Mathematics previous year question Of Derivatives for NDA

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previous year question Of Derivatives for NDA

Previous Year Questions

Set 1

Q 2703691548

If `F(x) = sqrt(9-x^2)` then what is `lim_( x→ 1) (F(x) - F(1))/(x-1)` equal to?
NDA Paper 1 2017

(A)

`-1/(4 sqrt2)`

(B)

`1/8`

(C)

`-1/(2 sqrt2)`

(D)

`1/(2 sqrt2)`

Solution:

`F(x) = sqrt(9-x^2)`

`lim_(x-> 1) (F(x) -F(1))/(x-1) = F'(1)`

`[((-2x))/(2 sqrt(9-x^2))]_(x=1)=(-1) /(2 sqrt 2)`

Correct Answer is `=>` (C) `-1/(2 sqrt2)`

Q 2176112076

Consider the equation `x+ | y |= 2y`.

What is the derivative of y as a function of x with respect
to x for x < 0?
NDA Paper 1 2016

(A)

`2`

(B)

`1`

(C)

`1/2`

(D)

`1/3`

Solution:

Given `x+ | y | = 2y`.

When `x < 0, y < 0`

`:. y = 1/3quad x`

`:.` Derivative of y as a function of x w.r. t. x for x < 0 is `1/3` .

`[∵ dy/dx = 1/3 (1) = 1/3 ]`

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

Q 2177423386

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)

`-6`

(B)

`-5`

(C)

`1`

(D)

`0`

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) + 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`

We have, `f(x) = x^(3) - 5x^(2) + 2x + 6`

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

`f'(x) = 3x^(2) -10x+ 2`

At ` x= 1, f'(1) = 3(1)^(2) - 10(1) + 2`

`=3-10+2= -5 `

Correct Answer is `=>` (B) `-5`

Q 2147723683

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`.

Consider the following
1. `f(2) = f(1)- f(0)`
2. `f"(2) - 2f'(1)= 12`

Which of the above is/are correct?
NDA Paper 1 2016

(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

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`

We have,` f(x) = x ^(3) - 5x^(2) + 2x + 6`

Now, `f(1)= 1^(3) -5(1)^(2) +2 xx 1+6 = 4, f(0)= 6`

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

`= 8 - 20 + 4 + 6 = - 2`

Now, `f(1) - f(0) = 4 - 6 = -2`

`:. f(2) = f(1)- f(0)`

2. Here,

`f''(x) = 6x- 10`

`=> f''(2) = 12- 10 = 2`

and `f'(x) = 3x^(2) -10 x+ 2`

`f ' (1) = 3(1 )^(2) -10 xx 1 + 2 = -5`

` f''(2)- 2f'(1) '= 2 - 2 xx (-5) = 12`

Here, both `1` and `2` are correct.

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

Q 2117878789

Consider the function `f(x) = | x^(2) - 5x + 6 |`

What is ` f' ( 4)` equal to?
NDA Paper 1 2016

(A)

`-4`

(B)

`-3`

(C)

`3`

(D)

`2`

Solution:

Given, `f(x) =| x^(2) - 5x + 6 |`

`=> f(x) = | (x- 2) (x- 3) |`

At `x = 4`, we take

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

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

`f'(x) = 2x - 5`

At`x = 4`,

`f'(4) = 2 xx 4 - 5 = 3`

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

Q 2211478329

The derivative of In `( x + sin x)` with respect to `( x + cos x)` is
NDA Paper 1 2015

(A)

` (1 + cos x) /( (x + sin x) ( 1 - sin x) )`

(B)

` (1 - cos x) /( (x + sin x) ( 1 + sin x) )`

(C)

` (1 - cos x) /( (x - sin x) ( 1 + cos x) )`

(D)

` (1 + cos x) /( (x - sin x) ( 1 - cos x) )`

Solution:

Let `u =` In `(x + sin x)` and `v = x + cos x`

Now, `(du)/(dx) = 1/(x +sin x) ( 1 + cos x)`

and ` (dv)/(dx) = 1- sin x`

Now, we can find derivative of u w.r.t. v, we get

`(du// dx)/ (dv//dx) = ( (1 + cos x)// (x + sin x) )/(1-sin x)`

` => (du)/(dv) = (1 + cos x)/( (x + sin x) (1 - sin x))`

Correct Answer is `=>` (A) ` (1 + cos x) /( (x + sin x) ( 1 - sin x) )`

Q 2251578424

If `y = cot^(-1) [ ( sqrt(1 + sin x) + sqrt(1 - sin x))/( sqrt(1 + sin x) - sqrt(1 - sin x))]`, where `0 < x < pi/2`,
then `(dy)/(dx) ` is equal to
NDA Paper 1 2015

(A)

` -1/2`

(B)

`2`

(C)

`sin x + cos x`

(D)

`sin x - cos x`

Solution:

Given, `y = cot^(-1) [ ( sqrt(1 + sin x) + sqrt(1 - sin x))/( sqrt(1 + sin x) - sqrt(1 - sin x))]`

` = cot^(-1) [ [( | sin \ x/2) + cos (x/2) | + | sin (x/2) - cos (x/2) | )/( | sin \ x/2 + cos \ x/2 | - | sin \ x/2 - cos \ x/2 |)]`

As ` 0 < x < pi/2`

`:. y = cot^(-1) [ ( ( sin \ x/2 + cos \ x/2 ) + ( sin \ x/2 - cos \ x/2 ) )/( ( sin \ x/2 + cos \ x/2 ) - ( sin \ x/2 - cos \ x/2 ))]`

` = cot^(-1) ( tan \ x/2 ) = cot^(-1) [ cot ( pi/2 - x/2 )] = ( pi/2 - x/2)`

`:. (dy)/(dx) = -1/2`

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

Q 1658080804

What is the derivative of `tan^(-1) ( (sqrt (1+x^2)-1)/x)` with
respect to `tan ^(-1) x?`
NDA Paper 1 2015

(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`x`

Solution:

Let `u = tan^(-1) ( (sqrt (1+x^2)-1)/x)` and ` v = tan^(-1) x`

To find `(du)/(dv)`

Consider,`u = tan^(-1) ( (sqrt (1+x^2)-1)/x)`

On putting `x = tan theta` . we get

` u = tan^(-1) ( (sqrt (1+tan^2 theta)-l)/ (tan theta))`

` = tan^(-1) ( (sec theta -1)/(tan theta)) = tan^(-1) ( (1 - cos theta )/(sin theta))`

` = ( (2 sin ^2 (theta/2))/ (2 sin (theta/2) cos (theta/2) ))= tan^(-1) ( tan (theta /2)) = theta /2`

`=> u = 1/2 tan^(- 1) x`

`=> (du)/(dx) = 1/2 (1/(1 + x^2) )` ... (i)

Now, consider `v = tan^(-1) x`

`=> (dv)/(dx) = 1/(1 + x^ 2)`... (ii)

From Eqs. (i) and (ii), we have

`(du)/(dv) = (du)/( dx) xx (dx)/(dv)`

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

Correct Answer is `=>` (B) `1/2`

Q 1669567415

`x = a (cos theta + theta sin theta )` and `y = a(sin theta - theta cos theta )`.

What is `(dy)/(dx)` equal to?
NDA Paper 1 2014

(A)

`tan theta `

(B)

`cot theta `

(C)

`sin 2 theta `

(D)

`cos 2 theta`

Solution:

We have, `x = a(cos theta + theta sin theta)`

and `y = a(sin theta - theta cos theta)`

`=> (dx)/(d theta) = a (- sin theta + theta cos theta +sin theta)`

`=> (dx)/(d theta) = a theta cos theta`

and ` (dy)/(d theta) = a(costheta + theta sin theta -cos theta)`

`=> (dy)/(d theta) = a theta sin theta`

`:. (dy)/(dx) = ((dx)/(d theta))/((dx)/(d theta)) = (a theta sin theta)/(a theta cos theta) = ten theta`

Correct Answer is `=>` (A) `tan theta `

Q 1639667512

`x = a (cos theta + theta sin theta )` and `y = a(sin theta - theta cos theta )`.

