Mathematics Must Do Problems of Applications of Derivatives for NDA

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Must Do Problems of Applications of Derivatives for NDA

Problems Set - 1

Q 2563123945

The displacement of a particle at time `t` is `x`, where `x = t^4 - kt^3` If the velocity of the particle at time `t = 2` is minimum, then
WBJEE 2010

(A)

`k = 4`

(B)

`k = -4`

(C)

`k = 8`

(D)

`k = -8`

Solution:

given `x = t^4 - kt^3`

`=> (dx)/(dt) = 4t^3-3kt^2`

`=> (dv)/(dt) = 12t^2-6kt`

At `t = 2 , (dv)/(dt) = text(minimum) = 0`

`therefore 12xx2^2-6kxx2 = 0`

`=> 48-12k = 0`

`=> k = 4`

Correct Answer is `=>` (A) `k = 4`

Q 2815301260

Let `f(x) = x^3 - 3 (7 - a )x^2 - 3 (9 - a^2) x + 2`
The value of parameter a, if `f(x)` has a negative point of local minima is

(A)

`phi`

(B)

`(-3,3)`

(C)

`( - oo , (58)/(14) )`

(D)

None of these

Solution:

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

or ` f '(x) = 3x^2 - 6 ( 7 - a) x - 3 ( 9 - a^2)`

For real root,` D >= 0`

or `49 + a^2 - 14 a + 9 - a^2 >= 0`

or ` a <= (58)/(14)` ........(i)

When point of minima is negative, point

of maxima is also negative.

Hence, equation

`f' (x) = 3x^2 - 6 (7 - a) x - 3 (a - a^2 ) = 0`,

has both roots negative.

Sum of roots `= 2(7 - a) < 0` or `a > 0`,

which is not possible as form (i), ` a <= (58)/(14)`

When point of maxima is positive, point

of minima is also positive.

Hence, equation

`f ' (x) = 3x^2 - 6(7- a) x - 3(9 - a^2 ) = 0`

has hoth roots positive.

Also , some of roots

` = 2 (7 - a) > a ` or ` a < 7` ......(ii)

product of roots is positive

or `- ( 9 - a^2) > 0` or `a^2 > 9`

or `a in ( - oo , - 3 ) cup (3 , oo)` .........(iii)

From Egs. (i), (ii) and (iii),

` a in ( - oo , -3) cup ( 5 , (58)/(14) )`

For points of extreme of opposite sign,

Eq. (i) has roots of opposite sign.

Thus, `a in (- 3 , 3)`

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

Q 1543191043

If `y=x^3-ax^2+48x+7` is an increasing function for all real values of `x`, then a lies in:
BITSAT 2012

(A)

`(-14, 14)`

(B)

`(-12, 12)`

(C)

`(-16, 16)`

(D)

`(-21, 21)`

Solution:

`y=x^3-ax^2+48x+7`

`y' =3x^2-2ax+48`

For given function to increasing for all `x`,

Discriminant of above quadratic `< 0`

`\Rightarrow (-2a)^2-4\cdot 3\cdot 48 < 0`

`\Rightarrow a^2-144 < 0`

`\Rightarrow (a+12)(a-12) < 0`

`\Rightarrow a\in(-12,12)`

Correct Answer is `=>` (B) `(-12, 12)`

Q 1522091831

Length of the subtangent to the curve `y=e^{\frac xa}` is:
BITSAT 2013

(A)

`e^{ x/a}`

(B)

`a`

(C)

`2/a`

(D)

None of these

Solution:

Length of the subtangent is given by `= frac{y}{\ frac{dy}{dx}}`

Here, `y = e^{\frac{x}{a}}`

`\ frac{dy}{dx} = \ frac{1}{a} \times e^{\frac{x}{a}}`

Length of the subtangent `= \ frac{y}{\ frac{dy}{dx}} = \ frac{e^{\frac{x}{a}}}{\ frac{1}{a} \times e^{\frac{x}{a}}} = a`

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

Q 1572791636

`f(x) = x^3- 6x^2 + 12x- 16` is strictly decreasing for
BITSAT 2013

(A)

`x \in R`

(B)

`x\in R - \{1\}`

(C)

`x\in R^+`

(D)

`x\in \phi`

Solution:

A function is strictly decreasing in the range when `\ frac{d(f(x))}{dx} < 0`

`\ frac{d(f(x))}{dx} = 3x^2 - 12x + 12 = 3(x^2 - 4x + 4) = 3(x - 2)^2`

Since the square of any number is non negative, we don't find any value of `x` for which the function would be strictly decreasing.

