Test: Maxima Minima (Derivative) - JEE MCQ

# Test: Maxima Minima (Derivative) - JEE MCQ

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## 10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Maxima Minima (Derivative)

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Test: Maxima Minima (Derivative) - Question 1

### The maximum and minimum values of f(x) =  are

Detailed Solution for Test: Maxima Minima (Derivative) - Question 1

f(x) = sinx + 1/2cos2x
⇒ f'(x) = cos x – sin2x
Now, f'(x) = 0 gives cosx – sin2x = 0
⇒ cos x (1 – 2 sinx) = 0
⇒ cos x = 0, (1 – 2 sinx) = 0
⇒ cos x = 0, sinx = 1/2
⇒ x = π/6 , π/2
Now, f(0) = 1/2,
f(π/6) = 1/2 + 1/4 = 3/4,
f(π/2) = 1 – 1/2 = 1/2
Therefore, the absolute max value = 3/4 and absolute min = 1/2

Test: Maxima Minima (Derivative) - Question 2

### ​Find the points of local maxima or minima for the function f(x) = x3.ex.

Test: Maxima Minima (Derivative) - Question 3

### The maximum value of f (x) = sin x in the interval [π,2π] is​

Detailed Solution for Test: Maxima Minima (Derivative) - Question 3

f(x) = sin x
f’(x) =cosx
f”(x) = -sin x
f”(3pi/2) = -sin(3pi/2)
= -(-1)
=> 1 > 0 (local minima)
f(pi) = sin(pi) = 0
f(2pi) = sin(2pi) = 0
Hence, 0 is the maxima.

Test: Maxima Minima (Derivative) - Question 4

The maximum and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 1 on the interval [1,4]​

Test: Maxima Minima (Derivative) - Question 5

The maximum value of  is​

Detailed Solution for Test: Maxima Minima (Derivative) - Question 5

For every real number (or) valued function f(x), the values of x which satisfies the equation f1(x)=0 are the point of it's local and global maxima or minima.
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating  dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e

Test: Maxima Minima (Derivative) - Question 6

Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24 x – 18x2

Detailed Solution for Test: Maxima Minima (Derivative) - Question 6

p’(x) = -24 - 36x
p”(x) = -36
Now, p’(x) = 0  ⇒ x = (-24)/36
x = -⅔
Also, p”(-⅔) = -36 < 0
By the second derivative test,  x = -⅔
Therefore, maximum profit = p(-⅔)
= 41 - 24(-⅔) - 18(-⅔)^2
= 41 +16 - 8
⇒ 49

Test: Maxima Minima (Derivative) - Question 7

If f (x) = a log |x| + bx2 + x has extreme values at x = –1 and at x = 2, then values of a and b are

Detailed Solution for Test: Maxima Minima (Derivative) - Question 7

f(x) = alog|x| + bx2 + x
f’(x) = a/x + 2bx + 1
f’(-1) = - a - 2b + 1
-a - 2b + 1 = 0
a = 1 - 2b
f’(2) = a/2 + 4b + 1 = 0
a + 8b = -2
Put the value of a in eq(1)
(1 - 2b) + 8b = - 2
6b = -3
b = -½, a = 2

Test: Maxima Minima (Derivative) - Question 8

The maximum and minimum values of f (x) = x50-x20 in the interval [0, 1] are

Test: Maxima Minima (Derivative) - Question 9

Find the maximum and minimum values of f (x) = 2x3 – 24x + 107 in the interval [1, 3].​

Detailed Solution for Test: Maxima Minima (Derivative) - Question 9

f(x)=2x³-24x+107 x ∈ [1, 3]
f'(x) = 6x^2 - 24
To find the points equate f'(x) = 0
in closed interval x= -2 doesn't lies,so discard x = -2
now find the value of function at x = 1, 2 ,3
f(1) =2(1)³-24(1)+107
= 2-24+107 = 85
f(2) =2(2)³-24(2)+107
= 16-48+107 = 75
f(3) =2(3)³-24(3)+107
= 54-72+107 = 89
So, the function has maximum value in close interval at x= 3, Maximum value= 89.
minimum value at x= 2, minimum value = 75

Test: Maxima Minima (Derivative) - Question 10

Find the point of local maxima or minima for the function

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests