi (r^2 -h^2/2)h` [using Eq. (i))

`=> V(h) = pi (r^2h - h^3/4)`

`=> V' (h) = pi (r^2 - (3h^2)/4)` ......... (ii)

For maximum, put `V' (h) = 0`

`=> r^2 - (3h^2)/4` `= h^2 = (4r^2)/3`

` => h = (2r)/sqrt(3)`

Again, differentiating Eq. (ii) w.r.t. h, we get

` V '' (h) = pi ((-6h)/4 )`

`=> V '' ((2r)/sqrt(3) ) = pi ((-6)/4 xx (2r)/sqrt(3)) < 0`

Thus, the volume is maximum when `h = (2r)/sqrt(3)`.

Correct Answer is `=>` (A) ` (2r)/sqrt(3)`

Q 1711501429

A cylinder is inscribed in a sphere of radius `r`.

What is the radius of the cylinder of maximum volume?
NDA Paper 1 2014

(A)

` (2r)/sqrt(3)`

(B)

` sqrt(2r)/sqrt(3)`

(C)

`r`

(D)

`sqrt(3) r`

Solution:

Clearly, volume of cylinder is maximum when

` h = (2r)/sqrt(3)`.

By using the relation `r^2 = R^2 + (h/2)^2 ` we have

` R^2 = r^2 - h^2/4 = r^2 - (4r^2)/(12)`

` = (8r^2)/(12) = (2r^2)/3`

` => R = sqrt((2r^2)/3) = sqrt(2r)/sqrt(3)`

Correct Answer is `=>` (B) ` sqrt(2r)/sqrt(3)`

Q 2322091831

A rectangular box is to be made from a sheet of cutting
out identical squares of side length x from the four
corners and turning up the sides.

What is the value of x for which the volume is
maximum?
NDA Paper 1 2014

(A)

1 inch

(B)

1.5 inch

(C)

2 inch

(D)

2.5 inch

Solution:

Let V be the volume of the box.
`:. V = (24- 2x) * (9- 2x) x` `quad quad quad ` ( `because` height of box `= x` inch)

`= (216- 48x -18x + 4x2 )x`

`V(x) = 4x^3 - 66x^2 + 216x`

`=> V'(x) = 12x^2 -132x + 216`

For maximum, put `V' (x) = 0`

`=> 12x^2 -132x + 216= 0`

`=> x ^2 - 11x + 18 = 0`

`=> (x - 9) (x- 2) = 0`
`=> x = 9` or `x =2`

Now `V" (x) =24x -132`

`:. V" (9) = 216-132 = 84 > 0`

and `V" (2)=48-132= 84 < 0`

Thus, volume is maximum when `x = 2` inch.

Correct Answer is `=>` (C) 2 inch

Q 2348856703

The minimum value of the function `f(x) = |x- 4|`
exists at
NDA Paper 1 2013

(A)

`x=0`

(B)

`x=2`

(C)

`x=4`

(D)

`x=-4`

Solution:

Given function `f(x) = |x- 4|`

Graph of `f(x)`,

From graph, we observe that `f(x)` has minimum value at `x = 4.`

Correct Answer is `=>` (C) `x=4`

Q 2338156902

The function `f(x) = x^2 - 4x, x in [0, 4]` attains
minimum value at
NDA Paper 1 2013

(A)

`x=0`

(B)

`x=1`

(C)

`x=2`

(D)

`x=4`

Solution:

Function,

`f(x)= x ^2- 4x, x in [0, 4]`

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

`f'(x)= 2x- 4`

For max or min of `f(x)`,

`f'(x) =0`

`=> 2x-4=0`

`:. x=2`

Again, differentiating w.r.t. `x`, we get

`f"(x)=2 > 0` (minimum)

Hence, `f(x)` is minimum at `x= 2`.

Alternate Method

Given, `f(x)= x^ 2 - 4x , x in [0, 4]`

At `x= 0, f(0)= 0-0=0`

At `x= 1, f(1)= (1)^2- 4*1=1-4=- 3`

At `x = 2, f(2)=(2)^2- 4*2 = 4- 8= -4` (minimum)

At `x=3, f(3)=(3)^2 -4*2=9-12=-3`

At `x = 4, f(4)= (4)^2- 4*(4)=16-16= 0`

Hence, `f(x), x in [0, 4]` attains minimum value at `x= 2`.'

Correct Answer is `=>` (C) `x=2`

Q 2368167005

The curve `y = xe^x` has minimum value equal to
NDA Paper 1 2013

(A)

`-1/e`

(B)

`1/e`

(C)

`-e`

(D)

`e`

Solution:

Given curve, `y=xe^x`

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

`(dy)/(dx) =x * e^x +e^x *1= xe^x +e^x`

For max and min of `y`,

`(dy)/(dx) = 0 => e^x(x+ 1)= 0`

`=> x=-1`

Again, differentiating w.r.t. `x`, we get

`(d^2y)/(dx^2) =x*e^x+e^x*1+e^x`

`=xe^x +2e^x`

`((d ^2y)/(dx^2))_(x -1 ) = (-1)e^(-1) +2e^(-1) =-1/e >0`

So, `f(x)` have minimum value at `x = -1`.

Hence, its minimum value is

`y(-1)=(-1)e^(-1) = (-1)/e`

Correct Answer is `=>` (A) `-1/e`

Q 2318367209

Consider the following statements

I. The derivative, where the function attains
maxima or minima be zero.

II. If a function is differentiable at a point, then it
must be continouos at that point.

Which of the above statements is/are correct?
NDA Paper 1 2013

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. The derivative, where the function attains maxima or
minima must be zero.

II. If a function is differentiable at a point, then it must be
continuous at that point but if a function is continuous at a
point, then it is not necessarily that function is differentiable at
that point.

So, both statements are correct.

Correct Answer is `=>` (C) Both I and II

Q 2328823701

Which one of the following statement is correct?
NDA Paper 1 2012

(A)

`e^x` is an increasing function

(B)

`e^x` is a decreasing function

(C)

`e^x` is neither increasing nor decreasing function

(D)

`e^x` is a constant function

Solution:

Let `f(x) =e^x`

Now, differentiate w.r.t. x, we get

`f'(x)=e^x > 0, AA x in R`

So, `f(x) =e^x` is an increasing function.

Correct Answer is `=>` (A) `e^x` is an increasing function

Q 2378467306

What is the minimum value of `|x |` ?
NDA Paper 1 2012

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Let ` y =| x|`

Redefined the function,

`y= {tt((x, x ge 0),(-x, x <0))`

So from curve, we abserve that, the
minimum value of `|x |` is zero.

Correct Answer is `=>` (B) `0`

Q 2318667509

If `f(x) = xln x`, then `f(x)` attains the minimum
value at which one of the following points?
NDA Paper 1 2011

(A)

`x=e^(-2)`

(B)

`x=e`

(C)

`x=e^(-1)`

(D)

`x=2e^(-1)`

Solution:

Given, `f(x) = x log x`

`f'(x) = x* 1/x + log x * 1`

`=1 +log x`

For maximum or minimum value of `f(x )`,

`f'(x) = 0`

`=> 1 + log_e x = 0`

`=> log_e x = -1`

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

Now `f"(x) =1/x`

At `(x = e^(-1)), f"(x) = 1/(e^(-1))= e >0` (minimum)

So at ` (x = e^(-1)), f(x)` attains minimum value.

Correct Answer is `=>` (C) `x=e^(-1)`

Q 2318667500

If the rate of change in volume of spherical soap
bubble is uniform, then the rate of change of
surface area varies as
NDA Paper 1 2011

(A)

square of radius

(B)

square root of radius

(C)

inversely proportional to radius

(D)

cube of the radius

Solution:

Let the volume of spherical soap bubble is `V=4/3 pi r^3`

where `r ->` radius

`=> (dV)/(dt)=4 pi r^2 (dr)/(dt)=C` (c:onstant) (given) ... (i)

and the surface area, `S = 4pi r^2`

`=> (dS)/(dt)= 4 pi * 2 * r (dr)/(dt)`

`=> (dS)/(dt)=8 pi r xx C/(4 pi r^2)` [from Eq. (i)]

`=> (dS)/(dt)=(2C)/r`

`=> (dS)/(dt) prop 1/r`

So, the rate of change of surface area varies as inversely
proportional to radius.

Correct Answer is `=>` (C) inversely proportional to radius

Q 2358867704

The point in the interval `(0, 2pi)`, where
`f(x) =e^x sin x` has maximum slope, is
NDA Paper 1 2011

(A)

`pi/4`

(B)

`pi/2`

(C)

`pi`

(D)

`(3 pi)/2`

Solution:

Given, `f(x) =e^x sin x`

Let slope of `f(x)` is

`M = f'(x) =e^x cos x+ e^x sin x`

`=> M =e^x (sin x +cos x)`............(i)

Now, `(dM)/(dx)=e^x(sin x+cos x)+e^x(cos x-sin x)`

`=2e^x cos x`

For maximum or minimum value of slope,

`=> 2 e^x cos x=0`

`=> e^x=0` or `cos x=0`

`=> e^x=e^(-oo)` or `cos x= cos pi/2`

`=> x ne -oo` or `x= pi/2` and `(3 pi)/2` in `x in (0,2 pi)`

Correct Answer is `=>` (B) `pi/2`

Q 2368067805

What is the interval over which the function
`f(x) = 6x- x^2` ,where `x > 0` is increasing'?
NDA Paper 1 2010

(A)

`(0, 3)`

(B)

`(3, 6)`

(C)

`(6, 9)`

(D)

None of these

Solution:

We have, `f(x) = 6x - x^ 2`

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

`f'(x) = 6- 2x`

`f(x)` will be increasing function, if

`f'(x) > 0 => 6-2x > 0 => x < 3`

Thus, the required interval is `(0, 3)`.

Correct Answer is `=>` (A) `(0, 3)`

Q 2318167909

If `f` and `g` are two increasing functions such that
fog is defined, then which one of the following is
correct'?
NDA Paper 1 2010

(A)

fog is always an increasing function

(B)

fog is always a decreasing function

(C)

fog is neither an increasing nor a decreasing function

(D)

None of the above

Solution:

We know that, the composition of two increasing
functions is an increasing function by property of composition of
functions.

Hence, fog is always an increasing function.

Correct Answer is `=>` (A) fog is always an increasing function

Q 2308178008

For a point of inflection of `y = f(x)`, which one of
the following is correct'?
NDA Paper 1 2010

(A)

`(dy)/(dx)` must be zero

(B)

`(d^2y)/(dx^2)` must be zero

(C)

`(dy)/(dx)` must be non-zero

(D)

`(d^2y)/(dx^2)` must be non-zero

Solution:

For a point of inflection `(d^2y)/(dx^2)` must be equal to zero.
(by definition)

Correct Answer is `=>` (B) `(d^2y)/(dx^2)` must be zero

Q 2221191021

Consider the function
`f(x) = - 2x^3 - 9x^2 - 12x + 1`

The function `f(x)` is an increasing function in the interval
NDA Paper 1 2015

(A)

`(- 2, -1)`

(B)

`(-oo,- 2)`

(C)

`(- 1, 2)`

(D)

`(-1, oo)`

Correct Answer is `=>` (A) `(- 2, -1)`

Q 2211191029

Consider the function
`f(x) = - 2x^3 - 9x^2 - 12x + 1`

The function `f( x)` is a decreasing function in the interval
NDA Paper 1 2015

(A)

`(- 2,- 1)`

(B)

`(- oo,- 2)`

(C)

`(-1,oo)`

(D)

`(- oo,- 2) cup (- 1, oo)`

Correct Answer is `=>` (D) `(- oo,- 2) cup (- 1, oo)`

Q 1762191035

Consider the function
` f(x) = (x^2 - x + 1)/(x^2 + x + 1)`

What is the maximum value of the function?

NDA Paper 1 2014

(A)

`1/2`

(B)

`1/3`

(C)

`2`

(D)

`3`

Correct Answer is `=>` (D) `3`

Q 1702291138

Consider the function
` f(x) = (x^2 - x + 1)/(x^2 + x + 1)`

What is the minimum value of the function?

NDA Paper 1 2014

(A)

`1//2`

(B)

`1//3`

(C)

`2`

(D)

`3`

Solution:

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

Now,` f' (x) = ([ (x^2 + x + 1) d/(dx) (x^2 - x + 1) - (x^2 - x + 1) d/(dx) (x^2 + x + 1) ])/(x^2 + x + 1)^2`

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

` => f' (x) = [ (2x^3 + 2x^2 + 2x - x^2 - x - 1 - 2x^3 + 2x^2 - 2x - x^2 + x -1])/(x^2 + x + 1)^2`

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

For maximum or minimum value of `f(x)`,

put `f' (x) = 0`

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

`:. x = pm 1`

Again ,

` f '' (x) = ((x^2 + x + 1)(4x)- (2x^2 - 2) 2(x^2 + x + 1)(2x + 1))/(x^2 + x + 1)^4`

` = ( 4(x^2 + x + 1)[x- (x^2 - 1) (2x + 1)])/(x^2 + x + 1)^4`

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

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

At ` x = 1`

` f '' (x) = ( 4(-2 -1 + 3+ 1))/(1 + 1 + 1)^3 = 4/(27) > (min)`

So, function `f(x)` is minimum at `x = 1`.

At ` x = -1`

` f' (-1) = ( 4(2- 1- 3 + 1))/((1 - 1 + 1)3)`

` = ( 4 xx (-1))/(1)^3 = - 4 < 0 (max)`

So, function `f(x)` is maximum at `x = -1`.

Now, minimum 'value of the function

At `x = 1, f(1) = ( 1 - 1 + 1)/(1 + 1 + 1) = 1/3`

Correct Answer is `=>` (B) `1//3`

Q 1713812740

What is the slope of the tangent to the curve
`y = sin^(-1) (sin^2 x)` at `x = 0`?
NDA Paper 1 2014

(A)

`0`

(B)

`1`

(C)

`2`

(D)

None of these

Correct Answer is `=>` (A) `0`

Q 1713012849

Consider the curve `y = e^(2x)`.

What is the slope of the tangent to the curve at `(0, 1)`?
NDA Paper 1 2014

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`4`

Correct Answer is `=>` (C) `2`

Q 1773112946

Consider the curve `y = e^(2x).`

Where does the tangent to the curve at `(0, 1)` meet the
`X` -axis?
NDA Paper 1 2014

(A)

`(1, 0)`

(B)

`(2, 0)`

(C)

`(-1//2, 0)`

(D)

`(1//2, 0)`

Correct Answer is `=>` (C) `(-1//2, 0)`

Q 1741501423

A cylinder is inscribed in a sphere of radius `r`.

What is the height of the cylinder of maximum volume?
NDA Paper 1 2014

(A)

` (2r)/sqrt(3)`

(B)

` 1/sqrt(3)`

(C)

`2r`

(D)

` sqrt(3)r`

Correct Answer is `=>` (A) ` (2r)/sqrt(3)`

Q 1711501429

A cylinder is inscribed in a sphere of radius `r`.

What is the radius of the cylinder of maximum volume?
NDA Paper 1 2014

(A)

` (2r)/sqrt(3)`

(B)

` sqrt(2r)/sqrt(3)`

(C)

`r`

(D)

`sqrt(3) r`

Correct Answer is `=>` (B) ` sqrt(2r)/sqrt(3)`

Q 2322091831

(A)

1 inch

(B)

1.5 inch

(C)

2 inch

(D)

2.5 inch

Correct Answer is `=>` (C) 2 inch

Q 2348856703

The minimum value of the function `f(x) = |x- 4|`
exists at
NDA Paper 1 2013

(A)

`x=0`

(B)

`x=2`

(C)

`x=4`

(D)

