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Q.22. The curve y = ax3 + bx2 + cx + 5, touches the x-axis at P(–2, 0) and cuts the y axis at a point Q, where its gradient is 3. Find a, b, c.           (1994 - 5 Marks)

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

Solution. Given that y = ax3 + bx2 + cx + 5 touches the x-axis at P (–2, 0)

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

an d – 8a + 4 b – 2c + 5 = 0 ...(2)

[∴ (– 2, 0) lies on curve]

Also the curve cuts the y-axis at Q

∴ For x = 0, y = 5 ∴ Q (0, 5) At Q gradient of the curve is 3

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

⇒ c = 3 ...(3)

Solving (1), (2) and (3), we get a = – 1/2, b = – 3/4 and c = 3.


Q.23. The circle x2 + y2 = 1 cuts the x-axis at P and Q. Another circle with centre at Q and variable radius intersects the first circle at R above the x-axis and the line segment PQ at S.

Find the maximum area of the triangle QSR.            (1994 - 5 Marks)

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

Solution. The given circle is x2 + y2 = 1 ...(1)

which intersect x-axis at P (– 1, 0) and Q (1, 0).

Let radius of circle with centre at Q (1, 0) be r, where r is variable.

Then equation of this circle is,

(x – 1)2 + y2 = r2        ...(2)

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

Subtracting (1) from (2) we get

(x – 1)2 – x2 = (r2 – 1)

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Substituting this value of x in (2), we get
Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
A will be  max. if A2 is max.
Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Differentiating A2 w.r. to r, we get Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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


Q.24. Let (h, k) be a fixed point, where h > 0, k > 0. A straight line passing through this point cuts the positive direction of the coordinate axes at the points P and Q. Find the minimum area of the triangle OPQ, O being the origin.              (1995 - 5 Marks)

Ans. 2 kh

Solution. Let the given line be  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE so that it makes an intercept of a units on x-axis and b units on y-axis. As it passes through the fixed point (h, k), therefore we must have

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

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

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


Q.25. A curve y = f(x) passes through the point P(1,1). The normal to the curve at P is a(y – 1) + (x – 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also obtain the area bounded by the y-axis, the curve and the normal to the curve at P.             (1996 - 5 Marks)

Ans. y = ea (x -1) ; 1 sq. unit

Solution. The normal to the curve at P is a (y – 1) + (x – 1) = 0

First we consider the case when a ≠ 0

Slope of normal at P (1 , 1) is  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

⇒ Slope of the tangent at (1, 1) is = a

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

But we are given that

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Where A is constant. As the curve passes through (1, 1)

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

∴ y = ea(x – 1) which is the required curve.
Now the area bounded by the curve, y-axis and normal to curve at (1, 1) is as shown the shaded region in the fig.

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

∴ Req. area = ar (PBC) = ar (OAPBCO) – ar (OAPCO)

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

Now we consider the case when a = 0. Then normal at (1, 1) becomes x – 1 = 0 which is parallel to y-axis, therefore tangent at (1, 1) should be parallel to x-axis. Thus

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE                       … (3)

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

From (3) and (4), we get k = 0 and required curve becomes y = 1

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

In this case the required area = shaded area in fig. = 1 sq. unit.


Q.26. Determine the points of maxima and minima of the function Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE where b > 0 is a constant.           (1996 - 5 Marks)

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

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

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

f '(x) = 0 ⇒ 16x2 - 8bx + 1=0 (for max. or min.) 

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

Above will give real values of x if b2 – 1 > 0 i.e. b > 1 or b < -1 . But b is given to be +ve. Hence we choose b > 1

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Its sign will depend on Nr, 16x2 – 1 as 8x2 is +ve. We shall consider its sign for Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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


Q.27. Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Where a is a positive constant. Find the interval in which f ' (x) is increasing.                  (1996 - 3 Marks)

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

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

Differentiating both sides, we have

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

Again differentiating both sides, we have

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

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

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

It is clear from number line that

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


Q.28. Let a + b = 4, where a < 2, and let g(x) be a differentiable function.             (1997 - 5 Marks)

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEEfor all x, prove that Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Solution.

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

Since g (x) is an increasing function (given)

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

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

Hence φ (t) increase as t increases.

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


Q.29. Suppose f(x) is a function satisfying the following conditions                       (1998 - 8 Marks)

(a) f(0) = 2, f(1) = 1,
 (b) f has a minimum value at x = 5/2, and
 (c) for all x,

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

where a, b are some constants. Determine the constants a, b and the function f(x).