What is `(d^2y)/(dx^2)` equal to?
NDA Paper 1 2014

(A)

`sec^2 theta`

(B)

`- cosec^2 theta`

(C)

`(sec^3 theta)/(a theta)`

(D)

None of these

Solution:

We have,` (dy)/(dx) = tan theta`

` => (d^2y)/(dx^2) = sec^2 theta (d theta)/(dx)`

` => (d^2y)/(dx^2) = sec^2 theta ( 1/( a theta cos theta)), \ \ \ \ \( ∵(dx)/(d theta) = a theta cos theta )`

`=> (d^2y)/(dx^2) = (sec^3 theta)/(a theta)`

Correct Answer is `=>` (C) `(sec^3 theta)/(a theta)`

Q 2460378215

What is the derivative of `x^3` with respect to `x^2`?
NDA Paper 1 2013

(A)

`3x^2`

(B)

`(3x)/2`

(C)

`x`

(D)

`3/2`

Solution:

Let `f(x) = x^3` and `g(x) = x^2`

Now, `(df(x))/(dx)= 3x^2` and `(dg(x))/(dx) = 2x`

`:.` Derivative of `x^3` w.r.t. `x^2`,

`= (df(x))/(dg(x)) = { (df (x))/(dx)} xx {(dx)/(dg (x)) }`

`= 3x^2 xx 1/(2x) = (3x)/2 .`

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

Q 1619112019

Consider the parametric equation `x = ( a(1 - t^2) )/( 1 +t^2) , y = (2at)/( 1 +t^2)`

What is `(dy)/(dx)` equal to?
NDA Paper 1 2015

(A)

`y/x`

(B)

`-y/x`

(C)

`x/y`

(D)

`-x/y`

Solution:

We have, `x = ( a(1 - t^2) )/( 1 +t^2) ,` and ` y = (2at)/( 1 +t^2)`

Consider `x = ( a(1 - t^2) )/( 1 +t^2)`

` =>(dx)/(dt) = a [ ( (1 + t^2) (-2t) - (1 - t^2) (2t) )/ (1 + t^2)^2 ]`

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

and ` y = (2at)/ (1 + t^2) => (dy)/(dt) = 2a [((1+t^2).1-t (2t))/(1 + t^2)^2]`

Now,`( dy)/(dx) = (dy)/(dt) . (dt)/(dx) = 2a ((1-t^2 )/(1 + t^2)^2 . (1 + t^2)^2/(-4at))`

` = (1-t^2 )/(2t) = -x/y` ...............(3)

Correct Answer is `=>` (D) `-x/y`

Q 2177223186

If `y = log_(10) x + log_x 10 + log_x x + log_(10) 10`,
then what is `((dy)/(dx))_( x = 10)` equal to?

NDA Paper 1 2016

(A)

`10`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

Given,

`y = log_(10) x + log_x 10 + log_x x + log_(10) 10`

`= (log x)/(log 10) + (log 10)/(log x) + 1 + 1`

` y = (log x)/(log 10) + (log 10)/(log x) + 2`

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

` (dy)/(dx) = 1/(log 10) . 1/x + log 10 (- 1/(log x)^2 * 1/x )`

At `x = 10, ( (dy)/(dx) )_( x = 10) = 1/(10 log 10) - (log 10)/(10( log 10)^2)`

` = 1/(10 log 10) - 1/( 10 log 10)`

` => ( (dy)/(dx) )_( x = 10) = 0`

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

Q 2231778622

If `x^a y^b = (x- y)^(a+ b)`, then the value of `(dy)/(dx) - y/x`. is equal to
NDA Paper 1 2015

(A)

` a/b`

(B)

` b/a `

(C)

`1`

(D)

`0`

Solution:

We have, `x^a y^b = (x- y)^(a + b)`

Taking log on both sides, we get

`a log x + b log y = (a+ b) log (x- y)`

Differentiating on both sides, we get

`a/x + b/y quad (dy)/(dx) = (a+ b) 1/ (x - y) ( 1 - (dy)/(dx))`

`=> (dy)/(dx) ( b/y + (a+b)/(x-y) )= (a+b)/(x-y) - a/x`

` => (dy)/(dx) = y/x . (bx +ay)/(bx +ay) = y/x => (dy)/(dx) - y/x = 0`

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

Q 2410778619

What is the differential coefficient of `log_x x`?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`1/x`

(D)

`x`

Solution:

Let `y= log_x x`

`y= 1` (by logarithm property, `log_a a = 1`)

Now, differentiating w.r.t. `x` on both sides, we get

`(dy)/(dx) =0`

Hence, the differential coefficient of `log_x x` is `0`.