Correct Answer is `=>` (D) `x\in \phi`

Q 2529078811

Let `f(x) = 2x^3 - 9x^2 + 12x + 6`. Discuss the global maxima
and global minima of `f(x)` in `(1, 3)`.

Solution:

`f(x) = 2x^3- 9x^2 + 12x + 6`

`=> f'(x) = 6x^2 -18x + 12 => f '(x) = 6 (x-1)(x-2)`

Let `f'(x) =0 => x = 1, 2`

`:. f (1) = 11` and `f (2) = 10` ... (i)

Let us consider the open interval `(1, 3)`.

Clearly, `x =2` is the only point in `(1, 3)`.

And `f(2) = 10` [from Eq. (i)]

Now, `lim_(x->1^+) f(x) = 11` and `lim_(x->3^-) f(x) = 15`

Thus, `x = 2` is the point of global minima in `(1, 3)` and global maxima does
not exist in `(1, 3)`.

Q 2302812738

The number of real roots of the equation

`3x^5 + 15x = 0`, greater than `1` is equal to
BITSAT Mock

(A)

`0`

(B)

`1`

(C)

`3`

(D)

`5`

Solution:

`f(x) = 3x^5 + 15x`

`=> f'(x) = 15(x^4 + 1) > 0` for all real `x`.

`=> f(x)` is an increasing function of `x`.

`:. f(1) = 18 => f(x) ge f(1) = 18 AA x ge 1`

`=> f(x)` does not have real roots in

the interval ` [ 1, oo )`

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

Q 2258167004

The curve `y + x = e^(xy)` has a vertical
tangent at the point
BITSAT Mock

(A)

`(1, 1)`

(B)

`(1, 0)`

(C)

`(0, 1)`

(D)

at no point

Solution:

`y + x = e^xy`

Differentiating with respect to `y`,
we get

`1 + (dx)/(dy) = e^(xy) (y (dx)/(dy) +x)`

For vertical tangent (parallel to `y`-axis),

`(dx)/(dy) =0`

`xe^(xy) = 1`, which is satisfied by the

point `(1, 0)`.

Correct Answer is `=>` (B) `(1, 0)`

Q 2875101066

In a circle of radius 'r', a right circular cone is drawn.
What will be the maximum height of cone having maximum valume?

(A)

`(4r)/3`

(B)

`(3r)/4`

(C)

`sqrt3/3 r`

(D)

`2/3 r`

Solution:

Let the radius of the cone be 'R' and height be 'h'.

Now, in `Delta OAB , r^2 = R^2 + (b - r)^2`

`=> r^2 = R^2 + b^2 + r^2 - 2rb`

`=> R^2 = 2rb - b^2`

Volume of cone, `V = 1/3 pi R^2 b`

`= 1/3 pi b (2 r b - b^2 ) = 1/3 pi (2 pi b^2 - b^3)`

Or differentiate with respect to b'.

` (dV)/(db) = 1/3 pi ( 4 rb - 3b^2)`

` (dV)/(dx) = 0 ` , for maximum and minimum value

`=> 4rb = 3b^2 => 4r = 3b`

`:. b = (4r)/3 `

Now ` (d^2 V)/(db^2) = 1/3 pi (4 r = 6 b)`

At ` b = (4r)/2 , ((d^2V)/(db^2)) _(b = (4r)/3) = 1/3 pi ( 4r - 6 xx (4r)/3 )`

` = pi/3 ( 4 r - 8 r) = (-4 pi)/3 < 0`

When ` b = (4r)/3` , then V will be

maximum. Thus, volume of cone will be

maximum at `b = (4 r)/3 ` which is the height

of the cone.

