AP CALCULUS PRACTICE TEST

Problem 1 :

Solution :

Integrating x^-3, we get

=  x^(-3+1)/(-3+1)

=  -x^-2/2

Applying the upper and lower limits, we get

=  -(2^-2)/2 + (1^-2/2)

=  -1/8 + (1/2)

=  (-1+4)/8

=  3/8

Option D is correct.

Problem 2 :

Solution :

f(x)  =  (2x+1)⁴

f'(x)  =  4(2x+1)³ (2)  ==>  8(2x+1)³

f''(x)  = 48(2x+1)²

f'''(x)  = 192(2x+1)

f'^v(x)  = 384

Option E is correct.

Problem 3 :

Solution :

y  =  3/(4+x²)

y  =  3(4+x²)^-1

dy/dx  =  -3(4+x²)^-2(2x)

dy/dx  =  -6x/(4+x²)²

Option A is correct.

Problem 4 :

Solution :

dy/dx  =  cos 2x

∫dy/dx  =  cos 2x

Integrating on both sides, we get

y  =   sin2x/2 + C

Option C is correct.

Problem 5 :

Solution :

=  lim n->∞   n²(4)/n²(1+10000/n)

=  lim n->∞ 4/(1+10000/n)

Applying the limit, we get

=  lim n->∞ 4/(1+10000/n)

=  4

So, option D is correct.

Problem 6 :

Solution :

y  =  ln(x/2)

Differentiating with respect to x, we get

dy/dx  =  [1/(x/2)] ⋅ (1/2)

dy/dx  =  1/x

Slope at x  =  4

dy/dx  =  1/4

Slope of the tangent at the point x = 4 is 1/4.

So, the correct option is B.

Problem 7 :

Solution :

Since the given function is even.

So, option D is correct.

Problem 8 :

Solution :

Given that,

s(t)  =  t²+4t+4

s'(t)  =  2t+4

a(t)  =  s''(t)  =  2

Acceleration is 2.

So, option B is correct.

Problem 9 :

Solution :

Let g(x)  =  1/√(x²+4)

More clearly f(x)  =  ln(x²)

f(g(x))  =  ln(1/√(x²+4))²

It is not true. So, let us apply the next option.

Let g(x)  =  √(x²+4)

f(g(x))  =  ln(√(x²+4))²

f(g(x))  =  ln((x²+4)

It is true. 

So, option C is correct.

Problem 10 :

Solution :

f(x)  =  x³-3x²

f'(x)  =  3x²-6x

f'(x)  =  3x(x-2)

Using first derivative, let us find critical points.

f'(x)  =  0

3x(x-2)  =  0

x  =  0 and x  =  2

Note :

If the first derivative is negative to the left of the critical point and positive to the right of critical point, it has relative minimum at c.

If the first derivative is positive to the left of the critical point and negative to the right of critical point, it has relative maximum at c.

Let x  = -1  ∈ (-∞, 0)

f'(x)  =  3x(x-2)

f'(-1)  =  9 > 0 (positive)

Let x  = 1  ∈ (0, 2)

f'(x)  =  3x(x-2)

f'(1)  =  -3 < 0 (negative)

So, relative maximum at x  =  0.

Option B is correct.

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