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

Q 2431301222

If `y = log (e^(mx) + e^(-mx))`, then what is the value of

`(dy)/(dx)` at `x= 0`?
NDA Paper 1 2012

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`y = log (e^(mx) +e^(-mx))`

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

`(dy)/(dx) = 1/(e^(mx) + e^(-mx)) * d/(dx) (e^(mx) + e^(-mx))`

`= 1/(e^(mx) + e^(-mx)) (me^(mx) - me^(-mx))`

`= [ (m (e^(mx) - e^(-mx)) )/(e^(mx) + e^(-mx)) ] `

`:. ((dy)/(dx))_(text (at)x=0 ) = (me^0 - e^(-0) m )/(e^0 + e^(-0))`

`= (m-m)/(1+1) = 0/2 =0`

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

Q 2411001829

If `y = log sqrt (tan x)` , then what is the value of `(dy)/(dx)` at

`x= pi/4`?
NDA Paper 1 2011

(A)

`0`

(B)

`-1`

(C)

`1/2`

(D)

`1`

Solution:

Given, `y = log sqrt (tan x) =1/2 log tan x`

`=> (dy)/(dx) = 1/2 * 1/(tan x) * sec^2 x`

`=> ((dy)/(dx))_(text (at) x= pi/4 ) =1/2 * (sec^2 (pi/4) )/(tan (pi/4) )`

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

Correct Answer is `=>` (D) `1`

Q 2431423322

What is the derivative of `log_x 5` with respect to
`log_5 x`?
NDA Paper 1 2009

(A)

`- (log_5 x)^(-2)`

(B)

`(log_5 x)^(-2)`

(C)

`- (log_x 5)^(2)`

(D)

`(log_x 5)^(-2)`

Solution:

Let `u = log_x 5` and `v = log_5 x`

Then, `(du)/(dx) = (-log 5)/(log x)^2 * 1/x ` and `(dv)/(dx) = 1/(x log 5)`

`(du)/(dv) = ((du)/(dx))/((dv)/(dx)) = ( (-log 5)/(log x)^2 * 1/x ) /((1/(log 5)) * 1/x) = - ((log 5)/(log x))^2 = - (log_x 5)^2 `

Correct Answer is `=>` (C) `- (log_x 5)^(2)`

Q 2411134029

If `f(x) = log_e (log_e x)`, then what is `f ' (e)` equal to?
NDA Paper 1 2008

(A)

`e^(-1)`

(B)

`e`

(C)

`1`

(D)

`0`

Solution:

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

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

`f'(x) = 1/(log_e x) * 1/x`

`=> f' (e) = 1/(log_e e) * 1/e = 1/e = e^(-1)`

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

Q 2471334226

If `y= 1/(log_(10 ) x)`, then what is `(dy)/(dx)` equal to?
NDA Paper 1 2008

(A)

`x`

(B)

`x log_e 10`

(C)

`- ( (log_x 10)^2 (log_(10) e) )/x`

(D)

`x log_(10) e`

Solution:

`:. y = 1/(log_(10) x)`

`:. (dy)/(dx) = -1/ (log_(10) x)^2 * 1/x log_(10) e ` `( :. log_(b) a = 1/(log_(a) b) )`

`= -( (log_x 10)^2 * log_(10)e )/x`

Correct Answer is `=>` (C) `- ( (log_x 10)^2 (log_(10) e) )/x`

Q 2431134022

If `y = sin^(-1) x + sin^(-1) sqrt (1-x^2)`, then what is `(dy)/(dx)` equals to?
NDA Paper 1 2008

(A)

`cos^(-1) x + cos^(-1) sqrt (1-x^2)`

(B)

`1/(cos x) +1/(cos sqrt (1-x^2) )`

(C)

`pi/2`

(D)

`0`

Solution:

`:. y = sin^(-1)x + sin^(-1) sqrt (1-x^2)`

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

`(dy)/(dx) = 1/(sqrt (1-x^2) ) + 1/(sqrt (1-1 +x^2) ) * 1/(2 sqrt (1-x^2) ) (-2x)`