Correct Answer is `=>` (A) `(4r)/3`

Q 2561267125

The maximum and minimum values of `cos^6 theta + sin^6 theta` are respectively
WBJEE 2013

(A)

1 and `1/4`

(B)

1 and 0

(C)

2 and 0

(D)

1 and `1/2`

Solution:

Let `f(theta) = cos^6 theta + sin^6 theta`

`=> f(theta) = (sin^2 theta )^3 + (cos^2 theta)^3`

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

`[because a^3+b^3 = (a+b)(a^2+b^2-ab)]`

` = 1 . { sin^2 theta + cos^2 theta)^2 - 3 sin^2 theta . cos^2 theta}`

` = 1 . (1-3/4 . 4 sin^2 theta . cos^2 theta)`

` = 1-3/4 (sin2theta)^2` `(because sin2A = 2sinAcosA)`

` = 1-3/8+3/8cos theta`

`f(theta) = 5/8+3/8 cos theta` ........(i)

`because -1 le cos4theta le 1`

`=> -3/8 le 3/8 cos4theta le 3/8`

`=> 5/8 -3/8 le 5/8+3/8 cos 4 theta le 5/8+3/8`

`=> 1/4 le f(theta) le 1` [from Eq. (i)]

so the maximum value is 1 and minimum value is `1/4`

Correct Answer is `=>` (A) 1 and `1/4`

Q 2512012839

The minimum value of the function `f(x) =2|x-1|+|x-2|` is
WBJEE 2013

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

Given, `f(x) 2|x-1|+|x-2|`

`=> f(x) = {tt((-2(x-1)-(x-2) , x < 1 ) , (2(x-1)-(x-2), 1 le x le 2 ) , (2(x-1)+(x-2) , x ge 2))`

`f(x) = {tt((-3x+4 , x < 1) , (x , 1 le x le 2) , (3x-4 , x ge 2))`

`f'(x) = {tt((-3 , x < 1 ) , (1 , 1 le x le 2) , (3 , x ge 2))`

So, f(x)will be minimum at x = 1 and the minimum value is 1.

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

Q 1685291167

The interval in which `2x^3 + 5` increases less rapidly than
`9x^2 -12x,` is
BITSAT 2016

(A)

`(- oo, 1)`

(B)

`(2, oo)`

(C)

`(1, 2)`

(D)

none of these

Solution:

Let `f(x) = 2x^3 + 5` and `g(x) = 9x^2 - 12x.`

Then, `f(x)` increases less rapidly than `g(x)` means that

`d/(dx) (f(x)) < d/(dx) (g(x))`

` => d/(dx) (f(x) - g(x)) < 0`

` => f'(x) - g'(x) < 0`

` => 6x^2 - (18x - 12) < 0`

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

` => x in (1,2)`

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

Q 1505223168

The function `f ( x ) = x ^{ 1/x }` is:
BITSAT 2011

(A)

increasing in ` ( 1,\infty )`

(B)

decreasing in `( 1,\infty )`

(C)

increasing in `(1,e)`, decreasing in `( e,\infty )`

(D)

decreasing in ` (1,e)`, increasing in `( e,\infty )`

Solution:

Let `y= x ^{ \frac { 1 }{ x } }`

`\Rightarrow \log y=\frac { 1 }{ x } \log x `

`\Rightarrow \quad \frac { 1 }{ y } \frac { dy }{ dx } =\frac { 1 }{ x ^{ 2 } } -\frac { \log _{ e } x }{ x ^{ 2 } }`

`\Rightarrow \quad \frac { dy }{ dx } = x ^{ 1/x } ( \frac { 1-\log _{ e } x }{ x ^{ 2 } } )`

For ` 1 < x <\infty , x ^{ 1/x } > 0`

and `\frac { 1-\log _{ e } x }{ x ^{ 2 } } > 0\quad in\quad (1,e)`

and `\frac { 1-\log _{ e } x }{ x ^{ 2 } } < 0\quad in\quad (e,\infty )`

Hence, ` f(x)` is increasing in `(1, e)` and decreasing in `(e,\infty )`

Correct Answer is `=>` (C) increasing in `(1,e)`, decreasing in `( e,\infty )`

Set - 2

Q 2553801744

Let `f(x) = x^( 3) e^(-3x), x > 0`. Then, the maximum
value of `f(x)` is
WBJEE 2011

(A)

`e^(-3)`

(B)

`3e^(-3)`

(C)

`27 e^(-9)`

(D)

`oo`

Solution:

Given, `f(x) = x^(3) e^(-3x)`

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

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

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

`= 3e^(-3x) (x^2- x^3)`

for maxima and minima, put `f' (x) = 0`

`=> 3x^2 e^(-3x) (1 - x) = 0`

`=> x= 0 , 1`

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

`- 9e^(-3x) (x^2 - x^3)`

`= 3e^(-3x) (3x^3 - 6x^2 +2x)`

At `x =1 , f'' (1) = 3 e^(-3) (-1) < 0`, maxima.

