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Q. 1. Prove that the minimum value of  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEESubjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE             (1979)

Solution. 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 2. Let x and y be two real variables such that x > 0 and xy = 1. Find the minimum value of x+y.            (1981 - 2 Marks)

Ans. 2

Solution. Given that x and y are two real variables such that x > 0 and xy = 1.

To find the minimum value of x + y.

Let S = x + y

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 3. For all x in [0, 1], let the second derivative f ¢¢ (x) of a function f(x) exist and satisfy | f '' (x)| < 1. If f (0) = f (1), then show that | f '(x)| < 1 for all  x in [0, 1].    (1981 - 4 Marks)

Solution. We are given that

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Here f (x) is continuous on [0, 1], differentiable on (0, 1) and  f (0) = f (1)
∴ By Rolle's thm.,
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE… (1)  
Now there may be three cases for x Î[0,1]
(i) x = c (ii) x > c (iii) x < c

Case I : For x = c.
If x = c then f '(x) = 0<1 [from (1)]
Hence the result | f ' (x) |<1 is obtained in this case.

Case II : For x > c
Consider the interval [c, x].

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
As f '(x) is continuous on [c, x] and differentiable on (c, x)
∴ | f ' (x) |<1 . Hence the result in this case.

Case III : For x < c Consder the interval [x, c].
As f '(x) is continuous on [x , c] and differentiable on (x, c)

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

∴ | f "(x) |<1 hence the result in this case.
Combining all the three cases we get

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 4. Use the function f (x) =x1/ x , x > 0. to determine the bigger of the two numbers eπ and πe            (1981 - 4 Marks)

Ans. eπ 

Solution.

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
⇒ Raising to the power pe on both sides we get
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 5. If f(x) and g(x) are differentiable function for 0 < x < 1 such that f(0) = 2, g(0) = 0,  f(1) = 6; g(1) = 2, then show that there exist c satisfying 0< c < 1 and f ' (c) = 2g' (c).           (1982 - 2 Marks)

Solution. Given that f (x) and g (x) are differentiable for x ∈ [0,1] such that  f (0) = 2 ; f (1) = 6,   g (0) = 0 ; g (1) = 2

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Let us consider h (x) = f (x) – 2g (x) Then h (x) is continuous on [0, 1] and differentiable on (0, 1)
Also h (0) = f (0) – 2g (0) = 2 – 2 × 0 = 2
h (1) = f (1) – 2g (1) = 6 – 2 × 2 = 2
∴ h (0) = h (1)
∴ All the conditions of Rolle's theorem are satisfied for h(x) on [0, 1]

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 6. Find the shortest distance of the point (0, c) from the parabola y = x2 where 0 < c < 5.  (1982 - 2 Marks)

Ans. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Solution. (0, c), y = x2, 0 < c < 5 .

Any point on parabola is (x, x2)

Distance between (x, x2) and (0, 1) is

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

To minimum D we consider

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

which is minimum when  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 7. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEEfor all positive x where a > 0 and b > 0 show that 27ab2 > 4c3 .      (1982 - 2 Marks)

Solution. 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

To show that 27 ab2 > 4c.
Let us consider the function f (x) = ax2 + b/x
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

As a, b are +ve, cubing both sides we get

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 8. Show that 1 + x Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE                      (1983 - 2 Marks)

Solution. To show

  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Hence f (x) is increasing function.

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 9. Find the coordinates of the point on the curve Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEEwhere the tangent to the curve has the greatest slope.             (1984 - 4 Marks)

Ans. (0, 0)

Solution. Equation of the curve is given by

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE             ...(1)

Differentiating with respect to x, we get

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
For the greatest value of slope, we have
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Again we find,
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Thus, second order derivative at x = 0 is negative and second order derivative at x =±√3 is positive.

Therefore, the tangent to the curve has maximum slope at (0, 0).


Q. 10. Find all the tangents to the curve y = cos(x + y), - 2p < x < 2π , that are parallel to the line x + 2y = 0.                 (1985 - 5 Marks)

Ans. 2x + 4y - π = 0  

2x + 4y + 3π = 0

Solution. Equation of given curve y = cos (x + y), -2p < x <

Differentiating with respect to x

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE                                ...(1)

Since the tangent to given curve is parallel to x + 2y = 0

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ 2sin (x + y) = 1 + sin (x + y)
⇒ sin (x + y) = 1
Thus, cos (x + y) = 0
Using equation of curve and above result, we get, y = 0

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Thus the points on curve at which tangets are parallel to given line are (π/2, 0) and (– 3π/2, 0)

The equation of tangent at (π/2, 0) is

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Thus the required equations of tangents are

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 11. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEEFind the intervals in which λ should lie in order that f(x) has exactly one minimum and exactly one maximum.         (1985 - 5 Marks)

Ans. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Solution. The given function is,

f (x) = sin3 x + λ sin2 x for – π/2 <  x  < π/2

∴ f ' (x) = 3 sin2 x cos x + 2λ sin x cos x

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

So, from f ' (x) = 0, we get x = 0 or 3 sin x + 2λ = 0

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

f " (x) = 3 sin x cos2 x = – 2λ cos2 x

Now, if 0  < x < π/2, then – 3/2 < λ < 0 and therefore f " (x) > 0.
⇒ f (x) has one minimum for this value of λ.

Also for x = 0, we have f " (0) = 2λ < 0, That is f (x) has a maximum at x = 0

Again if – p/2 < x < 0, then 0 <  λ <  3/2 and therefore f " (x) = – 2λ cos2 x < 0.

So that f (x) has a maximum.

Also for x = 0, f " (a) = 2λ > 0 so that f (x) has a minimum.

Thus, for exactly one maximum and minimum value of f (x),λ must lie in the interval

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 12. Find the point on the curve 4x2 + a2 y= 4a2, 4 < a2 < 8 that is farthest from the point (0, – 2).                (1987 - 4 Marks)

Ans. (0, 2)

Solution. The equation of given curve can be expressed as

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Clearly it is the question of an ellipse

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Let us consider a point P (a cos θ, 2 sin θ) on the ellipse.

Let the distance of P (a cos θ, 2 sin θ) from (0, – 2) is L.

Then, L2 = (a cos θ – 0)2 + (2 sin θ + 2)2

⇒ Differentiating with respect to θ, we have

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
For max. or min. value of L we should have

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

∴ L is max. at θ = π/2 and the farthest point is (0, 2).


Q. 13. Investigate for maxima and minima the function Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Ans. f is min at x = 7/5

Solution. We have,

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Then using the theorem, 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

We get, 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

For extreme values f ' (x) = 0 ⇒ x = 1, 2, 7/5
Now, f " (x) = (x – 2)2 (5x – 7) + 2 (x – 1) (x – 2) (5x – 7) + 5 (x – 1) (x – 2)2
At x = 1,  f " (x) = 1 (– 2) = – 2 < 0
∴ f is max. at x = 1
At  x = 2 f " (x) = 0
∴ f is neither maximum nor minimum at x = 2.

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 14. Find all maxima and minima of the function y = x( x - 1)2 , 0 < x < 2     (1989 - 5 Marks)

Ans. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Solution. 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Now the curve cuts the axis x at (0, 0) and (1, 0). When x increases from 1 to 2, y also increases and is +ve.
When y = 2, x (x – 1)2 = 2
⇒ x = 2
Using max./min. values of y and points of intersection with x-axis, we get the curve as in figure and shaded area is the required area.
∴ The required area = Area of square  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 15. Show that 2sin x + tan x >3x Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE                (1990 -  4 Marks)

Solution. Let f (x) = 2 sin x + tan x – 3x on 0 < x < π /2

then f ' (x) = 2 cos x + sec2 x – 3 and f " (x)

= – 2 sin x + 2 sec2 x tan x  

= 2 sin x [sec3 x – 1]

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ f ' (x) is an increasing function on 0 < x <π / 2.

