Partial Derivatives, Gradient- 1 - Free MCQ Practice Test with solutions,

∴ f (x )has minimum at x= 1 / 2 It Shows that a maximum value that will be at x = 4  or x = - 4 

At x = 4, f (x )= 10

∴ At x= −4, f (x ) = 18

∴ At x= −4, f (x ) has a maximum.

Option A is correct: the function y attains a maximum at x = e.

Write y = (ln x)/x.

Differentiate: dy/dx = (1 - ln x)/x².

Critical points satisfy dy/dx = 0, so ln x = 1 and hence x = e.

Compute the second derivative: d²y/dx² = (2 ln x - 3)/x³.

At x = e, d²y/dx² = (2·1 - 3)/e³ = -1/e³ < 0.

Because the second derivative is negative at the critical point, the function has a maximum at x = e. Thus option A is correct.

or pr – q² = 4 – 1 = 3 > 0 and r = +ve

so f(xy) is minimum at (0,0)

Hence, minimum value of d² at (0,0)

d2 = x² + y² + xy + 1 = (0)² + (0)² + (0)(0) + 1 = 1

Then the nearest point is

z² = 1 + xy = 1+ (0)(0) = 1

or z = 1

So maximum two extrema and three zero crossing  

B. 4 is the maximum value.

Rewrite the expression as y = (x-3)².

Because (x-3)² ≥ 0 for all real x, the minimum value of y is 0, attained at x = 3.

Evaluate y at the interval endpoints: at x = 2, y = 1; at x = 5, y = 4.

Comparing these values, the largest is 4, attained at x = 5; hence the maximum on the closed interval [2, 5] is 4.

If the interval were intended to be open, i.e. (2, 5), the supremum would still be 4 but it would not be attained (so there would be no maximum value in that open interval).

Let's solve it using partial diffraction:

Now comparing both sides:

A + B = 1  ----- ---Eq(ii)

-3A + 2B = 0  ----Eq(iiI)

after solving the above equations, we get:

after putting these value in Eq(i), we will get:

Concept

Calculation:

Given:

Similarly,

Now,