`:. f(1) = e^(-3)`

Hence, it is maximum at `x = 1`.

Correct Answer is `=>` (A) `e^(-3)`

Q 2532812732

If `f` is a real-valued differentiable function such that `f(x) f' (x) < 0` for all real `x`, then
WBJEE 2012

(A)

`f(x)` must be an increasing function

(B)

`f (x)` must be a decreasing function

(C)

`| f(x) |` must be an increasing function

(D)

`| f(x) |` must be a decreasing function

Solution:

Given, `f(x) f ' (x) < 0`

`=> f (x)` and `f ' (x)` must be of opposite sign .

(i) Let `f(x) = e^(-x)`

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

`=> f(x) > 0` and `f' (x) < 0, AA x in R`

(ii) Let `f(x) =-e^(-x)`

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

`=> f(x) < 0` and `f' (x) > 0, AA x in R`

But ` | f(x)| = | ± e^(-x) | = e^(-x)` in both cases

`:. d/(dx) | f(x) | = -e^(-x) < 0` in both case, `AA x in R`

` => | f(x) |` must be a decreasing function.

Correct Answer is `=>` (D) `| f(x) |` must be a decreasing function

Q 1653134044

The slope of the tangent to the curve `y = e^x cos x` is
minimum at `x = a, 0 le a le 2pi`, then the value of `a` is
BITSAT 2015

(A)

`0`

(B)

`pi`

(C)

`2pi`

(D)

`3pi//2`

Solution:

Let `m` be the slope of the tangent to the curve
`y = e^x cosx.`

Then, `m = (dy)/(dx) = e^x (cosx - sinx)`

` => (dm)/(dx) = e^x (cosx- sinx) +e^x (-sin x- cos x)`

` = -2e^x sinx`

and `(d^2m)/(dx^2) = -2 [ e^x sinx + e^x cosx ]`

`:. (dm)/(dx) = 0 => sinx = 0`

` => x = 0, pi , 2pi`

Clearly, `(d^2m)/(dx^2) > 0` for `x = pi`

Thus, y is minimum at `x = pi.`

Hence, `a = pi.`

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

Q 2325378261

The minimum value of `4^(sin x) + 4^(cos x)` is
BITSAT Mock

(A)

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

(B)

`2^(1- sqrt(2))`

(C)

`4^(1 - 2 sqrt (2) )`

(D)

`4^(1- sqrt (2))`

Solution:

`4^(sin x)` and `4^(cos x)` are both positive.

Their A.M. `ge` G.M.

`:. (4^(sin x) + 4^(cos x))/2 ge (4^(sin x + cos x))^(1/2)`

Minimum value `= 2^1 * 2^(sqrt(2) sin (pi/4 +x) )`

`=>` Minimum value of `4^(sin x) + 4^(cos x)` is

`2^(1 -sqrt (2) )`

`( :. ` minimum value of `sin (pi/4 +x)` is `-1 )`.

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

Q 2569580415

Let `f : R -> R` be defined by
` f(x) = { tt ((k - 2x , text(if) x <= -1),(2x + 3 , text(if) x > - 1) )`
If `f` has a local minimum at `x = - 1`, then the possible value of `k` is
BCECE Mains 2015

(A)

`- 1//2`

(B)

`-1`

(C)

`1`

(D)

`0`

Solution:

If `f( x)` has a local minimum at `x = -1`, then

`lim_(x -> 1^+) f(x) = lim_(x -> 1^-) f(x)`

`=> lim_(x -> -1^+) 2x + 3 = lim_(x -> -1^-) k - 2x`

`=> -2 + 3 = k + 2 => k = -1`

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

Q 2461134925

If `f(x) =80/(3x^4+8x^3-18x^2+60)` then the points of local maxima for the function `f(x)` are
UPSEE 2015

