Mathematics Tricks & Tips OF APPLICATION OF DERIVATIVEs FOR NDA
Click for Only Video

Finding Rate of Change Of Quantities

Q 2844080853

Given two squares of sides `x` and `y` such that `y = x + x^2`. What is the rate of change of area of the second square with respect to the area of the first square ?

(A)

`1 + 3x + 2x^2`

(B)

`1 + 2x + 3x^2`

(C)

`1- 2x + 3x^2`

(D)

`1 - 2x - 3x^2`

Solution:

`∵ ` Area of first square, `A_1 = x^2`

and area of second square,

`A_2 = y^2 = (x + x^2)^2`

`= x^2 + x^4 + 2x^3`

Now , `(dA_1)/(dx) = 2x`

and ` (dA_2)/(dx) = 2x + 4x^3 + 6x^2`

Hence, ` (dA_2)/(dA_1) = (2x + 4x^3 + 6x^2)/(2x)`

` = 1 + 2x^2 + 3x`
Correct Answer is `=>` (A) `1 + 3x + 2x^2`
Q 2932223132

The length `x` of a rectangle is decreasing at the rate of `3` cm/minute and the width `y` is increasing at the rate of 2cm/minute. When `x =10` cm and `y = 6` cm, find the rates of change of

(a) the perimeter and
(b) the area of the rectangle.

Solution:

Since the length `x` is decreasing and the width `y` is increasing with respect to time, we have

`(dx)/(dt) =-3` cm/min and `(dy)/(dx) = 2` cm/min

(a) The perimeter `P` of a rectangle is given by

`P = 2 (x + y)`

Therefore `(dP)/(dt) = 2((dx)/(dt) +(dy)/(dt))= 2 (-3+2)=-2` cm/min

(b) The area A of the rectangle is given by

`A= x * y`

Therefore `(dA)/(dt)= (dx)/(dt) * y + x * (dy)/(dt)`

`=-3 (6) + 10 (2) ` (as `x = 10` cm and `y = 6` cm)

`= 2 cm^2//min`
Q 2814680559

The rate of change of the surface area of a sphere of radius `r`, when the radius is increasing at the rate of `2` cm/s is proportional to

(A)

`1/r`

(B)

`1/r^2`

(C)

`r`

(D)

`r^2`

Solution:

Surface area of sphere, `S = 4 pi r^2`

and `(dr)/(dt) = 2`

`:. (dS)/( dt) = 4 pi xx 2 r (dr)/(dt) = 8 pi r xx 2 = 16 pi r`

` => (dS)/( dt) alpha r`
Correct Answer is `=>` (C) `r`
Q 1980334217

A balloon, which always remains spherical on inflation, is being inflated by pumping in `900` cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is `15` cm.
Class 12 Exercise 6.1 Q.No. 8
Solution:

Volume of the spherical balloon,

`V= 4/3 pi r^3`,

`(dV)/(dt) =d/(dr) ( 4/3 pi r^3) =3 xx 4/3 pi r^2 * (dr)/(dt)`

`(dV)/(dt) = 4 pi r^2 * (dr)/(dt)`

It is given that the balloon is inflated by pumping in `900` cubic em of gas per sec.

i.e., `(dV)/(dt) = 900 (cm)^(3)/sec` `:. 900 = 4 pi r^2 (dr)/(dt)`

When radius is `15` cm, putting `r = 15`

We have, `900 = 4 xx (22/7) xx 15 xx 15 xx (dr)/(dt)`

`:. (dr)/(dt) = (900 xx 7 )/(4 xx 22 xx 15 xx 15) =7/22`

Hence, the radius of the balloon is increasing at

the rate of `7/22` cm/sec i.e. `1/pi` cm/sec.
Q 2120667511

A ladder `5` m long is leaning against a wall. The
bottom of the ladder is pulled along the ground,
away from the wall, at the rate of `2`cm/s. How
fast is its height on the wall decreasing when
the foot of the ladder is `4`m away from the well ?
Class 12 Exercise Q.No. 0
Solution:

Let `AB` be the ladder and `OB` be the wall. At an

instant, let

`OA = x, OB = y, x^2 + y^2 = 25` ......(i)

On differentiating , `2x (dx)/(dt) + 2y (dy)/(dt) = 0`

`=> x (dx)/(dt) + y (dy)/(dt) = 0` ... (ii)

When `x = 4`, then from ( i), we have `16 + y^2 = 25`

`= y^2 = 9 => y = 3`

Now, `(dx)/( dt) = 0·02` m/sec.

