JEE Advanced Maths Differentiability Previous Year Questions with Solutions

Differentiability JEE Advanced Maths previous year questions with solutions are given on this page. Chain rule of differentiation, Continuity, Leibnitz’s theorem, Rolle’s theorem, Lagrange’s mean value theorem, etc are the important topics in differentiability. As far as the JEE Advanced exam is concerned, differentiability has great importance. So students are recommended to revise these solutions so that they can easily score better ranks in the JEE Advanced exam.

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Question 1: Let [t] denote the greatest integer ≤ t and lim_{x→ 0} x[4/x] = A. Then the function, f(x) = [x²] sin πx is discontinuous, when x is equal to

(a) √(A + 1)

(b) √(A + 5)

(c) √(A + 21)

(d) √A

Answer: a

Solution:

lim_x→0 x[4/x] = A

lim_{x→ 0} x[4/x – {4/x}] = A

⇒ lim_x→0 4 – x{4/x} = A

⇒ 4-0 = A

As, f(x) = [x²] sin (πx) will be discontinuous at non-integers

And when x = √(A + 1)

⇒ x = √5, which is not an integer.

Hence, f(x) is discontinuous when x is equal to √(A + 1).

Hence, option a is the answer.

Question 2: Let f be a twice differentiable function such that g’(x) = -f(x) and f’(x) = g(x), h(x) = {f(x)}² + {g(x)}². If h(5) = 11, then h(10) is equal to

(a) 22

(b) 11

(c) 0

(d) 20

Answer: b

Solution:

Given g’(x) = -f(x) and f’(x) = g(x)

h(x) = {f(x)}² + {g(x)}²

Differentiate w.r.t x

h’(x) = 2f(x) f’(x) + 2g(x)g’(x)

= 2f(x) g(x) + 2g(x)(-f(x))

= 0

So h(x) is a constant.

h(5) = 11

So h(10) = 11

Hence, option (b) is the answer.

Question 3: A differentiable function f(x) is defined for all x > 0, and satisfies f(x³) = 4x⁴, for all x> 0. The value of f’(8) is

(a) 16/3

(b) 32/3

(c) 16√2/3

(d) 32√2/3

Answer: b

Solution:

f(x³) = 4x⁴

Differentiate w.r.t.x

f’(x³) 3x² = 16x³

f’(x³) = 16x/3

To find f’(8) put x = 2

f’(2³) = 16×2/3

= 32/3

Hence, option (b) is the answer.

Question 4: Let f: R→ R and g: R→ R be two non-constant differentiable functions. If f’(x) = (e^{(f(x) – g(x))}) g’(x) for all x ∈ R, and f(1) = g(2) = 1, then which of the following statement (s) is (are) true?

(a) f(2) < 1 – log_e 2

(b) f(2) > 1 – log_e 2

(c) g(1) > 1 – log_e 2

(d) g(1) < 1 – log_e 2

Answer: b, c

Solution:

Given f’(x) = (e^{(f(x) – g(x))}) g’(x)

⇒ e^-f(x) f’(x) = e^-g(x) g’(x)

Integrating both sides, we get

-e^-f(x) = -e^-g(x) + c

⇒ -e^-f(x)+e^-g(x) = c

⇒ -e^-f(1) + e^-g(1) = -e^-f(2) + e^-g(2)

So -e^-1 + e^-g(1) = -e^-f(2) + e^-1 (Since f(1) = g(2) = 1)

e^-f(2) + e^-g(1) = 2/e

⇒ e^-f(2) < 2/e and e^-g(1) < 2/e

⇒ -f(2) < ln 2 – 1 and -g(1) < ln 2 – 1

⇒ f(2) > 1 – ln 2 and g(1) > 1 – ln 2

Question 5: Let a, b ∈ R and f: R → R be defined by f(x) = a cos (|x³ – x|) + b|x| sin (|x³ + x|). Then f is

(a) differentiable at x= 0 if a = 0 and b = 1

(b) differentiable at x= 1 if a = 1 and b = 0

(c) not differentiable at x= 0 if a = 1 and b = 0

(d) not differentiable at x= 1 if a = 1 and b = 1

Answer: a, b

Solution:

f(x) = a cos (|x³ – x|) + b|x| sin (|x³ + x|)

If a = 0, b = 1

⇒ f(x) = |x| sin (|x³ + x|)

= x sin(x³ + x), which is differentiable everywhere.