`= 1/(sqrt (1-x^2) ) - 1/(sqrt (1-x^2) ) =0`

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

Q 2372312236

Given, that `d/(dx) ((1 + x^2 + x^4)/(1 + x +x^2)) = Ax +B`

What is the value of `A` ?
NDA Paper 1 2015

(A)

`-1`

(B)

`1`

(C)

`2`

(D)

`4`

Solution:

We have , `d/(dx)

((1 + x^2 + x^4)/(1 + x +x^2)) = Ax +B` .........(i)

Let us first divide `x^4 + x^2 +1 ` by `x^2 +x +1`

we get

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

`:. ` From (i), we have

`d/(dx) (x^2 - x +1 ) = Ax +B`

`=> 2x - 1 = Ax + B` .................(ii)

On comparing, the coefficient of A, we get

`=> A=2`

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

Q 2332412332

Given, that `d/(dx) ((1 + x^2 + x^4)/(1 + x +x^2)) = Ax +B`

What is the value of `B` ?
NDA Paper 1 2015

(A)

-1

(B)

(C)

(D)

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

Q 2281880727

Consider `f' (x) = x^2/x - kx + 1` such
that `f(0) = 0` and `f(3) = 15`.

The value of `k` is

NDA Paper 1 2015

(A)

`5/3`

(B)

`3/5`

(C)

`- 5/3`

(D)

`- 3/5`

Solution:

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

On integrating, we get

` int f'(x) dx = int ( x^2/2 -kx + 1) dx`

`=> f(x) = x^3/6 - (kx^2)/2 + x + C` ............(i)

Given, `f(0) = 0`

`:. 0=0-0+0+C => C = 0`

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

Given, `f(3) = 15`

`:. 15 = (3)^3/6 - (k(3)^2)/2 + 3`

`=> 15 = (27)/9 - 9/2 k + 3 => k = - 5/3`

Correct Answer is `=>` (C) `- 5/3`

Q 2241080823

Consider `f' (x) = x^2/x - kx + 1` such
that `f(0) = 0` and `f(3) = 15`.

`f' ' (- 2/3)` is equal to
NDA Paper 1 2015

(A)

` -1`

(B)

`1/3`

(C)

`1/2`

(D)

`1`

Solution:

`∵ f ' (x) = x^2/x - kx + 1`

`=> f '(x) =x^2/x - ( (-5)/3) x + 1`

` = (x^ 2)/2 + 5/3 x + 1`

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

` f''(x) = (2x)/3 + 5/3 = x + 5/3`

`:. f '' (-2/3) = -2/3 + 5/3 =1`

Correct Answer is `=>` (D) `1`

Q 1669167015

Consider the function `f(x) = {tt ( (x^2 - 5 , x <= 3) ,( sqrt(x + 13) , x > 3) )`

What is the differential coefficient of `f(x)` at `x = 12`?
NDA Paper 1 2014

(A)

`5/2`

(B)

`5`

(C)

`1/5`

(D)

`1/(10)`

Solution:

We have, `f(x) = sqrt(x + 13), x > 3`

` => f' (x) = 1/( 2 sqrt(x + 13) )`

` => f' (12) = 1/( 2 sqrt(12 + 13) ) = 1/(2 xx 5) = 1/(10)`

Correct Answer is `=>` (D) `1/(10)`

Q 1669767615

If `y = x ln x + xe^x`, then what is the value of `(dy)/(dx)` at `x = 1`?
NDA Paper 1 2014

(A)

`1 + e`

(B)

`1 - e`

(C)

`1 + 2e`

(D)

None of these

Solution:

Given, `y = x log x + xe^x`

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

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

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

`:. ( (dy)/(dx) )_(x = 1) = 1 + log 1 + 1. e^ 1 + e^ 1`

` = 1 + 2e quad ( ∵ log 1 = 0)`

Correct Answer is `=>` (C) `1 + 2e`

Q 2308123008

If `f(x) = 2x^2 + 3x- 5`, then what is `f' (0) + 3f' (-1)`
equal to?
NDA Paper 1 2013

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Given function,

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

Now, `f'(x) = 4x + 3`

At `x= 0; f'(0) = 4 xx 0 + 3 = 3`

and at `x=-1; f'(-1)=4xx (-1)+ 3`

`=-4+3=-1`

`f'(0) + 3f'(-1) = 3 + 3(-1)`

`=3-3=0`

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

Set 2

Q 2430478312

What is the derivative of `sin(sinx)`?
NDA Paper 1 2013

(A)