Put these values in (ii) , `4 xx 0·02 + 3 xx (dy)/(dt) = 0`

`=> (dy)/(dt) = (-0·08)/3 = (- 8)/(300) = (-2)/(75)`

Hence, the height of the ladder on the wall is

decreasing at the rate of `2/(75)` m/sec. = `8/3` cm/sec.
Q 1970423316

The volume of a cube is increasing at the rate of `8 (cm)^(3)//s`. How fast is the surface area increasing when the length of an edge is `12` cm?
Class 12 Exercise 6.1 Q.No. 2
Solution:

Let `x` be the length of the cube

volume `V = x^3`, surface Area `S = 6x^2`

Rate of change of volume `= (dV)/(dt)`

`V= x^3 `, `:. (dV)/(dt) = (dV)/(dx) xx (dx)/(dt)`

Now, we have `(d V)/(dt) = 8 cm^(3) /sec, (dx)/ (dt) = 8/(3x^2)`

Further, `S= 6x^2`

`:. (dS)/(dt) = (dS)/(dx) xx (dx)/(dt) = (32)/x (cm)^(2)//sec`

When `x =12 cm , (dS)/(dt) = 32/12 = 8/3 = 2 (2/3) (cm)^(2) //sec`.
Q 1960123915

A stone is dropped into a quiet lake and waves move in circles at the speed of `5` cm/s. At the instant when the radius of the circular wave is `8` cm, how fast is the enclosed area increasing?
Class 12 Exercise 6.1 Q.No. 5
Solution:

Let `r` be the radius of a wave circle we have

`(dr)/(dt)= 5 ` cm/sec.

(A) Area =` pi r^2, (dA)/(dt) = (dA)/(dr) xx (dr)/(dt) = 2 pi r xx 5`

`(dA)/(dt) =10 pi r`

When `r = 8`, Rate of increase in area `= (dA)/(dt)`

`= 10 pi r = 10 pi xx 8 + 80 pi (cm)^(2)/sec`.

Finding whether a Function is Monotonic, Increasing , Decreasing , Strickly Increasing or Strickly Decreasing

Monotonic function `f` in an interval `I` means `f` is either is either increasing or decreasing.

To check increasing, decreasing , strickly Increasing, strickly decreasing find `f'(x)`

(i) `f` is strickly increasing in `(a,b)` if `f'(x) > 0` for each `x in (a,b)`

(ii) `f` is strickly decreasing in `(a, b)` if `f ′(x) < 0` for each `x ∈ (a, b)`


1. Let f be continuous on `[a, b]` and differentiable on the open interval `(a,b)`. Then

(a) `f` is increasing in `[a,b]` if `f ′(x) > 0` for each `x ∈ (a, b)`
(b) `f` is decreasing in `[a,b]` if `f ′(x) < 0` for each `x ∈ (a, b)`
(c) `f` is a constant function in `[a,b]` if `f ′(x) = 0` for each `x ∈ (a, b)`

(a) Let `x_1, x_2 ∈ [a, b]` be such that `x_1 < x_2`.

Then, by Mean Value Theorem , there exists a point `c` between `x_1` and `x_2` such that

`f (x_2) – f (x_1) = f ′(c) (x_2 – x_1)`

i.e. `f (x_2) – f (x_1) > 0` (as `f ′(c) > 0` (given))

i.e. `f (x_2) > f (x_1)`

Thus, we have

`x_1 < x_2 ⇒ f (x_1) < f (x_2 )`, for all `x_1, x_2 ∈[a,b]`

Hence, `f` is an increasing function in `[a,b]`.

The proofs of part (b) and (c) are similar. It is left as an exercise to the reader.

Q 2952623534

Show that the function f given by `f (x) = x^3 – 3x^2 + 4x, x ∈ R` is strictly increasing on R.