If a = 1, b = 0

⇒ f(x) = cos (|x³ – x|)

= cos(x³-x), which is differentiable everywhere.

When a = 1, b = 1, f(x) = a cos (|x³ – x|) + b|x| sin (|x³ + x|), which is differentiable at x = 1.

Question 6: The derivative of tan^-1 (√(1 + x²) – 1)/x) with respect to tan^-1(2x√(1 – x²)/(1 – 2x²)) at x = 1/2 is

(a) 2√3/5

(b) √3/12

(c) 2√3/3

(d) √3/10

Answer: d

Solution:

Let u = tan^-1 ((√(1 + x²) – 1)/x)

Put x = tan θ

⇒ θ = tan^-1 x

So u = tan^-1(sec θ – 1)/tan θ

= tan^-1 tan (θ/2)

= θ/2

= ½ tan^-1 x

du/dx = ½ (1/(1 + x²))

Let v = tan^-1(2x√(1 – x²)/(1 – 2x²))

Put x = sin φ

φ = sin^-1x

v = tan^-1(2 sin φ cos φ/cos 2φ)

= tan^-1 tan 2φ

= 2φ

= 2sin^-1x

dv/dx = 2/√(1 – x²)

du/dv = (du/dx)/(dv/dx)

= √(1-x²)/4(1+x²)

du/dv at x = ½ = √3/10

Hence option d is the answer.

Question 7: If (a +√2 b cos x)(a – √2 b cos y) = a² – b², where a > b > 0, then dx/dy at (π/4, π/4) is

(a) (a – 2b)/(a + 2b)

(b) (a – b)/(a + b)

(c) (a + b)/(a – b)

(d) (2a + b)/(2a – b)

Answer: c

Solution:

(a + √2 b cos x)(a – √2 b cos y) = a² – b²

Differentiating both sides

-√2b sin x(a – √2b cos y) + (a + √2b cos x)√2b sin y (dy/dx) = 0

dy/dx = (√2b sin x)(a – √2 b cos y)/(a + √2 b cos x)√2b sin y

So dy/dx at (π/4, π/4) = (a – b)/(a + b)

Therefore dx/dy = (a + b)/(a – b).

Hence option c is the answer.

Question 8: Let f: [0, 2] → R be a function which is continuous on [0, 2] and its differentiable on (0, 2) with f(0) = 1.

(a) e² – 1

(b) e⁴ – 1

(c) e – 1

(d) e⁴

Answer: b

Solution:

F’(x) = f(x) 2x

F’(x) = f’(x) for all x belongs to (0, 2)

⇒ f(x) 2x = f’(x)

⇒ f’(x)/f(x) = 2x

On integrating w.r.t.x,we get

ln f(x) = x² + c

Since f(0) = 1

⇒ e^c = 1

Hence,

F(2) = e⁴ – 1

Hence option b is the answer.

Question 9: If y is a function of x and log(x + y) – 2xy = 0, then the value of y’(0) is equal to

(a) 1

(b) -1

(c) 2

(d) 0

Answer: a

Solution:

log(x + y) – 2xy = 0

When x = 0, y = 1

Differentiating w.r.t.x

⇒ dy/dx = ([1/(x + y)] – 2y)/(2x – 1/(x + y))

y’(0) = (1 – 2)/(0 – 1)

= 1

Hence, option a is the answer.

Question 10: If y = (sin x)^{tan x}, then dy/dx is equal to

(a) (sin x)^{tan x}(1 + sec²x log sin x)

(b) tan x(sin x)^{tan x – 1} cos x

(c) (sin x)^{tan x} sec²x log sin x

(d) tan x (sin x)^{tan x- 1}

Answer: a

Solution:

Given y = (sin x)^{tan x}

log y = tan x log sin x

Differentiate w.r.t.x

(1/y) dy/dx = sec²x log sin x + tan x .(1/sin x) cos x

dy/dx = (sin x)^{tan x} [ 1 + sec²x log sin x]

Hence option a is the answer.
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