`cos(cosx)`

(B)

`cos(sinx)`

(C)

`cos(sinx)cosx`

(D)

`cos (cosx) cosx`

Solution:

Let `f(x) = sin(sinx)`

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

`f'(x) = cos (sin x) * d/(dx) (sin x)`

`= cos x * cos (sin x)`

Correct Answer is `=>` (C) `cos(sinx)cosx`

Q 2450478314

What is the derivative of `| x - 1 | ` at `x = 2`?
NDA Paper 1 2013

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

Does not exist

Solution:

Let `f(x) = | x- 1 |`

Redefined the function `f(x)`

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

`=> f' (x) = { tt ( (-1; x <1), (1; x > 1) )`

`:. f'(2) =1`

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

Q 2400478318

If `2x^3 - 3y^2 = 7`, what is `(dy)/(dx)` equal to

(where , `y ne 0`) ?
NDA Paper 1 2013

(A)

`x^2/(2y)`

(B)

`x/(2y)`

(C)

`x^2/y`

(D)

None of these

Solution:

Given curve is `2x^3 - 3y^2 = 7` ` ( :. yne 0)`

Now, differentiating both sides w.r.t. `x`, we get

`2* 3 * x^2 - 3* 2 * y * (dy)/(dx) =0 => x^2 -y (dy)/(dx) =0`

`:. (dy)/(dx)=x^2/y .`

Correct Answer is `=>` (C) `x^2/y`

Q 2430678512

If `y= x^x`, then what is the value of `(dy)/(dx)` at `x=1`?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

`2`

Solution:

Given curve, `y=x^x`

Taking log on both sides, we get

`log y= x log x`

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

`1/y * (dy)/(dx) = x * 1/x + log x * 1`

`=> (dy)/(dx)= (1+ log x) * y`

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

`:. ((dy)/(dx))_(text (at) x=1 ) = (1)^1 (1+ log 1) =1 (1+0) =1`

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

Q 2410878710

The derivative of `sec^2 x` with respect to `tan^2 x` is
NDA Paper 1 2013

(A)

`1`

(B)

`2`

(C)

`2 sec x tan x`

(D)

`2 sec^2 x tan x`

Solution:

Let `u = sec^2 x` and `v = tan^2 x`

Now, `(du)/(dx) = 2 sec x * sec x * tan x`

`= 2 sec^2 x * tan x`

and `(dv)/(dx) = 2tan x * sec^2 x`

`:. (du)/(dv)= (d (sec^2 x) )/(d (tan^2 x) ) = ((du)/(dx))/((dv)/(dx))`

`= (2 sec^2 x * tan x)/(2 tan x * sec^2 x) =1`

Correct Answer is `=>` (A) `1`

Q 2470878716

If `x^m + y^m = 1`, such that `(dy)/(dx) = - x/y`, then what should be the value of `m`?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`2`

(D)

None of these

Solution:

Given, `x^m + y^m = 1`

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

`mx^(m-1) + my^(m-1) (dy)/(dx) = 0`

`=> (dy)/(dx) = - (mx^(m-1))/(my^(m-1))`

`= (-x^(m-1))/(y^(m-1)) = - ((x^m)/(y^m)) (y/x)`

Given , `(-x)/y = - ((x^m)/(y^m)) (y/x)`

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

which is true when `m = 2`.

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

Q 2410878719

Consider the following statements

I. If `y = ln (sec x +tan x)`, then `(dy)/(dx) = sec x`

II. If `y = ln (cosec x - cot x)`, then `(dy)/(dx) = cosec x`

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:

Given, `y = In (sec x + tan x)`

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

`(dy)/(dx) = 1/(sec x + tan x) d/(dx) (sec x + tan x)`

`= 1/(sec x + tan x) * (sec x * tan x + sec^2 x)`

`=> = 1/(sec x + tan x) sec x (tan x + sec x) = sec x`

II. Given , `y = log (cosec x - cot x)`

`(dy)/(dx) = 1/(cosec x - cot x) d/(dx) (cosec x - cot x)`

`= 1/(cosec x - cot x) * (- cosec x * cot x + cosec^2 x)`

`= cosec x* ((cosec x - cot x)/(cosec x - cot x))= cosec x`

So, Statements I and II both are true.