(A)

`1,3`

(B)

`-3,1`

(C)

`-1,3`

(D)

`-1,-3`

Solution:

We have,

`f(x)=80/(3^4+8x^3-18x^2+60)`

`=> f'(x) =(-80 (12x^3+24x^2-36 x))/((3x^4+8x^3-18x^2+60)^2)`

`=((-80)(12)(x)(x^2+2x-3))/((3x^4+8x^3-18x^2+60)^2)`

Put `f'(x) = 0`

`=> x=0,x=1` and `x=-3`

Clearly, the sign scheme of `f'( x)` is

Hence, `x = - 3` and `x = 1` are the points of maxima

Correct Answer is `=>` (B) `-3,1`

Q 2511356220

The angle between the tangents drawn from the origin to the circle `(x -7 )^2 + ( y + 1 )^2 = 25` is
BCECE Stage 1 2015

(A)

`pi/3`

(B)

`pi/6`

(C)

`pi/2`

(D)

`pi/8`

Solution:

Equation of circle is

`(x -7)^2 + (y + 1)^2 = 25`
. . Centre is (7, -1) and radius is 5.

Let y = m x be the tangent on the circle.
`:. ` Length of perpendicular from centre is equal to the radius of circle

`=> (7 m+1) /sqrt(1 +m^2) = pm 5`

`=> 49m^2 + 1 + 14m = 25 (1 + m^2 )`
`=> 24m^2 + 14m- 24 = 0`

`=>12m^2 +7m-12=0`

`=> 12m^2 + 16m- 9m -12= 0`

`=> 4m(3m+4)-3(3m+4)=0`

`=> (3m+4)(4m - 3)=0`

`=> m_1 = -4/3` and `m_2 =3/4`

`:. m_1m_2 = -4/3* 3/4 =-1`

`=> ` Tangent is perpendicular.

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

Q 2884580457

Let `f(x) = (1 + b^2)x^2 + 2bx + 1` and `m(b)` be the minimum value of `f(x)`. As `b` varies, the range of `m(b)` is

(A)

`[0, 2]`

(B)

`[ 0 , 1/2]`

(C)

`[ 1/2 , 1]`

(D)

`[ 0 ,1]`

Solution:

`f ' (x) = 2x (1 + b^2 ) + 2b = 0`

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

` f '' (x) = 2( 1 + b^2)` is always positive, so

that `f(x)` is minimum when

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

`:.` min `f(x) = (1 + b^2)(- b/(1+ b^2))^2 + 2b (- b/(1+ b^2))+ 1`

` = ((1 + b^2) - b^2)/(1 +b^2) = 1/(1 + b^2)`

` :. m (b) = 1/(1 + b^2) = + ve`

Clearly, `m( b)` is always greater than zero

and less than or equal to `1`.

So, the range of `m(b)` is `(0, 1]`.

Correct Answer is `=>` (D) `[ 0 ,1]`

Q 2854480354

The function `f(x) = log_e (x^3 + sqrt(x^6 + 1) )` is

(A)

even

(B)

odd

(C)

decreasing

(D)

None of these

Solution:

`f(x) = log { sqrt(x^6 + 1 + x^3 }`

`f (-x) = log { sqrt( x^6 + 1) - x^3}`

`:. f (-x) + f(x) = log {x^6 + 1 - x^6}`

`= log 1 = 0`

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

Hence, `f(x)` is odd.

Again,

` (dy)/(dx) = 1/( sqrt(x^6 + 1) + x^3) [ 3x^2 + (6x^5)/(2 sqrt (x^6 + 1 )) ]`

` = (3x^2)/sqrt(x^6 + 1) = ` Positive , ` AA x in R`

Hence , `f(x)` is an increasing function.

Correct Answer is `=>` (B) odd

Q 2427701681

The greatest value of `f(x)=(x+1)^(1//3)-(x-1)^(1//3)` on `[0,1]` is
UPSEE 2010

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`1/3`

Solution:

We have, `f(x) = (x + 1)^(1//3) - (x- 1)^(1//3)`

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

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

Clearly, `f' (x)` does exist at `x = ± 1`

`:. f'(x)= 0`

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

Clearly, `f' (x) ne 0` for any other value of

`x in [0, 1]`. The value of `f(x)` at `x = 0` is `2`.