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ f (x) is an increasing function on Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE  Hence proved


Q. 16. A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the triangle PQR.     (1990 -  4 Marks)

Ans. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Solution. As QR || XY diameter through P is ⊥ QR. 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Now area of ΔPQR is given by  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

But QR = 2.QA = 2r sin 2θ and PA = OA + OP = r cos 2θ + r

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
= r2. 2 sin θ cos θ . 2 cos2 θ = 4 r2 sin q cos3 θ

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

∴ A is maximum at q = 30°

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 17. A window of perimeter P (including the base of the arch) is in the form of a rectangle surmounded by a semi circle. The semi- circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass transmits three times as much light per square meter as the coloured glass does.

What is the ratio for the sides of the rectangle so that the window transmits the maximum light ? (1991 -  4 Marks)


Ans. 6 + π : 6


Solution. Let ABCEDA be the window as shown in the figure and let 

AB = x m
 BC = y m

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Then its perimeter including the base DC of arch

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE            ...(1)

Now, area of rectangle ABCD = xy

and  area of arch Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Let λ be the light transmitted by coloured glass per sq. m. Then 3λ will be the light transmitted by clear glass per sq. m.

Hence the area of light transmitted  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE          ...... (2)

Substituting the value of y from (1) in (2), we get

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

For A to be maximum  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

[Using value of P from (1)]

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

∴The required ratio of breadth to length of the rectangle

= 6 + π : 6


Q. 18. A cubic f (x) vanishes at x = 2 and has relative minimum / maximum at  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE cubic f(x).    (1992 - 4 Marks)

Ans. x3 + x2 -x+ 2


Solution. Let f (x) = ax3 + bx2 + cx + d

ATQ, f (x) vanishes at x = – 2

⇒ – 8a + 4b – 2c + d = 0     ...(1)

f ' (x) = 3ax2 + 2bx + c

Againg ATQ, f (x)  has relative max./min at

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ f ' (– 1) = 0 = f ' (1/3)

⇒ 3a – 2b + c = 0           ... (2)

and a + 2b + 3c = 0          ... (3)

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ b + 3d = 7            ...(4)

From (1), (2), (3), (4) on solving, we get

a = 1, b = 1, c = – 1, d = 2

∴ The required cubic is x3 + x2 – x + 2.


Q. 19. What normal to the curve y = x2 forms the shortest chord?    (1992 - 6 Marks)

Ans. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Solution. The given curve is y = x2 ...(1)

Consider any point A (t, t2) on (1) at which normal chord drawn is shortest.

Then eq. of normal to (1) at A (t, t2) is

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE            ...(2)

This normal meets the curve again at point B which can be obtained by solving (1) and (2) as follows :

Putting y = x2 in (2), we get

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
For shortest chord, we have to minimize Z, and for that dZ/dt =0 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 20. Find the equation of the normal to the curve y = (1 + x)y+ sin -1 (sin2x) at x = 0         (1993 - 3 Marks)

Ans. x +y = 1

Solution. The given curve is y = (1+ x)y + sin–1 (sin2 x)

Here at x = 0, y = (1 + 0)y + sin–1 (0) ⇒ y = 1

∴  Point at which normal has been drawn is (0, 1).

For slope of normal we need to find dy/dx, and for that we consider the curve as  Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
∴ Equation of normal to given curve at (0, 1) is

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE


Q. 21. Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Find all possible real values of b such that f(x) has the smallest value at x = 1.   (1993 - 5 Marks)

Ans. b ∈ (-2, - 1) ∪ (1,∞)

Solution. 

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

We can see from definition of the function, that

f (1) = 2(1) – 3 = – 1

Also f (x) is increasing on [1, 3], f ' (x) being 2 > 0.
∴ f (1) = – 1 is the smallest value of f (x)
Again f ' (x) = – 3x2 for x ∈ [0, 1] such that f ' (x) < 0
⇒ f (x) is decreasing on [0, 1]
∴ For fixed value of b, its smallest occur when x → 1

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

As given that the smallest value of f (x) occur at x = 1
∴ Any other smallest value > f (1)

Subjective Type Questions: Applications of Derivatives - 1 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ b ∈ (-2, - 1) ∪ (1,∞)

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