Solution:

Note that

`f ′(x) = 3x^2 – 6x + 4`

`= 3(x^2 – 2x + 1) + 1`

`= 3(x – 1)^2 + 1 > 0`, in every interval of `R`

Therefore, the function f is strictly increasing on `R`.
Q 2982623537

Prove that the function given by `f (x) = cos x` is

(a) strictly decreasing in `(0, π)`
(b) strictly increasing in `(π, 2π)`, and
(c) neither increasing nor decreasing in `(0, 2π)`.

Solution:

Note that `f ′(x) = – sin x`

(a) Since for each `x ∈ (0, π), sin x > 0`, we have `f ′(x) < 0` and so f is strictly decreasing in `(0, π)`.

(b) Since for each `x ∈ (π, 2π), sin x < 0`, we have `f ′(x) > 0` and so `f` is strictly increasing in `(π, 2π)`.

(c) Clearly by (a) and (b) above, `f` is neither increasing nor decreasing in `(0, 2π)`.
Q 2902823738

Find the intervals in which the function `f` given by `f (x) = x^2 – 4x + 6` is

(a) strictly increasing
(b) strictly decreasing

Solution:

We have

`f (x) = x^2 – 4x + 6`

or `f ′(x) = 2x – 4`

Therefore, `f ′(x) = 0` gives `x = 2`. Now the point `x = 2` divides the real line into two
disjoint intervals namely, `(– ∞, 2)` and `(2, ∞)` . In the interval `(– ∞, 2)`,

`f ′(x) = 2x – 4 < 0`.

Therefore, f is strictly decreasing in this interval. Also, in the interval `(2,∞) , f ′(x) > 0`

and so the function `f` is strictly increasing in this interval.
Q 2982023837

Find the intervals in which the function `f` given by `f (x) = 4x^3 – 6x^2 – 72x + 30`
is
(a) strictly increasing
(b) strictly decreasing.

Solution:

We have

`f (x) = 4x^3 – 6x^2 – 72x + 30`

or `f ′(x) = 12x^2 – 12x – 72`

`= 12(x^2 – x – 6)`

`= 12(x – 3) (x + 2)`

Therefore, `f ′(x) = 0` gives `x = – 2, 3`. The points `x = – 2` and `x = 3` divides the real line into
three disjoint intervals, namely, `(– ∞, – 2), (– 2, 3)` and `(3, ∞)`.

In the intervals `(– ∞, – 2)` and `(3, ∞), f ′(x)` is positive while in the interval `(– 2, 3),
f ′(x)` is negative.
Consequently, the function f is strictly increasing in the intervals `(– ∞, – 2)` and `(3, ∞)` while the function is strictly decreasing in the interval `(– 2, 3)`.

However, f is neither increasing nor decreasing in R.
Q 2982123937

Find intervals in which the function given by `f (x) = sin 3x in [ 0, pi/2]` is
(a) increasing
(b) decreasing.

Solution:

We have

`f (x) = sin 3x`

or `f ′(x) = 3cos 3x`

Therefore, `f ′(x) = 0` gives `cos 3x = 0` which in turn gives `3x= pi/2, (3 pi)/2` (as `x in [ 0, pi/2]` implies `3 x in [0, (3 pi)/2]`)

so, `x= pi/6` and `pi/2` The point `x=pi/6` divides the interval `[ 0, pi/2]` into two disjoint intervals `[0, pi/6)` and `(pi/6 , pi/2]`

Now, `f ′(x) > 0` for all `x in [ 0, pi/6)` as `0 le x < pi/6 => pi/6 => 0 le 3x < pi/2` and `f'(x) < 0` for all `x in (pi/6 , pi/2)` as `pi/6 < x < pi/2 => pi/2 < 3x < (3 pi)/2)`

Therefore, `f` is strictly increasing in `[0, pi/6)` and strictly decreasing in `(pi/6 , pi/2)` .