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

Solution. Applying R3 → R3 - R1- 2R2 we get

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

⇒ f '(x) = 2ax+b

Integrating, we get , f (x) = ax2 + bx + C where C is an arbitrary constant. Since f has a maximum at x = 5/2,

f '(5 / 2) = 0 ⇒ 5a +b=0 … (1)

Also f (0) = 2 ⇒ C = 2

and f (1) = 1 ⇒ a + b + c = 1

∴ a + b = – 1 … (2)

Solving (1) and (2) for a, b we get,

a = 1/4, b = – 5/4

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


Q. 30. A curve C has the property that if the tangent drawn at any point P on C meets the co-ordinate axes at A and B, then P is the mid-point of AB. The curve passes through the point (1, 1). Determine the equation of the curve.               (1998 - 8 Marks)

Ans. xy = 1

Solution. Equation of the tangent at point (x, y) on the curve is

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

This meets axes in

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

Mid-point of AB is  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

We are given

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

Intergrating both sides,

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

Put x = 1, y = 1,
⇒ log 1 = – log 1 + c  ⇒ c = 0 ⇒ log y + log x = 0
⇒ log yx = 0 ⇒ yx = e0 = 1

Which is a rectangular hyperbola.


Q. 31. Suppose p(x) = a0 + a1x + a2x2 +....... + anxn. If Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEESubjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE             (2000 - 5 Marks)

Solution. Given that,  p (x) = a0 + a1 x + a2 x2 + … + anxn … (1)

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

It can be clearly seen that in order to prove the result it is sufficient to prove that | p '(1) | < 1

We know that,

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

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

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE   [Using equation (2) for x = 1]

⇒ | p(1) | < 0

But being absolute value, | p(1) | > 0 .
Thus we must have | p(1) | =0

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


Q. 32. Let - 1 < p < 1. Show that the equation 4x3 - 3x - p = 0 has a unique root in the interval[1/2, 1] and identify it.           (2001 - 5 Marks)

Solution. Given that -1 < p < 1 .
Consider f (x) = 4x3 – 3x – p = 0

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

∴  f (x) has at least one real root between [1/2, 1].
Also f '(x) = 12x2 - 3>0 on [1/2, 1]
⇒ f is increasing on [1/2, 1]
⇒ f has only one real root between [1/2, 1]

To find the root, we observe f (x) contains 4x3 – 3x which is multipe angle formula of cos 3θ if we put x = cos θ.
∴ Let the req. root be cos θ then,

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


Q. 33. Find a point on the curve x2 + 2y2 = 6 whose distance from the line x + y = 7, is minimum.             (2003 - 2 Marks)

Ans. (2, 1)

Solution. The given curve is  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Any parametric point on it is  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Its distance from line x + y = 7 is given by 

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

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

Q. 34. Using the relation 2(1 – cos x) < x2, x ≠ 0 or otherwise, prove that sin (tan x) Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE            (2003 - 4 Marks)

Solution. Given that 2 (1– cos x) < x2, x ≠ 0

To prove sin (tan x) > x, x ∈ [0,π / 4).

Let us consider f (x) = sin (tan x) – x

⇒ f '( x) = cos (tan x) sec2 x- 1

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

∴ f '(x) > 0 ⇒ f (x) is an increasing function.

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


Q. 35. If the function f : [0,4] → R is differentiable then show that             (2003 - 4 Marks)

(i) For a, b ∈ (0,4), (f(4))2 – (f(0))2 = 8f '(a) f(b)

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

Solution. Given that f is a differentiable function on [0, 4]

∴ It will be continuous on [0, 4]

∴ By Lagrange's mean value theorem, we get

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

Again since f is continuous on [0, 4] by intermediate mean value theorem, we get

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

[If f (x) is continuous on  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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

Multiplying (1) and (2) we get

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

or[f (4)]2 – [f (0)]2 = 8 f ' (a) f (b) 

Hence Proved.

(ii) To prove

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

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

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

Then clearly F (x) is differentiable and hence continuous on [0, 2]
By LMV theorem, we get some, μ∈ (0, 2)

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

Again by intermediate mean value theorem,

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

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


Q. 36. Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEEfor all x > 1 then prove that P(x) > 0 for all x > 1.                 (2003 - 4 Marks)

Solution. We are given that,

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

Multiplying by e–x, we get,

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

⇒ e–x P(x) is an increasing function.

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


Q. 37. Using Rolle’s theorem, prove that there is at least one root in (451/100, 46) of the polynomial          (2004 - 2 Marks)

P(x) = 51x101 – 2323(x)100 – 45x + 1035.

Solution. We are given,

P(x) = 51x101 – 2323(x)100 – 45x + 1035.

