JEE Main Maths Differentiation Previous Year Questions with Solutions

JEE Main Maths Differentiation previous year questions with solutions are available on this page. Students are recommended to revise and learn the differentiation JEE Main questions with solutions. Revising the previous years’ questions on differentiation will help students to score higher marks in the JEE Main exam. They can easily download the PDF of the differentiation question bank here for free.

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Question 1: If log_e (x+y) = 4xy, find (d²y)/(dx²) at x = 0.

Solution:

Given that log_e (x+y) = 4xy
Differentiating with respect to x, we get

(1/(x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]
1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)
If x = 0, then y = 1.
From (i), we get

1 + dy/dx = 4
⇒ dy/dx = 3
Again differentiate (i) w.r.t. x.
d²y/dx² = 4(x + y)[x (d²y)/(dx²) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))
At x = 0, y = 1, dy/dx = 3
d²y/dx² = 4(0 + 1)[0 + 2x³]+4[0 + 1](1 + 3)

= 40
So, d²y/dx² = 40.

Question 2: If P is a point on the parabola y = x² + 4, which is closest to the straight line y = 4x – 1, then the coordinates of P are:

(a) (-2, 8)

(b) (1, 5)

(c) (3, 13)

(d) (2, 8)

Solution:

Given that P is a point on the parabola, y = x² + 4

The tangent at P is parallel to the given line.

(dy/dx)_P = 4
=> 2x₁ = 4

x₁ = 2

The required point is (2, 8).

Hence, option (d) is the answer.

Question 3: d/dx ( log_e x) ( log_a x) =

(a) (1/x) log_a x

(b) (1/x) log_x x

(c) (2/x) log_a x

(d) (2/x) log x

Solution:

Let y = ( log_e x) ( log_a x)

= (log x/log e) ( log x/ log a)

= (log x)²/log a

Differentiating with respect to x, we get;

dy/dx = 2 log x (1/x)/log a

= (2/x) log_a x

Hence, option (c) is the answer.

Question 4: If y = sin² α + cos² (α+β) + 2 sin α sin β cos (α+β), then d³y/dα³ is, (keeping β as constant):

(a) 1

(b) 0

(c) cos (α + 3β)

(d) none of these

Solution:

Given y = sin² α + cos² (α+β) + 2 sin α sin β cos (α+β)

Using the identity, 2 sin A sin B = cos (A – B) – cos (A+B), we get;

y = sin² α + cos² (α+β) + [ cos (α-β) – cos (α+β)] cos (α+β)

= sin² α + cos² (α+β) + cos (α-β) cos (α+β) – [cos (α+β)]² …(i)

Differentiating with respect to α,

dy/dα = 2 sin α cos α – 2 cos (α+β) sin (α+β) – sin (α-β) cos (α+β) – cos (α-β) sin (α+β) + 2 cos (α+β) sin (α+β)

= sin 2α – 2 sin (α+β) – [sin (α-β+α+β)] + sin 2(α+β) (since sin (A + B) = sin A cos B + cos A sin B)

= sin 2α – 2 sin (α+β) – sin 2α + sin 2(α+β)

= 0

So, d³y/dα³ = 0

Hence, option (b) is the answer.

Question 5: Let the curve y = y(x) be the solution of the differential equation, dy / dx = 2 (x + 1). If the numerical value of the area bounded by the curve y = y(x) and the x-axis is 4√8/3, then the value of y(1) is equal to:

Solution:

y = x² + 2x + c

Area of rectangle ABCD = |(c – 1)√(1 – c)|
Area of parabola and x-axis = 2 ×(⅔) ×[(1 – c)^3/2]) = 4√8 / 3
1 – c = 2
c = -1
Equation of f (x) = x² + 2x – 1
f(1) = 1 + 2 – 1

= 2

Hence the value of y(1) = 2.

Question 6: If f (x) = sin [cos^-1 (1 – 2^2x) / (1 + 2^2x)] and its first derivative with respect to x is (- b / a) log_e 2 when x = 1, where a and b are integers, then the minimum value of |a² – b²| is:

Solution:

Given that f (x) = sin [cos^-1 (1 – 2^2x) / (1 + 2^2x)]

cos^-1 [1 – 4^x] / [1 + 4^x]

Let 2^x = t > 0

cos^-1 [(1 – t²) / (1 + t²)], t > 0 and t = tan θ

cos^-1 (cos 2θ) = 2θ ∈ (0, π), θ ∈ π / 2, 2θ ∈ (0, π)

⇒ 2θ

sin {cos^-1 [[1 – 4^x] / [1 + 4^x]]} = sin2θ

So y = [2 tanθ / (1 + tan² θ)]

= 2t / (1 + t²)

= [2 × 2^x] / [1 + 4^x]

dy/dx = [20 ln 2 – 32 ln 2] / 25

= – 12 ln 2 / 25

a = 25, b = 12

|a² – b²|_min = |25² – 12²| = 481

Hence, the minimum value of |a² – b²| is 481.

