JEE Main Maths Differential Calculus Previous Year Questions With Solutions

JEE Main Maths Differential Calculus previous year questions with solutions are provided on this page. The JEE Main Maths previous year questions on differential calculus given here will aid students to be well prepared to face the upcoming exam eventually. The solutions are given in a detailed manner. Students are advised to download the PDF of Differential Calculus JEE Main previous year’s questions with solutions.

Question 1:

A curve is represented parametrically by x = t + e^at and y = -t + e^at, t belongs to R and a>0. Then the curve touches the x-axis at

(a) (1, 0)

(b) (1/e, 0)

(c) (e, 0)

(d) (2e, 0)

Solution:

Given curve is x = t + e^at and y = -t + e^at.

Let the curve touches x-axis at (x₁,0), then the slope of the tangent at this point is O.

Now, dx/dt = 1 + ae^at

dy/dt = -1 + ae^at

dy/dx = (-1+ae^at)/(1+ae^at)

At (x₁, 0), (-1+ae^at1)/(1+ae^at1) = 0

ae^at1 = 1 …(1)

Also -t₁ + e^at1 = 0

⇒ e^at1 = t₁

From (1), we get

at₁ = 1

So t₁ = 1/a

Also, we get; ae = 1 (from (1))

⇒ a = 1/e

⇒ t₁ = e

Therefore x₁ = t₁ + e^at1

= e + e

= 2e

So the point of contact is (2e, 0).

Hence, option (d) is the answer.

Question 2:

The minimum value of α for which the equation (4/sin⁡x) + 1/(1-sin ⁡x) = α has at least one solution in (0, π/2) is

Solution:

f(x⁡) = 4/sin⁡x + 1/(1 – sin⁡x )

Let sin⁡x = t

Since x∈ (0, π/2) ⇒ 0 < t < 1

f(t⁡) = 4/t + 1/(1-t)

f'(t⁡) = (-4/t²) + 1/(1 – t⁡)² = 0

⇒ t = 2/3

f_min at t = 2/3

α_min = f(2/3)

= 4/(2/3) + 1/(1 – 2/3)

= 6 + 3

= 9

Hence, the minimum value of α is 9.

Question 3:

Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x∈R. If

(a) (9, 12)

(b) (6, 9)

(c) (3, 6)

(d) (0, 3)

Solution:

Given

⇒ f(x) f”(x) – f'(x)² =0

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

Then h'(x) = 0

⇒ h(x) = k

⇒ f(x)/f'(x) = k

⇒ f(x) = k f’(x)

⇒ f(0) = k f'(0)

⇒ k = 1/2

f(x) = ½ f'(x)

⇒∫ 2dx = ∫ f'(x)/f(x) dx

⇒ 2x = ln |f(x)| + C

As f(0) = 1 ⇒ C = 0

⇒ 2x = ln |f(x)|

⇒ f(x) = ±e^2x

As f(0) = 1

⇒ f(x) = e^2x

Therefore, f(1) = e² ≈ 7.38

Hence, option (b) is the answer.

Question 4:

If the curve y = ax² + bx + c, x ∈ R passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are:

(a) a = 1, b = 1, c = 0

(b) a = -1, b = 1, c = 1

(c) a = 1, b = 0, c = 1

(d) a = 1/2, b = 1/2 ,c = 1

Solution:

Given that y = ax² + bx + c

The curve passes through the point (1, 2).

⇒ 2 = a + b + c ..(i)

dy/dx = 2ax + b

(dy/dx)_(0,0) = b

Comparing y = x with y = mx+c, we get m = 1.

⇒ b = 1

Put b = 1 in (i)

⇒ a + c = 1

Since (0, 0) lies on curve,

c = 0, a = 1

⇒ a = 1, b = 1, c = 0.

Hence, option (a) is the answer.

Question 5:

If the curves,

(a) a + b = c + d

(b) a – b = c – d

(c) ab = (c+d)/(a+b)

(d) a – c = b + d

Solution:

Differentiating wrt x, we get;

Now,

Differentiating wrt x, we get;

m₁m₂ = –1

⇒

⇒ bdx² = – acy²…….(5)

(1)-(3) ⇒ (1/a – 1/c)x² + (1/b – 1/d) y² = 0

Solve above equation using equation (5), we get;

⇒ (c – a) – (d – b) = 0

⇒ c – a = d – b

⇒ c – d = a – b

Hence, option (b) is the answer.