Hence, the greatest value of `f(x)` is `2`.

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

Q 2322180031

The normal to the curve
`x = a (cos θ + θ sin θ), y = a (sin θ − θ cos θ)`
at any point `θ` is such that :
BITSAT Mock

(A)

it is at a constant distance from the origin

(B)

it makes a constant angle with x-axis

(C)

it passes through the origin

(D)

None of these

Solution:

`x = a (cos θ + θ sin θ)`

`(dx)/(dθ) = a (− sin θ + θ cos θ + sin θ)`

` (dx)/(dθ) = a θ cos θ` ...(1)

`y = a (sin θ − θ cos θ)`

`⇒ (dy)/(dθ) = a (cos θ + θ sin θ − cos θ)`

`(dy)/(dθ) = a θ sin θ` ...(2)

from (1) and (2)

`(dy)/(dx) = (dy)/(dθ) × (dθ)/(dx) = (a θ sin θ)/(a θ cos θ) = tan θ`

`∴` Slope of the normal

`= − 1/(tan θ) = − cot θ`

`∴` normal `= y − a (sin θ − θ cos θ)`

`= − cot θ [x − a (cos θ + θ sin θ)]`

[equation of normal `y − y_1 = m (x − x_1)]`

`⇒ y sin θ − a (sin^2 θ − θ sin θ cos θ)`

`= − cos θ (x − a cos θ − aθ sin θ)`

[multiplying both sides by `sin θ`]

`⇒ y sin θ − a sin^2 θ + aθ sin θ cos θ`

`= − x cos θ + a cos^2 θ + a θ sin θ cos θ`

`⇒ x cos θ + y sin θ − a (sin^2 θ + cos^2 θ) = 0`

`⇒ x cos θ + y sin θ − a = 0`

`∴` Length of perpendicular from `(0, 0)`

upon the normal

`= | (-a)/sqrt(cos^2θ + sin^2θ) | = a = ` constant.

Correct Answer is `=>` (A) it is at a constant distance from the origin

Q 2446223173

Let `f(x) = x(x-1)^2`, the point at which `f(x)`
assumes maximum and minimum are
respectively
UPSEE 2014

(A)

` 1/3 , 1`

(B)

`1 , 1/3`

(C)

`3 , 1`

(D)

None of these

Solution:

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

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

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

`=> f' (x) = (x - 1) (3x - 1)`

Using number line rule for `f'(x)`, we have

adjoining figure which shows `f'(x)` changes

from +ve to - ve at `x = 1/3`

Hence, at `x = 1/3`, we have maximum and `f'(x)`

changes sign from -ve to + ve at `x = 1` Hence

`f(x)` minimum at `x = 1`

Correct Answer is `=>` (A) ` 1/3 , 1`

Q 2426323271

Rectangles are inscribed in a circle of radius r.
The dimensions of the rectangle which has
the maximum area, are
UPSEE 2014

(A)

`r, r`

(B)

`2r, 2r`

(C)

`sqrt(2)r, sqrt(2)r`

(D)

None of the above

Solution:

Let `ABCD` be the rectangle inscribed in a circle

of radius `r`.

Let `AB = x`

and `BC = y`

then ` x^2 + y^2 = 4r^2`

`=> y = sqrt(4r^2 - x^2)` ......(i)

Area of rectangle,

`A = xy`

`= x sqrt(4r^2 - x^2) `

Let `u = A^2 = x^2 (4r^2 - x^2)`

` => (du)/(dx) = 8r^2x - 4x^3`

Put `(du)/(dx) = 0` for maxima or minima.

`:. 4x (2r^2 - x^2 ) = 0`

`=> x = sqrt(2)r`

Also, `(d^2 u)/(dx^2) = 8r^2 - 12x^2`

` :. ( (d^2u)/(dx^2) )_( x = sqrt(2)r) = 8r^2 - 24r^2 < 0`

`:. u` and `A` are maximum at `x = sqrt(2)r`.

From Eq. (i), `y = sqrt(2) r = x`

`:.` Dimensions of the rectangle are `sqrt(2)r` and `sqrt(2)r`.

Correct Answer is `=>` (C) `sqrt(2)r, sqrt(2)r`