35 Year JEE Previous Questions: Applications of Derivatives with solutions

Distance of origin from  

∴ Maximum distance from origin = a + b

Also f (x) is continuous and differentiable in [0, 1] and [0, 1]. So by Rolle’s theorem, f ' (x) = 0.
i.e ax² + bx + c = 0 has at least one root in [0, 1].

f (x) = 2x³ - 9ax² + 12a²x +1
f '(x) = 6x² - 18ax + 12a²; f"(x) = 12x- 18a
For max. or min.
6x² - 18ax + 12a² = 0 ⇒ x² - 3ax + 2a² = 0
⇒ x = aorx = 2a.At x = a max. and at x = 2a min

∴ p = a and q = 2a

As per question p² = q 
∴ a = 2a ⇒ a = 2ora = 0
but a > 0, therefore, a = 2.

f ''(x) = 6(x - 1). Inegrating, we get
f '( x) = 3x² -6 x+c
Slope at (2, 1)  = f '(2) = c = 3
[∴ slope of tangent  at (2,1) is 3]
∴ f '( x) = 3x² - 6 x + 3 = 3(x- 1)2
Inegrating again, we get  f (x) = (x -1)³+D
The curve passes through (2, 1)
⇒1= (2-1)³ +D ⇒ D = 0
∴ f (x) = (x – 1)³

∴ The slope of the normal  at θ = tan θ
∴ The equation of the normal at θ is
y -a sin θ = tan θ( x -a -a cosθ)

which always passes through (a, 0)

Let us defin e a function

Being  polynomial, it is continuous and differentiable, also,

∴ f (x) satisfies all condition s of Rolle’s theor em therefore f ’(x) = 0 has a root in (0, 1)

i.e. ax² + bx +c= 0 has at lease one root in (0, 1)

Area of rectangle ABCD = 2a cosθ (2b sin θ) = 2ab sin2θ

⇒ Area of greatest rectangle is equal to 2ab When sin 2θ = 1 .

x =a (cos θ + θ sinθ

   .....(1)

From equations (1) and (2), we get

Equation of normal at ' θ ' is y – a (sin θ – θ cos θ)

Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.

Given that

Again f (x) has root a,  ⇒ f (a)=0
∴ f (0) = f (a)
∴ By Roll’s theorem f'(x) = 0 has root between (0, α)

Hence f '(x) has a positive root smaller than a.

 has local min at x = 2.

Maximum value of sinq is 1 at 

Using Lagrange's Mean Value Theorem
Let f (x) be a function defined on [a, b]

Given f (x) = tan^–1 (sin x + cos x)

if f ’ (x) > O then f (x) is increasing function.
Hence f (x) is increasing, if 

Hence, f (x) is increasing when  

Given that p² + q² = 1
∴ p = cosθ and q = sinθ
Then p +q = cos θ + sinθ
We know that

Let f (x) = x⁷ + 14x⁵ + 16x³ + 30x –560

⇒The curve y = f (x) crosses x-axis only once.
∴ f (x) = 0 has exactly  one real root.

Given that f (x) = x | x | and g (x) = sin x
So that go f (x) = g (f (x)) = g (x | x |) = sin x | x |

Here we observe L (go f )' (0) =  0  =  R (go f)' (0)
⇒ go f is differentiable at x = 0 and (go f)' is continuous at x = 0

∴ L(go f)''  (0) ≠ R (go f )''  (0)
⇒ go f (x) is not twice differentiable at x = 0.
∴ Statement - 1 is true but statement -2 is false.

We have P (x) = x⁴ + ax³ + bx² + cx + d
⇒ P' (x) = 4x³ + 3ax² + 2bx + c
But  P' (0) = 0 ⇒ c = 0
∴ P(x) = x⁴ + ax³ + bx² + d
As given that P (– 1) < P (a)
⇒ 1 – a + b + d    <   1 + a + b + d ⇒ a > 0
Now P ' (x) = 4x³ + 3ax² +2bx = x (4x² + 3ax + 2b)
As P' (x) = 0, there is only one solution x = 0, therefore 4x² + 3ax + 2b = 0 should not have any real roots i.e. D < 0

∴ P (x) is an increasing function on (0,1) ∴ P (0)  <  P (a)
Similarly we can prove P (x) is decreasing on (– 1, 0)
∴ P (– 1) >  P (0)
So we can conclude that
Max P (x) = P (1) and Min P (x) = P (0) ⇒ P(–1) is not minimum but P (1) is the maximum of  P.

Since tangent is parallel to x-axis,

Equation of tangent is y – 3 = 0 (x – 2) ⇒ y = 3

This is true where k = – 1

Shortest distance between two curve occurred along the common normal Slope of normal to y² = x at point P(t², t) is – 2t and slope of line y – x = 1 is 1.
As they are perpendicular to each other

and shortest distance 

So shortest distance between them is

Volume of spherical balloon 

Differentiating both the sides, w.r.t 't' we get,

          ...(ii)
On solving (i) and (ii) we get  

So maxima at x = –1,2
Hence both the statements are true and statement 2 is a correct explanation for 1.

Equation of a line passing through (x₁,y₁) having slope m is given by y – y₁ = m (x–x₁)
Since the line PQ is passing through (1,2) therefore its equation is
(y – 2) = m (x – 1)
where m is the slope of the line PQ.
Now, point P (x,0) will also satisfy the equation of PQ
∴ y –2 = m (x –1)
⇒ 0 – 2 = m (x – 1)

Similarly, point Q (0,y) will satisfy equation of PQ

∴ equation of tangent is

y – 2 = 2(x – 2) or y + 2 = 2(x + 2)

⇒ x-intercept = ± 1.

Since, f and g both are continuous functions on [0, 1] and differentiable on (0, 1) then