Applications of Derivatives | JEE Advanced - Free MCQ Practice Test

Consider the function f (x) = ax³ + bx² +  cx on [0, 1] then being a polynomial. It is continuous on  [0, 1], differentiable on (0, 1) and f (0) = f (1) = 0 [as given a + b + c = 0]
∴ By Rolle's theorem such that 2
f '(x) = 0 ⇒ 3ax² + 2bx +c = 0
Thus equation 3ax² + 2bx + c = 0 has at least one root in [0, 1].

which is max. when sin 2α = 1
i.e. α = 45º
∴ ΔABC is an isosceles triangle.

Dividing (2) by (1), we get

∴ Slope of normal = – cot θ
∴ Equation of normal is

⇒ y sin θ – a sin² θ + a sin θ cos θ
= – x cos θ + a cos² θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
As q varies inclination is not constant.
∴ (a) is not correct.
Clearly does not pass through (0, 0).

which is constant

y = a ln x + bx² + x
has its extremum values at x = – 1 and 2

has – 1 and 2 as its roots.
∴ 2b – 1 + a = 0 … (1)
8b + 2 + a = 0 … (2)
Solving (1) and (2) we get a = 2, b = – 1/2.

For y² = 4ax, y-axis is tangent at (0, 0), while for x² = 4ay, x-axis is tangent at (0, 0).
Thus the two curves cut each other at right angles. 

f '( x) = -(x+ 2)e^{- x} + e^{- x} = -( x+ 1)e^-x =0 ⇒ x = – 1
For x ∈ (-∞, -1), f '(x)>0 and for
x ∈ (-1, ∞), f '(x)< 0
∴ f (x) is increasing on (-∞,-1) and decr easing on (-1, -∞) 

Also at x = 0, y = 0, at x = 1, y = 0, and at x = 1/4, y > 0

∴ Max. value of y occurs at x = 1/4

Slope of tangent at (x, f (x)) is 2x + 1
⇒ f ' (x) = 2x + 1 ⇒ f (x) = x² + x + c
Also the curve passes through (1, 2)
∴ f (1) = 2
⇒ 2 = 1 + 1 + c ⇒ c = 0,  ∴ f (x) = x² + x

where sin² x is always +ve, when 0 < x < 1 . But to check Nr., we again let
h (x) = sin x – x cos x
⇒ h '(x) = x sin x > 0 for 0 < x<1  ⇒ h (x) is increasing
⇒ h (0) < h (x), when 0 < x<1
⇒ 0 <  sin x – x cos x, when 0 < x < 1
⇒ sin x – x cos x > 0, when 0 < x <1
⇒ f ' (x) > 0, x ∈ (0,1]

Here tan² x > 0 But to check Nr. we consider
p (x) = tan x – x sec² x
p'( x) = sec² x - sec² x- x.2 sec x.sec x tanx

⇒ p '(x) = -2 x sec² x tanx<0 for 0 < x <1
⇒ p (x) is decreasing, when 0 < x<1
⇒ p (0) > p (x)  ⇒ 0 > tan x – x sec² x
∴ g '(x) < 0

Hence g (x) is decreasing when 0 < x < 1.

We are given f (x) = sin⁴ x + cos⁴ x
⇒ f '(x) = 4 sin³ x cos x- 4 cos³ x sinx
= – 4 sin x cos x (cos² x – sin² x)  
= – 2. sin 2x cos 2x = – sin 4x
Now for f (x) to be increasing function
f '(x) > 0 ⇒ -sin4x >0 ⇒ sin4x < 0

Since, If f (x) increasing on (π /4,π /2)

It will be increasing on (π / 4, 3π / 8) .

From graph it is clear that both sin x and cos x in the interval (π / 2,π) are decreasing function.

∴ S is correct.
To disprove R let us consider the counter example :
f (x) = sin x on (0, π / 2) so that f '(x) = cosx
Again from graph it is clear that f (x) is increasing on (0, π / 2) but f '(x) is decreasing on (0, π / 2)
∴ R is wrong.

For decreasing function, f ' (x) < 0
⇒ e^x (x – 1) (x – 2) < 0  ⇒ (x – 1) (x – 2) < 0
⇒ 1 < x < 2,

It is clear from figur e that at x = 0, f (x) is n ot differentiable.

⇒ f (x) has neither maximum nor minimum at x = 0.

Let f ( x) = ex -1-x then f '(x) = e^x - 1>0 for x ∈ (0,1)
∴ f (x) is an increasing function.


∴ g (x) is decreasing on (0, 1) ∴ x > 0
⇒ g (x) < g (0)
⇒ log (1 + x) – x < 0 ⇒ log (1 + x) < x
∴ (b) holds. Similarly it can be shown that (c) and (d) do not hold.

f (x) = xe^{x (1–x)}

∴ f (x) is increasing on [–1/2, 1]

Tangent to y = x² + bx – b at (1, 1) is
y – 1 = (2 + b) (x – 1) ⇒ (b + 2) x – y = b + 1

⇒ b² + 2b + 1 = – 4 (b + 2) ⇒ b² + 6b + 9 = 0
⇒ (b + 3)² = 0 ⇒ b = – 3

f (x) = (1 + b²) x² + 2bx + 1
It is a quadratic expression with coeff. of x² = 1 + b² > 0.
∴ f (x) represents an upward parabola whose min value is   being the discreminant.

3 sin x – 4 sin³ x = sin 3x which increases for

The given curve is y³ + 3x² = 12y

There is only one function in option (a) whose critical point    for the rest of the parts critical point 0 ∉ (0, 1). It can be easily seen that functions in options (b), (c) and (d) are continuous on [0, 1] and differentiable in (0, 1).

∴ f is not differentiable at 1 / 2 ∈ (0,1)
∴ LMV is not applicable for this function in [0, 1]

Equation of tan gen t to the ellipse 

For Rolle's theorem in [a, b]
f (a) = f (b), ln [0, 1] ⇒ f (0) = f (1) = 0
∴ The function has to be continuous in [0, 1]

Applying L' Hospital's Rule

Let the polynominal be P (x) = ax² + bx + c
Given P (0) = 0 and P (1) = 1 ⇒ c = 0 and a +  b = 1
⇒ a = 1 – b
∴ P (x) = (1 – b) x² + bx
⇒ P' (x)  = 2 (1 – b) x + b

⇒ 2 (1 – b) x + b > 0
⇒ When x = 0, b > 0 and when x = 1, b < 2
⇒ 0 < b < 2
∴ S = {(1 - a) x² + ax,a∈ (0, 2)}

The equation of tangent to the curve y = e^x at (c, e^c) is

y - e^c = ec (x-c) … (1)

and equation of line joining (c –1, e^c–1) and (c +1, e^c+1) is

Subtracting equation (1) from (2), we get

∴ The two lines meet on the left of line x = c.

The given curves are
C₁ : y² = 4x ...(1)  and C₂ : x² + y² – 6 x + 1 = 0...(2)
Solving (1) and (2) we get
x² + 4x – 6x + 1 = 0 ⇒ x = 1 and ⇒ y = 2 or –2
∴ Points of intersection of the two curves are (1, 2) and  (1, –2).

∴ C₁ and C₂ touch each other at two points.

The graph of y = f (x) is as shown in the figure. From graph, clearly, there is one local maximum (at  x = –1) and one local minima (at x = 0)
∴ total number of local maxima or minima = 2.