Differential Calculus MCQ Level - 1

Let z = x² + xy + 3x + 2y + 5

For max. or min of z, we must have

at (–2, 1)
rt – s² = –1 < 0
∴  neither max. nor min.
The correct answer is: neither max nor min 

We get, x = ±cos x


 

∴ f(x) has maximum where x = cos x.

The correct answer is: cos x

Let l be the distance from (0, 0, 0) to (x, y, z) on the surface
z² = xy + 1

Solving, we get
x = 0, y = 0
r = 2, s = 1, t = 2
rt – s² = 3 > 0 and r > 0
Thus, minimum at (0, 0)
When, x = 0, y = 0, z = ±1
So, the points are (0, 0, 1) and (0, 0, –1)

The correct answer is: (0, 0, 1) and (0, 0, –1)

Since, x₁ < x₂
So, x₁ = –2  &  x₂ = 3

The correct answer is: –2, 3

Let the two numbers be x and y

and let z be the sum of squares.

= x² + (k – x)²

Hence, z is minimum at x = k/2 and its minimum value is k²/2.

The correct answer is: k²/2

 = –x log x

i.e., 1 + log x = 0
i.e., log x = –1

Hence, y is maximum where x = 1/e and maximum value is e^1/e.

The correct answer is: e^1/e

Let f(x, y) = x² + y²

For stationary points,  
⇒ (0, 0) is the only stationary point
r = f_xx = 2, t = f_yy = 2, s = f_xy = 0
rt – s² = 4 – 0 = 4 > 0 and r > 0
Thus, (0, 0) is a point of minima.

The correct answer is: minima

For stationary points, put dy/dx = 0
i.e. (x – a)²ⁿ(x – b)^2p+1 = 0
i.e. x = a, b
Nature at x = a : Since 2n is even, dy/dx does not change its sign while x passes through a.

∴  Neither maximum nor minima at x = a.

Nature at x = b : when x is slightly less than  is negative as 2p + 1 is odd.

Also, when x is slightly more than    is positive. Thus, y is minimum at x = b.

The correct answer is: neither maxima nor minima, minima

f(y) = (x – 1)(x – 2)(x – 3)
= x³ – 6x² + 11x – 6

Hence, f(x) has minima at   and f(x) has maxima at 
∴  the maximum value of f(x) is 

The correct answer is:  

For max. or min.,

Solving, we get
x = 1, y = 1
x = –1, y = –1

at (1, 1)     r = 2, s = 0, t = –2
rt – s² = –4 < 0

∴ neither max nor min at (1, 1)
at (–1, –1)  r = –2, s = 0, t = 2
rt – s² = –4 < 0
Thus, (–1, –1) is also a saddle point.

The correct answer is: both (1, 1) and (–1, –1)