Also, the given function is continuous at `x = 0` and `x= pi/6` Therefore, by Theorem `1`, `f` is increasing on `[0, pi/6]` and decreasing on `[ pi/6 , pi/2]`.
Q 2814291159

If `f(x) = 3x^2 + 6x - 9`, then

(A)

`f(x)` is increasing in `(-1 , 3)`

(B)

`f(x)` is decreasing in `(3, oo)`

(C)

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

(D)

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

Solution:

`∵ f (x) = 3x^2 + 6x - 9`

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

`f ' (x) = 6x + 6 => f ' (x) < 0 , AA (- oo , - 1)`

`:. f (x)` is decreasing in `( - oo , -1)`.
Correct Answer is `=>` (D) `f(x)` is decreasing in `(- oo , -1)`
Q 2354256154

The set of all points for which
`f (x) = x^2e^(−x)` strictly increases is :
BITSAT Mock
(A)

`(0, 2)`

(B)

`(2, ∞)`

(C)

`(−2, 0)`

(D)

`(− ∞, ∞)`

Solution:

`f (x) = x^2e^(−x)`

`⇒ f ′ (x) = 2xe^(−x) − x^2e^(−x)`

`= xe^(− x) (2 − x)`

Since `f(x)` is strictly increasing

`∴ f ′ (x) > 0`

`⇒ xe^(− x) (2 − x) > 0`

`⇒ x (2 − x) > 0`

`⇒ x > 0, 2 − x > 0`

`⇒ x > 0, 2 > x`

`⇒ 0 < x < 2`

`⇒ x ∈ (0, 2)`.
Correct Answer is `=>` (A) `(0, 2)`
Q 2824380251

If `f(x) = xe^(x(1 - x)) `, then `f(x)` is

(A)

increasing on ` [ -1/2 , 1]`

(B)

decreasing on `R`

(C)

increasing on `R`

(D)

decreasing on ` [ -1/2 , 1]`

Solution:

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

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

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

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

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

Now, exponential function is always

positive and the sign of `f ' (x)` will be

opposite to the sign of A which is

negative in ` [ -1/2 , 1]`

Hence, `f'(x)` is positive in ` [ -1/2 , 1]` , so

that `f(x)` is an increasing function in

this interval.
Correct Answer is `=>` (A) increasing on ` [ -1/2 , 1]`
Q 2814280159

The interval in which function `f(x) = x - e^x + tan (2 pi)/7 ` increases is equal to

(A)

`(0 , oo)`

(B)

`(1 , oo)`

(C)

`(2 , oo)`

(D)

`(- oo , -1)`

Solution:

`(dy)/(dx) = 1 - e^x ` is positive, if `e^x < 1`

`=> x , 0 => x in ( - oo , 0)`

So, the interval `( - oo , - 1)` is part of

interval `(- oo , 0)`.
Correct Answer is `=>` (D) `(- oo , -1)`

Tangents and Normals

Q 2962134035

Find the point at which the tangent to the curve `y = sqrt(4x-3)-1` has its slope `2/3`

Solution:

Slope of tangent to the given curve at `(x, y)` is `(dy)/(dx) = 1/2 ( 4x -3)^(-1/2) 4 = 2/( sqrt(4x-3))`

The slope is given to be `2/3`.

so `2/sqrt(4x-3) = 2/3`

or `4x-3 = 9`

or `x = 3`

Now `y = sqrt(4x-3) -1`.

So when `x = 3 , y = sqrt(4(3) -3)-1 = 2`
Therefore, the required point is `(3, 2).`
Q 2912234130

Find points on the curve `x^2/4+y^2/(25) = 1` at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Solution:

Differentiating `x^2/4+y^2/(25) = 1` with respect to `x`, we get

`x/2 + (2y dy)/(25 dx) = 0`

or `(dy)/(dx) = (-25)/4 x/y`

(i) Now, the tangent is parallel to the x-axis if the slope of the tangent is zero which gives `(-25)/4 x/y = 0`. This is possible if `x = 0`. Then `x^2/4+y^2/(25) = 1` for `x = 0` gives `y^2 = 25` i.e. `y = pm 5`.