`x=-4`

Solution:

Given function `f(x) = |x- 4|`

Graph of `f(x)`,

From graph, we observe that `f(x)` has minimum value at `x = 4.`

Correct Answer is `=>` (C) `x=4`

Q 2338156902

The function `f(x) = x^2 - 4x, x in [0, 4]` attains
minimum value at
NDA Paper 1 2013

(A)

`x=0`

(B)

`x=1`

(C)

`x=2`

(D)

`x=4`

Correct Answer is `=>` (C) `x=2`

Q 2368167005

The curve `y = xe^x` has minimum value equal to
NDA Paper 1 2013

(A)

`-1/e`

(B)

`1/e`

(C)

`-e`

(D)

`e`

Correct Answer is `=>` (A) `-1/e`

Q 2318367209

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Correct Answer is `=>` (C) Both I and II

Q 2328823701

Which one of the following statement is correct?
NDA Paper 1 2012

(A)

`e^x` is an increasing function

(B)

`e^x` is a decreasing function

(C)

`e^x` is neither increasing nor decreasing function

(D)

`e^x` is a constant function

Solution:

Let `f(x) =e^x`

Now, differentiate w.r.t. x, we get

`f'(x)=e^x > 0, AA x in R`

So, `f(x) =e^x` is an increasing function.

Correct Answer is `=>` (A) `e^x` is an increasing function

Q 2378467306

What is the minimum value of `|x |` ?
NDA Paper 1 2012

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Let ` y =| x|`

Redefined the function,

`y= {tt((x, x ge 0),(-x, x <0))`

So from curve, we abserve that, the
minimum value of `|x |` is zero.

Correct Answer is `=>` (B) `0`

Q 2318667509

If `f(x) = xln x`, then `f(x)` attains the minimum
value at which one of the following points?
NDA Paper 1 2011

(A)

`x=e^(-2)`

(B)

`x=e`

(C)

`x=e^(-1)`

(D)

`x=2e^(-1)`

Correct Answer is `=>` (C) `x=e^(-1)`

Q 2318667500

If the rate of change in volume of spherical soap
bubble is uniform, then the rate of change of
surface area varies as
NDA Paper 1 2011

(A)

square of radius

(B)

square root of radius

(C)

inversely proportional to radius

(D)

cube of the radius

Correct Answer is `=>` (C) inversely proportional to radius

Q 2358867704

The point in the interval `(0, 2pi)`, where
`f(x) =e^x sin x` has maximum slope, is
NDA Paper 1 2011

(A)

`pi/4`

(B)

`pi/2`

(C)

`pi`

(D)

`(3 pi)/2`

Correct Answer is `=>` (B) `pi/2`

Q 2368067805

What is the interval over which the function
`f(x) = 6x- x^2` ,where `x > 0` is increasing'?
NDA Paper 1 2010

(A)

`(0, 3)`

(B)

`(3, 6)`

(C)

`(6, 9)`

(D)

None of these

Correct Answer is `=>` (A) `(0, 3)`

Q 2318167909

If `f` and `g` are two increasing functions such that
fog is defined, then which one of the following is
correct'?
NDA Paper 1 2010

(A)

fog is always an increasing function

(B)

fog is always a decreasing function

(C)

fog is neither an increasing nor a decreasing function

(D)

None of the above

Solution:

We know that, the composition of two increasing
functions is an increasing function by property of composition of
functions.

Hence, fog is always an increasing function.

Correct Answer is `=>` (A) fog is always an increasing function

Q 2308178008

For a point of inflection of `y = f(x)`, which one of
the following is correct'?
NDA Paper 1 2010

(A)

`(dy)/(dx)` must be zero

(B)

`(d^2y)/(dx^2)` must be zero

(C)

`(dy)/(dx)` must be non-zero

(D)

`(d^2y)/(dx^2)` must be non-zero

Solution:

For a point of inflection `(d^2y)/(dx^2)` must be equal to zero.
(by definition)

Correct Answer is `=>` (B) `(d^2y)/(dx^2)` must be zero