JEE Advanced (Single Correct Type): Applications of Derivatives | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The side of an equilateral triangle is increasing at the rate of 2 cm/s. The rate at which area increases when the side is 10 is
(a) 10 cm²/s
(b) 10/3 cm²/s
(c) √3 cm²/s
(d) 10√3 cm²/s

Correct Answer is option (d)

Assume that x be the side of an equilateral triangle and A be its area.
Hence, A = (√3/4)x² square units
Differentiate both sides with respect to t, we get
dA/dt = (√3/4)²x(dx/dt) …(1)
Given that x = 10 cm and dx/dt = 2 cm/s
Now, substitute the values in (1), we get
dA/dt = (√3/4)²(10)(2)
Therefore, dA/dt = 10√3 cm2/s
Hence, option (d) is the correct answer.

Q.2. The equation of the normal to the curve y = sin x at (0, 0) is
(a) x = 0
(b) y = 0
(c) x+y =0
(d) x-y =0

Correct Answer is option (c)

Given: y = sin x

Hence, dy/dx = cos x

So, (dy/dx) at (0, 0) = cos 0 = 1

Thus, the slope of the normal = -1/(dy/dx) = -1/1 = -1.

Therefore, the equation of the normal at (0, 0) is

y-0 =-1(x-0)

y=-x

Hence, x+y =0 is the correct answer.

Q.3. If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is
(a) 1%
(b) 2%
(c) 3%
(d) 4%

Correct Answer is option (a)
We know that the formula for the time period of a pendulum is T = 2π × √(l/g)
dT/dl = π/√gl
Given that the percentage error in measuring the length “l” = 2%
Δl/l = 2/100
Δl = 2l/100
Therefore, the approximate error in T = dT = (dT/dl)Δl
dT = T/100 = 1% of T.
Hence, the percentage error in T is 1%.

Q.4. The absolute maximum value of y = x³ – 3x + 2 in 0 ≤ x ≤ 2 is
(a) 0
(b) 2
(c) 4
(d) 6

Correct Answer is option (c) 4

Given: y = x³ – 3x + 2

Therefore, y’ = 3x²-3.

For a point of absolute maximum or minimum, y’ = 0

Hence, x = ±1

Let y = f(x)

Therefore, f(0) = 03-3(0) +2 = 2

f(1) = 13– 3(1)+2 = 0

f(2) = 23 – 3(2) +2 = 4

Hence, f(x) achieves a maximum value of 4 when x = 2.

Hence, the correct answer is an option (c) 4.

Q.5. The line y = x + 1 is a tangent to the curve y² = 4x at the point
(a) (1, 2)
(b) (2, 1)
(c) (-1, 2)
(d) (1, -2)

Correct Answer is option (a)

y = x+1…(1)

y²= 4x …(2)

Substitute (1) in (2), we get

(x+1)² = 4x

x²+1+2x = 4x

x²-2x+1 = 0, which is equal to (x-1)²=0

⇒ x = 1

Now, substitute x = 1 in y=x+1, we get

y = 1+1 = 2.

Hence, the line y = x+1 is a tangent to the curve y² = 4x at the point (1, 2).

Therefore, option (a) (1, 2) is the correct answer.

Q.6. The function f(x) = x + cos x is
(a) Always increasing
(b) Always decreasing
(c) Increasing for a certain range of x
(d) None of these

Correct Answer is option (a) 

Given:

f(x) = x+cos x

f’(x) = 1 – sinx

f’(x)>0 for all vaues of x.

Since sin x is lying between -1 and +1, f(x) is always increasing

Q.7. Let the f: R → R be defined by f(x) = 2x + cos x, then f
(a) has a maximum, at x = 0
(b) has a minimum at x = 3t
(c) is an increasing function
(d) is a decreasing function

Correct Answer is option (c)

Given: f(x) = 2x + cos x

f’(x) = 2 – sin x

Now, f’(x) will be positive for all values of x

It means, f’ (x)>0 ∀ x∈R

Hence, the function is increasing for all values of x and it does not have any defined maximum and minimum. Hence, option (c) is the correct answer.

Q.8. The point(s) on the curve y = x², at which y-coordinate is changing six times as fast as x-coordinate is/are
(a) (6, 2)
(b) (2, 4)
(c) (3, 9)
(d) (3, 9), (9, 3)

Correct Answer is option (c)

dy/dt = (2x)(dx/dt)

⇒6. (dx/dt) = 2x.(dx/dt)

⇒x = 3

Now, substitute x = 3 in y=x², we get

y = (3)² = 9

Therefore, y = 9

Hence, the coordinate is (3, 9)

Therefore, the correct answer is an option (c).

Q.9. If y = x³ + x2 + x + 1, then y
(a) has a local minimum
(b) has a local maximum
(c) neither has a local minimum nor local maximum
(d) None of the above

Correct Answer is option (c)

If y = x³ + x² + x + 1, then y neither has a local minimum nor local maximum.

Q.10. The tangent to the curve y=e^2x at the point (0,1) meet the x-axis at
(a) (0, 1)
(b) (2, 0)
(c) (-2, 0)
(d) (-½, 0)

Correct Answer is option (d)

Given: y=e^2x

dy/dx = 2e^2x

Hence, dy/dx at (0, 1) is = 2e⁰ = 2(1) = 2

Therefore, the equation of a tangent is y-1 = 2(x-0)

⇒2x-y+1 = 0, if it meet the x-axis, then y=0

Hence, x = -½.

Hence, the point is (-½, 0).