Thus, the points at which the tangents are parallel to the x-axis are `(0, 5)` and `(0, – 5).`

(ii) The tangent line is parallel to y-axis if the slope of the normal is 0 which gives `(4y)/(25 x) = 0` i.e. `y = 0`. Therefore `x^2/4+y^2/(25) = 1` for `y = 0` gives `x = pm 2`. Hence, the points at which the tangents are parallel to the y-axis are (2, 0) and (–2, 0).
Q 2942234133

Find the equation of tangent to the curve given by `x = a sin^3 t , y = bcos^3 t` at a point where `t = pi/2`

Solution:

Differentiating given equation with respect to `t`, we get

`(dx)/(dt) = 3 a sin^2 t cost` and `(dy)/(dt) = -3 b cos^2 t sint`

or `(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (-3 b cos^2 t sin t )/(3 a sin^2 t cost) = (-b cost)/(a sin t)`

Therefore, slope of the tangent at `t = pi/2` is

`[(dy)/(dx)]_(t = pi/2) = (-b cos \ \ pi/2)/(a sin \ \ pi/2) = 0`

Also, when `t = pi/2 , x = a` and `y = 0`.
Hence, the equation of tangent to the given curve at `t = pi/2` i.e. at `(a , 0)` is `y-0 = 0 ( x-a)` , i.e. `y = 0`
Q 2446891773

If `y = 4x- 5` is a tangent to the curve
`y ^2 = px^3 +q` at `(2, 3)`, then
UPSEE 2010
(A)

`p=2, q=-7`

(B)

`p=-2, q=7`

(C)

`p=-2, q=-7`

(D)

`p=2, q=7`

Solution:

Since, `(2, 3)` lies on `y ^2 = px^3 + q`

Therefore, `9 = 8p + q` .......... (i)

`=> y^2 = px^3 + q`

`=> 2y (dy)/(dx) = 3px^2`

`(dy)/(dx) =(3px^2)/(2y)`

`=> ((dy)/(dx))_((2,3)) =(12p)/6=2p`

Since, `y = 4x -5` is tangent to `y^ 2 = px^3 + qat`
`(2, 3)`. Therefore,

`:. ((dy)/(dx))_(2,3)=` Slope of the line `y = 4x - 5`

`=> 2p =4 => p=2`
Correct Answer is `=>` (A) `p=2, q=-7`
Q 2814380259

The point at which the tangent to the curve `y = 2x^2 - x + 1` is parallel to `y = 3x + 9` will be

(A)

`(2, 1)`

(B)

`(1, 2)`

(C)

`(3, 9)`

(D)

`(-2, 1)`

Solution:

Given equation is `y = 2x^2 - x + 1`

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

`(dy)/(dx) = 4x - 1`

Since, this is parallel to the given line

`y = 3x + 9`

Slope of second line `= (du)/(dx) = 3`

Therefore, these slopes are equal.

`=> 4x - 1 = 3 => x = 1`

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

Thus, the point is `(1, 2)`,
Correct Answer is `=>` (B) `(1, 2)`
Q 2523145041

If the normal to the curve `y = f(x)` at the point `(3, 4)` make an angle `(3pi)/4` with the positive x-axis, then `f' (3)` is
WBJEE 2010
(A)

`1`

(B)

`-1`

(C)

`-3/4`

(D)

`3/4`

Solution:

`(dy)/(dx) = f'(x)` Slope of normal

` = -1/(f'(x)) , -1/(f'(3))`

` = tan((3pi)/4) = -1`


`f'(3) = 1`
Correct Answer is `=>` (A) `1`
Q 2449880713

The slope of the normal to the curve `x = 1- a sin theta, y = b cos^2 theta` at ` theta = pi/2` is
BCECE Stage 1 2014
(A)

`a/(2b)`

(B)

`(2a)/b`

(C)

`a/b`

(D)

`(-a)/(2b)`

Solution:

Given. `x = 1 - a sin theta` and `y = b cos ^2 theta`.