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

Q 2440480313

If `f(x) = 2^(sin x)`, then what is the derivative of `f(x)`?
NDA Paper 1 2012

(A)

`2^(sin x) ln 2`

(B)

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

(C)

`(cos x) 2^(sin x-1)`

(D)

none of the above

Solution:

Given, `f(x) = 2^(sin x)`

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

`=> d/(dx) f(x) = d/(dx) (2^(sin x))`

`= (2)^(sin x) log 2 *d/(dx) (sin x) = 2^(sin x) (cos x) * (log 2)`

Correct Answer is `=>` (D) none of the above

Q 2440580413

If `y = log (e^(mx) + e^(-mx))`, then what is the value of `(dy)/(dx)` at `x= 0`?
NDA Paper 1 2012

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`y =log (e^(mx) + e^(-mx))`

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

`(dy)/(dx) = 1/(e^(mx) + e^(-mx)) * d/(dx) (e^(mx) +e^(-mx))`

`= 1/( e^(mx) + e^(-mx)) (me^(mx) - me^(-mx))`

`= [ ( m (e^(mx) - e^(-mx)) )/(e^(mx) + e^(-mx)) ] `

`:. ((dy)/(dx))_(text (at) x =0) = (me^0 - e^(-0) (m ))/(e^0 + e^(-0))`

`= (m-m)/(1+1) = 0/2 =0`

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

Q 2471601526

If `y =cos t` and `x = sin t`, then what is the value of `(dy)/(dx)` ?
NDA Paper 1 2012

(A)

`xy`

(B)

`x/y`

(C)

`-y/x`

(D)

`-x/y`

Solution:

Given that, `y = cos t`

and `x= sin t`

Then, `(dy)/(dt)= -sin t ` and `(dx)/(dt)= cos t`

Now, `(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (- sin t)/(cos t) = (-x)/y`

Correct Answer is `=>` (D) `-x/y`

Q 2410491310

If `y =cos t` and `x = sin t`, then what is the value of `(dy)/(dx)` ?
NDA Paper 1 2012

(A)

`xy`

(B)

`x/y`

(C)

`-y/x`

(D)

`-x/y`

Solution:

Given that, `y = cos t`

and `x= sin t`

Then, `(dy)/(dt) = -sin t ` and `(dx)/(dt)= cos t`

Now, `(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (- sin t)/(cos t) = (-x)/y`

Correct Answer is `=>` (D) `-x/y`

Q 2441701623

What is the rate of change of `sqrt (x^2 + 16)` with respect to `x^2 ` at `x =3`?
NDA Paper 1 2012

(A)

`1/5`

(B)

`1/10``

(C)

`1/20`

(D)

None of these

Solution:

let `u = sqrt (x^2 +16)` and `v = x^2`

Now, `(du)/(dx)= 1/(2 sqrt (x^2 +16) ) xx 2x = x/(sqrt (x^2 +16) )`

and `(dv)/(dx) =2x`

`:. ` Now, rate of charge of `u` w.r.t. to `v`,

`(du)/(dv) = ((du)/(dx))/((dv)/(dx)) = x/(sqrt (x^2 +16) ) xx 1/(2x)`

`(du)/(dv) = 1/ (2 sqrt (x^2 +16) )`

`=> ((du)/(dv))(text(at) (x=3) ) = 1/(2 sqrt (9+16) ) =1/(2 sqrt (25) ) =1/(2 xx 5) = 1/10`

`:. (d (x^2 +16) )/(d (x^2) ) =1/10`

Correct Answer is `=>` (B) `1/10``

Q 2420591411

What is the rate of change of `sqrt (x^2 +16)` with respect to `x^2` at `x =3` ?
Paper 1 2012

(A)

`1/5`

(B)

`1/10`

(C)

`1/20`

(D)