Question 7: Let f : S → S where S = (0, ∞) be a twice differentiable function such that f (x + 1) = x f(x). If g : S → R be defined as g (x) = log_e f (x), then the value of |g″(5) – g″(1)| is equal to:

(a) 197/144

(b) 187/144

(c) 205/144

(d) 1

Solution:

Given that f(x+1) = x f(x)

g(x) = log_e f (x) ..(i)

g(x+1) = log_e(f(x+1))

g(x+1) = log_e x + log_e f(x)..(ii)

(ii) – (i) gives

g(x+1) – g(x) = log_e x

g″(x+1) – g″(x) = – 1 / x²

g″(2) – g″(1) = –1

g″(3) – g″(2) = – 1/4

g″(4) – g″(3) = – 1/9

g″(5) – g″(4) = – 1/16

g″(5) – g″(1) = – [1 + (1/4) + (1/9) + (1/16)]

|g″(5) – g″(1)| = [144 + 36 + 16 + 9] / [16 × 9]

= 205/144

Hence, option (c) is the answer.

Question 8: Find y(π) if y(0) = 7 and dy/dx = 2(y – 2 sin x – 10)x + 2 cos x.

Solution:

Given that dy/dx = 2(y – 2 sin x – 10)x + 2 cos x

dy/dx – 2 cos x = 2(y – 2 sin x – 10)x

(d/dx (y – 2 sin x – 10))/((y – 2 sin x – 10) ) = 2x

⇒∫d(y – 2 sin x – 10)/((y – 2 sin x – 10) ) = ∫2x dx

⇒ log |y – 2 sin x – 10| = x² + C

When x = 0, y = 7

⇒ log |7 – 0 – 10| = 0 + C

So C = log 3

When x = π

⇒ log |y – 2 sin π – 10| = π² + log 3

⇒ log ((y – 10)/3) = π²

Question 9:

If

(a) x dy/dx – y

(b) (x dy/dx – y)²

(c) y dy/dx – x

(d) (y dy/dx – x)²

Solution:

Given that y = x log (x/(a+bx))

y/x = log x/(a+bx)

y/x = log x – log (a + bx)

Differentiating with respect to x, we get

(x dy/dx – y)/x² = (1/x) – b/(a+bx)

(x dy/dx – y) = x – bx²/(a+bx)

(x dy/dx – y) = (ax + bx² -bx²)/(a+bx)

(x dy/dx – y) = ax/(a+bx)…(i)

Again differentiate w.r.t.x

dy/dx + x d²y/dx² – dy/dx = ((a+bx)a – abx)/(a+bx)²

x d²y/dx² = a²/(a+bx)²

Multiply both sides by x²

x³ d²y/dx² = a²x²/(a+bx)²…(ii)

Squaring both sides (i)

(x dy/dx – y)² = a²x²/(a+bx)²

= x³ d²y/dx² from (ii)

=> x³ d²y/dx² = (x dy/dx – y)²

Hence, option (b) is the answer.

Question 10:

If e^y + xy = e, then the value of d²y/dx² for x = 0 is

(a) 1/e

(b) 1/e²

(c) 1/e³

(d) none of these

Solution:

Given that e^y + xy = e

When x = 0, we get y = 1

Differentiate w.r.t.x

e^ydy/dx + x dy/dx + y = 0 …(i)

Put x = 0 and y = 1, we get

dy/dx = -y/(x+ e^y) = -1/e

Again differentiate (i) w.r.t.x

e^y(dy/dx)² + e^yd²y/dx² + dy/dx + x d²y/dx² + dy/dx = 0

d²y/dx² [ e^y + x] = -2dy/dx – e^y(dy/dx)² …(ii)

Put x = 0 and y = 1 and dy/dx = -1/e in (ii), we get;

d²y/dx² [ e + 0] = 2/e – e/e²

d²y/dx² e = 1/e

d²y/dx² = 1/e²

Hence, option (b) is the answer.

Question 11: If √(1 – x²) + √(1 – y²) = a(x – y), then dy/dx =

(a) √(1 – x²)/√(1 – y²)

(b) √(1 – y²)/√(1 – x²)

(c) √(x² – 1)/√(1 – y²)

(d) none of these

Solution:

Given that √(1 – x²) + √(1 – y²) = a(x – y) …(i)

Let x = sin A and y = sin B

A = sin^-1 x

B = sin^-1 y

So (i) becomes

√(1 – sin² A) + √(1 – sin² B) = a(sin A – sin B)

cos A + cos B = a(sin A – sin B) …(ii)

We have cos A + cos B = 2 cos (A + B)/2 cos (A – B)/2

Also, sin A – sin B = 2 cos (A + B)/2 sin (A – B)/2

Substitute the above 2 equations in (ii)

2 cos (A + B)/2 cos (A – B)/2 = a 2 cos (A + B)/2 sin (A – B)/2

cos (A – B)/2 = a sin (A – B)/2

cot (A – B)/2 = a

(A – B) = 2 cot^-1 a

Substitute the values of A and B.

sin^-1 x – sin^-1 y = 2 cot^-1 a

Differentiating with respect to x, we get;

1/√(1 – x²) – 1/√(1 – y²) dy/dx = 0

1/√(1 – x²) = 1/√(1 – y²) dy/dx

dy/dx = √(1 – y²)/√(1 – x²)

Hence, option (b) is the answer.

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