On differentiating w.r.t `theta`. we get

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

and `(dy)/(d theta) = (2b)/a sin theta`

`:.` Slope of normal at the point `theta = pi/2` is

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

`= -1/((2b)/a sin (pi/2) ) = -a/(2b)`
Correct Answer is `=>` (D) `(-a)/(2b)`
Q 1652701634

The tangents to curve `y =x^3-2x^2+x-2` which are
parallel to straight line ` y = x ` are
UPSEE 2016
(A)

`x - y = 2` and ` x+ y = (86)/( 27)`

(B)

`x + y = 2`and ` x +y = (27)/(86)`

(C)

`x + y = 2 `and `x- y = ( 86)/(27)`

(D)

`x-y =2` and ` x-y = (86)/(27)`

Solution:

Only in `(D)`option slopes of both lines(tangents) are `1` that is equal to slope of `y=x` line


alternative methode

`(dy)/(dx) = 3x^2-4x+1` ........(1)


`y = x => (dy)/(dx) =1` .......(2)


from equation (1) and eq(2) ` 1= 3x^2-4x+1 =>x =0, 4/3`

`x =0`, gives ` y =-2 ` and `x = 4/3`gives `y = (-50)/(27)`


Thus the tangents to the curve at the points` (0,-2)`and `(4 / 3,(-50) / (27))` are parallel to line `y = x` .The
equations of these tangents are `y - (-2) =1(x - 0)` and `y - ((-50) / (27)) =1(x - 4 / 3)`

so that `x-y =2` and `x-y = (86)/(27)`
Correct Answer is `=>` (D) `x-y =2` and ` x-y = (86)/(27)`
Q 2550180914

The points at which the tangent to the curve `y = x^3 - 3x^2 - 9x + 7` is parallel to the x-axis are
BCECE Stage 1 2013
(A)

(3,- 20) and (- 1, 12)

(B)

(3, 20) and (1, 12)

(C)

(1,- 10) and (2, 6)

(D)

None of these

Solution:

Tangent to the curve is parallel to the axis is

when slope of the tangent is 0.

`:.` Equation of the curve is

` y = x^3 - 3x^2 - 9x + 7` .. (i)

` :. (dy)/(dx) = 3x^2 - 6x - 9`

Now, the tangent is parallel to x-axis, then

slope the tangent is zero or we can say that

`(dy)/(dx) = 0`.

`=> 3x^2 - 6x - 9 = 0`

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

`=> (x- 3)(x + 1) = 0`

`=> x = 3, -1`

When `x = 3`, then from Eq. (i). we get

` y = (3)^3 - (3) . (3)^2 - 9. 3 + 7`

` = 27 - 27 - 27 + 7 = -20`

When `x = -1` then frorn Eq. (i), we get

` y = (-1)^3 - 3 (-1)^2 - 9 (-1) + 7`

` = - 1 - 3+ 9 + 7 = 12`

Hence, the points at which the tangent is

parallel to x-axis are `(3,-20)` and `(-1, 12)`.
Correct Answer is `=>` (A) (3,- 20) and (- 1, 12)
Q 2219056819

The equation of the normal at the point
‘`t`’ to the curve `x = at^2, y = 2at` is :
BITSAT Mock
(A)

`tx + y = 2at + at^3`

(B)

`tx + y = 2at`

(C)

`tx + y = at^3`

(D)

None of these

Solution:

`x = at^2, y = 2at`

`⇒ (dx)/(xt) = 2at, (dy)/(dt) = 2a`

`⇒ (dy)/(dx) = (dy)/(dt) × (dt)/(dx)`

`= 2a × 1/(2at)`

`= 1/t`

`∴ `Slope of tangent `= 1/t`

Also slope of normal

`= − 1/text(slope of tangent)`

`= −t`

`∴` equation of normal is

`y − y_1` = slope of normal `(x − x_1)`

`⇒ y − 2at = −t (x − at^2)`

`⇒ tx + y = 2at + at^3`.
Correct Answer is `=>` (A) `tx + y = 2at + at^3`

How to Find Maxima and Minima

(1) Use `2^(nd)` derivative test.

Let `f` be a function defined on an interval `I` and `c ∈ I`. Let `f` be twice differentiable at `c`. Then

(i) `x = c` is a point of local maxima if `f ′(c) = 0` and `f ″(c) < 0`

The value `f (c)` is local maximum value of `f` .

(ii) ` x = c` is a point of local minima if `f ′(c) = 0` and `f ″(c) > 0`

In this case, `f (c)` is local minimum value of `f` .

(iii) The test fails if `f ′(c) = 0` and `f ″(c) = 0`.

In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion.