None of these

Solution:

let `u = sqrt (x^2 + 16)` and `v= x^2`

Now, `(du)/(dx) = 1/(2 sqrt (x^2 +16) ) xx 2x = x/(sqrt (x^2 +16) )`

and `(dv)/(dx) =2x`

`:.` Now, rate of charge of u w.r.t. to `v`,

`(du)/(dv) = ((du)/(dx))/((dv)/(dx)) =x/(sqrt(x^2 +16) ) xx 1/(2x)`

`(du)/(dv) =1/(2 sqrt (x^2 +16) )`

`=> ((du)/(dv))_(text (at) x=3 ) = 1/(2 sqrt (9+16) ) =1/(2 sqrt (25)) = 1/(2 xx 5) =1/10`

`:. (d (x^2 +16) )/(d (x^2)) =1/10`

Correct Answer is `=>` (B) `1/10`

Q 2421801721

If `y= (x+1)/(x-1)`, then what is `(dy)/(dx)` equal to ?
NDA Paper 1 2012

(A)

`-2/(x-1)`

(B)

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

(C)

`2/(x-1)^2`

(D)

`2/(x-1)`

Solution:

we have, `y = (x+1)/(x-1)`

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

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

`=> (dy)/(dx)= ( (x-1)* 1- (x+1) * 1 )/(x-1)^2`

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

Correct Answer is `=>` (B) `-2/(x-1)^2`

Q 2460691515

If `y = (x+1)/(x-1)`, then what is `(dy)/(dx)` equal to ?

Paper 1 2012

(A)

`-2/(x-1)`

(B)

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

(C)

`2/(x-1)^2`

(D)

`2/(x-1)`

Solution:

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

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

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

`=> (dy)/(dx) = ( (x-1) * 1 - (x+1) * 1 )/(x-1)^2`

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

Correct Answer is `=>` (B) `-2/(x-1)^2`

Q 2461801725

If `y = (1 + x^(1/4) ) (1+ x^(1/2)) (1- x^(1/4))`, then what is `(dy)/(dx)` equal to ?
NDA Paper 1 2011

(A)

`1`

(B)

`-1`

(C)

`x`

(D)

`x^(1/2)`

Solution:

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

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

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

`= 1-x`

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

`(dy)/(dx)= -1`

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

Q 2431101922

If `f(x) = 2^x`, then what is `f''(x)` equal to ?
NDA Paper 1 2011

(A)

`2^x (log 2)^2`

(B)

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

(C)

`2^(x+1) (log 2)`

(D)

`2^x (log_(10) 2)^2`

Solution:

`:. f (x) = 2^x`

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

`f'(x) = 2^x (log 2)`

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

`f" (x) = 2^x (log 2)^2`

Correct Answer is `=>` (A) `2^x (log 2)^2`

Q 2421112021

What is the derivative of `sin^2 x` with respect to
`cos^2 x`?
NDA Paper 1 2010

(A)

`tan^2 x`

(B)

`cot^2 x`

(C)

`-1`

(D)

`1`

Solution:

let `u = sin^2 x` and `v = cos^2 x`

`=> (du)/(dx) =2 sin x cos x`

and `(dv)/(dx) = -2 sin x cos x`

`:. (du)/(dv) = ((du)/(dx))/((dv)/(dx)) = (2 sin x cos x)/(-2 sin x cos x) = -1`

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

Q 2471412326

If `sqrt(x) + sqrt (y) =2`, then what is `(dy)/(dx)` at `y = 1` equal to?
NDA Paper 1 2010

(A)

`5`

(B)

`4`

(C)

`2`

(D)

`-1`

Solution:

`:. sqrt(x) + sqrt (y) =2`

At `y =1 , sqrt (x) +1 =2 => x =1`

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

`1/(2 sqrt(x)) +1/(2 sqrt (y)) (dy)/(dx) =0`

At `y =1 , x =1`

`1/2 +1/2 (dy)/(dx) = 0 => (dy)/(dx)= -1`

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

Q 2461512425

If `x =cos 2t` and `y = sin^2 t`, then what is `(d^2 y)/(dx^2)` equal to?
NDA Paper 1 2010

(A)

`0`

(B)

`sin (2t)`

(C)

`-cos (2t)`

(D)

`-1/2`

Solution:

Given that, `x = cos 2t, y = sin^2 t` (given)

`(dx)/(dt) = -2 sin 2t ` and `(dy)/(dt) = 2 sin t cos t = sin 2t`

`=> (dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (sin 2t)/(-2 sin 2t) = -1/2`

`=> (d^2 y)/(dx^2) =0`

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

Q 2481123027