Question 6:

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is

(a) (4, 5)

(b) (5, 4)

(c) (4, 4)

(d) (5, 5)

Solution:

Let x – 2 = t ⇒ dx = dt

and y + 4 = u ⇒dy = du

dy/dx = du/dt

du/dt = t + u/t ⇒ du/dt – u/t = t

Here, IF = 1/t

u. (1/t) = ∫ t.(1\t) dt

⇒ u/t = t + c

⇒ (y+4)/(y-2) = (x – 2) + c

Passing through (0, 0)

c = 0

⇒ (y + 4) = (x – 2)²

Hence, option (d) is the answer.

Question 7:

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria has increased by 20% in 2 hours. If the population of bacteria is 2000 after k / log_e (6 / 5) hours, then (k / log_e 2)² is equal to

(a) 4

(b) 2

(c) 16

(d) 8

Solution:

Let x be the number of bacteria at time t.

dx/dt ∝ x

dx/dt = λx

∫₁₀₀₀^x dx/x = ∫₀^t λ dt

(ln x) – (ln 1000) = λt

ln (x / 1000) = λt

Put t = 2, x = 1200

(ln 12 / 10) = 2λ

λ = (1 / 2) ln (6 / 5)

Now, ln (x / 1000) = (t / 2) ln (6 / 5)

x = 1000e^{t/2 ln (6 / 5)}

Given, x = 2000 at t = k / log_e (6 / 5)

2000 = 1000 e^{[k/2 ln (6 / 5)] * [ln (6 / 5)]}

2 = e^k/2

⇒ ln 2 = k / 2

k / ln 2 = 2

[k / ln 2]² = 4

Hence, option (a) is the answer.

Question 8:

The maximum slope of the curve y = (1/2)x⁴ – 5x³ + 18x² – 19x occurs at the point

(a) (2, 9)

(b) (2, 2)

(c) (3, 21 / 2)

(d) (0, 0)

Solution:

Given y = (1/2)x⁴ – 5x³ + 18x² – 19x

dy/dx = 2x³ – 15x² + 36x – 19

Let f(x) = 2x³ – 15x² + 36x – 19

f’(x) = 6x² – 30x + 36 = 0

x² – 5x + 6 = 0

x = 2, 3

f’’(x) = 12x – 30

f’’(x) < 0 for x = 2

Hence, at x = 2, slope is maximum.

y = 8 – 40 + 72 – 38

= 2

Maximum slope occurs at (2, 2).

Hence, option (b) is the answer.

Question 9:

If f(x) = (2x – 3π)⁵ + (4/3)x + cos x and g is the inverse function of f, then g’(2π) is equal to

(a) 7/3

(b) 3/7

(c) 30

(d) none of these

Solution:

Given that f(x) = (2x – 3π)⁵ + (4/3)x + cos x

g is the inverse function of f.

So f^-1(x) = g(x) ..(i)

We know, f f^-1(x) = x

So f(g(x)) = x

Differentiating w.r.t.x

f’(g(x)) g’(x) = 1

So g’(x) = 1/f’(g(x))

g’(2π) = 1/f’(g(2π)) ..(ii)

Let y = f(x)

x = f^-1(y)

x = f^-1(2π)

f(x) = 2π

The value of g(2π) is the same as the value of x for which f(x) = 2π.

⇒ f(x) = (2x – 3π)⁵ + (4/3)x + cos x = 2π ..(iii)

This is possible if 2x = 3π (so that (2x-3π)⁵ = 0)

⇒ x = 3π/2

Put x = 3π/2 in (iii)

0 + 2π + 0 = 2π

Therefore, x = 3π/2

So g(2π) = 3π/2

⇒ g’(2π) = 1/f’(3π/2)

f’(x) = 10(2x – 3π)⁴ + (4/3) – sin x

And f’(3π/2) = (4/3) + 1

= 7/3

g’(2π) = 1/(7/3)

= 3/7

Hence, option (b) is the answer.

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