(2) Use `1^(st)` derivative method.
Q 2982234137

Find local maximum and local minimum values of the function `f` given by `f (x) = 3x^4+4x^3-12x^2+12`

Solution:

We have `f(x) = 3x^4+4x^3-12x^2+12`

or `f'(x) = 12x^3+12x^2-24x = 12x(x-1) (x+2)`

`f'(x) = 0` at `x = 0 , x = 1` and `x = -2`.
Now `f''(x) = 36x^2+24x-24 = 12 (3x^2+2x-1)`

or `{ tt ((f''(0) = -12 < 0) , (f''(1) = 48 > 0) , ( f''(-2) = 84 > 0))`

Therefore, by second derivative test, `x = 0` is a point of local maxima and local maximum value of `f` at `x = 0` is `f (0) = 12` while x = 1 and `x = – 2` are the points of local minima and local minimum values of `f` at `x = – 1` and `– 2` are `f (1) = 7` and `f (–2) = –20,`
respectively.
Q 2942334233

Find two positive numbers whose sum is `15` and the sum of whose squares is minimum.

Solution:

Let one of the numbers be `x`. Then the other number is `(15 – x)`. Let `S(x)` denote the sum of the squares of these numbers. Then

`S(x) = x^2+(15-x)^2 = 2x^2-30x+225`

`{ tt (( S' (x) = 4x-30) , (S' (x) = 4))`

Now `S′(x) = 0` gives `x = 15/2.` Also `S' (15/2) = 4 > 0`. Therefore, by second derivative test `x = 15/2` is the point of local minima of S. Hence the sum of squares of numbers is minimum when the numbers are `15/2` and `15 - 15/2 = 15/2`.
Q 2912434330

Find the shortest distance of the point `(0, c)` from the parabola `y = x^2`, where `0 le c le 5`.

Solution:

Let `(h, k)` be any point on the parabola `y = x^2`. Let `D` be the required distance between `(h, k)` and `(0, c).` Then

`D = sqrt((h-0)^2+(k-c)^2) = sqrt(h^2+(k-c)^2)` ..................(i)

Since `(h, k)` lies on the parabola `y = x^2,` we have `k = h^2`. So (i) gives

`D equiv D(k) = sqrt(k+(k-c)^2)`

or `D'(k) = (1+2(k-c))/(2 sqrt(k+(k-c)^2))`

Now `D'(k) = 0` gives `k = (2c - 1)/2`

Observe that when `k < (2c - 1)/2 ` , then `2 ( k-c) +1 < c` i.e. `D' (k) < 0`. Also when `k > (2c-1)/2` , then `D' (k) > 0`. So, by first derivative test `D(k)` is minimum at `k = (2c -1)/2`

Hence, the required shortest distance is given by `D ((2c-1)/2) = sqrt((2c-1)/2+((2c-1)/2 - c )^2) = sqrt(4c-1)/2`
Q 2902434338

Find absolute maximum and minimum values of a function `f` given by `f(x) = 12 x^(4/3) - 6 x^(1/3) , x in [-1, 1]`

Solution:

We have `f(x) = 12 x^(4/3) - 6 x^(1/3),`

or `f'(x) = 16^(1/3) - 2/(x^(2/3) =(2 (8x-1))/(x^(2/3))`

Thus, `f ′(x) = 0` gives `x= 1/8` Further note that `f ′(x) ` is not defined at `x = 0`. So the critical points are `x = 0` and `x=1/8` . Now evaluating the value of f at critical points `x=0, 1/8` and at end points of the interval `x = –1` and `x = 1`, we have

`f (–1) = 12(−1)^(4/3) − 6(−1)^(1/3) =18`

`f (0) = 12 (0) – 6(0) = 0`

`f(1/8) = 12(1/8)^(4/3) - 6(1/8) ^(1/3) =(-9)/4`

`f(1) = 12 (1)^(4/3) - 6 (1)^(1/3) = 6`

Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1

and absolute minimum value of f is `(-9)/4` that occurs at `x=1/8`
Q 2942534433

Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Solution:

Let `OC = r` be the radius of the cone and `OA = h` be its height. Let a cylinder with radius `OE = x` inscribed in the given cone Fig . The height `QE` of the cylinder is given by