To show that at least one root of P (x) lies in (451/100, 46), using Rolle's theorem, we consider antiderivative of P (x)

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

Then being a polynominal function F(x) is continuous and differentiable.

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

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

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

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

∴ Rolle's theorem is applicable.
Hence, there must exist at least one root of F '(x) = 0

i.e. P (x) = 0 in the interval  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
Q. 38. Prove th at for Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE Explain the identity if any used in the proof.           (2004 - 4 Marks)

Solution. Let us consider,

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

⇒  f '(x) is a decreasing function.   .... (1)

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

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE           …(3)

Equations (1), (2) and (3) shows that.
⇒ There exists a certain value of x ∈ [0,π / 2] for which f ' (x) = 0 and this point must be a point of maximum for f (x) since the sign of f ' (x) changes from +ve to –ve.

Also we can see that f (0) = 0 and

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

Let x = p be the point at which the max. of f (x) occurs.
There will be only one max. point in [0, π/2]. Since f '(x) = 0 is only once in the interval.
Consider ,  x ∈ [0,p]

⇒ f ' (x) > 0  ⇒ f (x) is an increasing function.

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

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE             ..... (4)

Also for x ∈[p,π / 2]

⇒ f ' (x) < 0  ⇒ f (x) is decreasing function.

Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE              ..... (5)

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

Hence from (4) and (5) we conclude that

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

Q. 39. If |f (x1) – f (x2)| < (x1 – x2)2, for all x1, x2 ∈ R. Find the equation of tangent to the cuve y = f (x) at the point (1, 2).          (2005 - 2 Marks)

Ans. y = 2

Solution. Given that,  | f (x1) – f (x2) | < (x1 – x2)2, x1 , x2 ∈ R

Let x1 = x + h and x2 = x then we get

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

Taking limit as h → 0 on both sides, we get

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

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

⇒ f (x) is a constant function. Let f (x) = k i.e., y = k

As f (x) passes through (1, 2) ⇒ y = 2

∴ Equation of tangent at (1, 2) is,

y – 2 = 0 (x – 1) i.e. y = 2


Q. 40. If p(x) be a polynomial of degree 3 satisfying p(–1) = 10, p(1) = –6 and p(x) has maxima at x = – 1 and p'(x) has minima at x = 1. Find the distance between the local maxima and local minima of the curve.             (2005 - 4 Marks)

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

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

p (– 1) = 10

⇒ – a + b – c + d = 10 ...... (i)

p (1) = – 6

⇒ a + b + c + d = – 6 ...... (ii)

p (x) has max. at x = – 1

∴ p' (– 1) = 0

⇒ 3a – 2b + c = 0 ...... (iii)

p' (x) has min. at x = 1

∴ p'' (1) = 0

⇒ 6a + 2b = 0 ...... (iv)

Solving (i), (ii), (iii) and (iv), we get

From (iv), b = – 3a

From (iii), 3a + 6a + c = 0 ⇒ c = – 9a

From (ii), a – 3a – 9a + d = – 6 ⇒ d = 11a – 6

From (i), – a – 3a + 9a + 11a – 6 = 10

⇒ 16a = 16 ⇒ a = 1 ⇒ b = – 3, c = – 9, d = 5

∴ p (x) = x3 – 3x2 – 9x + 5

⇒ p' (x) = 3x2 – 6x – 9 = 0

⇒ 3 (x + 1) (x – 3) = 0 ⇒ x = – 1 is a point of max. (given) and x = 3 is a point of min.

[∴ max. and min. occur alternatively]

∴ points of local max. is (– 1, 10) and local min. is (3, – 22).

And distance between them is

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


Q. 41. For a twice differentiable function f (x), g(x) is defined as g(x) = (f '(x)2 + f"(x)) f(x) on [a, e]. If for a < b < c < d < e, f (a) = 0, f (b) = 2, f (c) = –1, f (d) = 2, f (e) = 0 then find the minimum number of zeros of g(x).           (2006 - 6M)

Ans. 6

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

Let h (x) = f (x) f ' (x)

Then, f (x) = 0 has four roots namely a, a, b, e

where b < a < c and c < b < d.

And f ' (x) = 0 at three points k1, k2, k3

where a < k1 < α, α < k2 < β, β < k3 < e

[∴ Between any two roots of a polyn omial fun ction f (x) = 0 there lies atleast one root of f ' (x) = 0]

∴ There are atleast 7 roots of f (x) . f ' (x) = 0

⇒  There are atleast 6 roots of  Subjective Type Questions: Applications of Derivatives - 2 | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

i.e. of g (x) = 0 

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