`(QE)/(OA) = (EC)/(OC) \ \ ( text(since) DeltaQEC ~ Delta AOC)`

or `(QE)/h = (r-x)/r`

or `QE = (h ( r-x))/r`

Let `S` be the curved surface area of the given cylinder. Then

`S equiv S(x) = (2 pi x h(r-x))/r = (2pi h)/r ( rx - x^2)`

or `{ tt (( S' (x) = (2pi h)/r (r-2x)) , (S''(x) = (-4 pi h)/r))`

Now `S′(x) = 0` gives ` x = r/2` . Since `S″(x) < 0` for all `x, S'' (r/2) < 0.` So `x = r/2`. is a point of maxima of S. Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Q 2912434339

Let `AP` and `BQ` be two vertical poles at points `A` and `B`, respectively. If `AP = 16 m, BQ = 22 m` and `AB = 20 m,` then find the distance of a point `R` on `AB` from the point `A` such that `RP^2 + RQ^2` is minimum

Solution:

Let `R` be a point on `AB` such that `AR = x m.` Then `RB = (20 – x) m` (as `AB = 20 m`). From Fig.
We have `RP^2 = AP^2+AR^2`

and `RQ^2 = RB^2+BQ^2`

Therefore `RP^2+RQ^2 = AR^2+AP^2+RB^2+BQ^2`

` = x^2+ (16)^2+(20-x)^2+(22)^2`

`= 2x^2-40x+1140`

Let `S = S(x) = RP^2+RQ^2 = 2x^2-40x+1140`

Therefore `S'(x) = 4x-40`.
Now `S′(x) = 0` gives `x = 10.` Also `S″(x) = 4 > 0,` for all `x` and so `S″(10) > 0.` Therefore, by second derivative test, `x = 10` is the point of local minima of `S`. Thus, the distance of `R` from `A` on `AB` is `AR = x =10 m.`
Q 2932534432

Find the absolute maximum and minimum values of a function f given by

`f (x) = 2x^3 – 15x^2 + 36x +1` on the interval `[1, 5]`.

Solution:

We have

`f (x) = 2x^3 – 15x^2 + 36x + 1`

or `f ′(x) = 6x^2 – 30x + 36 = 6 (x – 3) (x – 2)`

Note that `f ′(x) = 0` gives `x = 2` and `x = 3.`

We shall now evaluate the value of f at these points and at the end points of the interval `[1, 5]`, i.e., at `x = 1, x = 2, x = 3` and at `x = 5`. So

`f (1) = 2(1^3) – 15 (1^2) + 36 (1) + 1 = 24`

`f (2) = 2(2^3) – 15 (2^2) + 36 (2) + 1 = 29`

`f (3) = 2(3^3) – 15 (3^2) + 36 (3) + 1 = 28`

`f (5) = 2(5^3) – 15 (5^2) + 36 (5) + 1 = 56`

Thus, we conclude that absolute maximum value of `f` on `[1, 5] ` is `56,` occurring at `x =5,` and absolute minimum value of f on `[1, 5]` is 24 which occurs at `x = 1`.
Q 2922434331

An open topped box is to be constructed by removing equal squares from
each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the
sides. Find the volume of the largest such box.

Solution:

Let `x` metre be the length of a side of the removed squares.

Then, the height of the box is `x`, length is `8 – 2x` and breadth is `3 – 2x` . If `V(x)` is the volume
of the box, then

`V(x) = x (3 – 2x) (8 – 2x)`

`= 4x^3 – 22x^2 + 24x`

Therefore , `{ tt (( V'(x) = 12x^ 2+ 44x +24 =4(x- 3)(3x- 2)),(V'(x) = 24 x -44))`

Now `V′(x) = 0` gives `x=3, 2/3`. But `x ≠ 3` (Why?)

Thus, we have `x= 2/3` . Now `V '' (2/3) = 24 (2/3) -44 =-28 < 0`

Therefore, `x = 2/3` is the point of maxima, i.e., if we remove a square of side `2/3` metre from each corner of the sheet and make a box from the remaining sheet, then the volume of the box such obtained will be the largest and it is given by `V (2/3) = 4 (2/3)^3 - 22 (2/3)^2 + 24 (2/3)`

`200/27 m^3`

 
SiteLock