In interval [1,e], the greatest value of ${x^2}\log x$ is

$\frac{1}{e}\log \frac{1}{{\sqrt e }}$

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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Easy

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Showing 1-30 of 58 questions

1 Easy

1 of 58

Analyzing +4 | -1

Let

f: R \rightarrow R

be a function defined by

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

. Then, which of the following is NOT true?

View answer and explanation

Correct option: C

For

n_1=3, n_2=5

, there exists

\alpha \in(3,5)

where

f

attains local maxima.

The correct answer is C. For

n_1=3, n_2=5

, there exists

\alpha \in(3,5)

where

f

attains local maxima is NOT true because with both powers odd, there will be a local minimum.

2 Easy

2 of 58

Application +4 | -1

The sum of two numbers is 20. If the product of the square of one number and cube of the other is maximum, then the numbers are

View answer and explanation

Correct option: A

12, 8

Option A (12, 8) is correct because these numbers maximize the product of the square of one number and the cube of the other, given their sum is 20.

3 Easy

3 of 58

Analyzing +4 | -1

The coordinates of the point on the curve

x^3=y(x-a)^2, a>0

, where the ordinate is minimum

View answer and explanation

Correct option: C

\left(3 a, \frac{27 a}{4}\right)

Option C

\left(3 a, \frac{{27 a}}{{4}}\right)

is correct because it represents the point where the first derivative is zero and the second derivative is positive, indicating a minimum ordinate.

4 Easy

4 of 58

Analyzing +4 | -1

On the interval

\left[\frac{5 \pi}{4}, \frac{4 \pi}{3}\right]

, the least value of the function

f(x)=\int_{5 \pi / 4}^x(3 \sin t+4 \cos t) d t

View answer and explanation

Correct option: A

\frac{3}{2}+\frac{1}{\sqrt{2}}-2 \sqrt{3}

Option A is correct because the function

f(x)

is decreasing on the given interval, so the least value occurs at the right endpoint,

x = \frac{4\pi}{3}

5 Easy

5 of 58

Analyzing +4 | -1

\left|a_2\right| > \left|a_1+a_3\right|

, then atleast one root of the equation

a_1 x^4+a_2 x^3+a_3=0

lies in the interval

View answer and explanation

Correct option: B

(-1,1)

Option B ((-1,1)) is correct because the condition |a_2| > |a_1+a_3| implies that there is a sign change in the function within the interval (-1, 1), indicating a root.

6 Easy

6 of 58

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left| {{x^3} + {x^2} + 3x + \sin x} \right|\left( {3 + \sin \frac{1}{x}} \right),\;x \ne 0}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 0} \end{array}} \right.

then number of points (where

f\left( x \right)

attains its minimum value) is

View answer and explanation

Correct option: A

Option A (1) is correct because the function attains its minimum value of 0 only at x=0, due to the absolute value and the term (3 + sin(1/x)).

7 Easy

7 of 58

Analyzing +4 | -1

If the absolute maximum value of the function

f(x)=

\left(x^2-2 x+7\right) e^{\left(4 x^3-12 x^2-180 x+31\right)}

in the interval

[-3,0]

f(\alpha)

, then

View answer and explanation

Correct option: B

\alpha=-3

Option B (

\alpha=-3

) is correct because the absolute maximum value of the function occurs at x = -3 within the given interval [-3, 0].

8 Easy

8 of 58

Application +4 | -1

For which value of

x

f(x)=\sum\limits_{{k=1}}(x-k)^2

minimum ?

View answer and explanation

Correct option: B

1008

Option B (1008) is correct because by taking the derivative of the function and setting it to zero, we find the value of x that minimizes the function.

9 Easy

9 of 58

Understanding +4 | -1

A function

f

such that

f^{\prime}(2)=f^{\prime \prime}(2)=0

and

f

has a local maximum of -17 at 2 is

View answer and explanation

Correct option: C

-17-(x-2)^{4}

Option C (

-17-(x-2)^{4}

) is correct because it satisfies the given conditions:

f'(2) = f''(2) = 0

and has a local maximum of -17 at x=2.

10 Easy

10 of 58

Application +4 | -1

The minimum value of

f(x)=|3-x|+|2+x|+|5-x|

View answer and explanation

Correct option: B

Option B (7) is correct because the minimum value of the function occurs at x=3, and f(3) = 7.

11 Easy

11 of 58

Analyzing +4 | -1

If the curve

x^2=-4(y-a)

does not intersect the curve

y=\left[x^2-x+1\right] (

where [.] denotes the greatest integer function) in

\left[0, \frac{1+\sqrt{5}}{2}\right]

, then

View answer and explanation

Correct option: C

\frac{1}{4} < a < 1

Option C (

\frac{1}{4} < a < 1

) is correct. By analyzing the intersection conditions of the parabola and the greatest integer function, we find the range of

a

to be

\frac{1}{4} < a < 1

12 Easy

12 of 58

Application +4 | -1

The function

f(x)=\frac{x}{2}+\frac{2}{x}

has a local minimum at

View answer and explanation

Correct option: A

x=2

Option A (

x=2

) is correct because the second derivative of the function is positive at this point, indicating a local minimum.

13 Easy

13 of 58

Understanding +4 | -1

If for a function

f\left( x \right),\;f'\left( a \right) = 0,\;f''\left( a \right) = 0,\;f'''\left( a \right) > 0

, then at

x = a,\;f\left( x \right)

View answer and explanation

Correct option: C

Not an extreme point

Option C (Not an extreme point) is correct because if the first and second derivatives are zero at a point, and the third derivative is non-zero, then the point is a point of inflection, not an extreme point.

14 Easy

14 of 58

Analyzing +4 | -1

\sin x+\cos x=y^2-y+a

has no value of

x

for any

y

, if

a

belongs

View answer and explanation

Correct option: D

(\sqrt{3}, \infty)

Option D (

(\sqrt{3}, \infty)

) is correct because the range of

\sin x + \cos x

[-\sqrt{2}, \sqrt{2}]

, and

y^2 - y + a

must be outside this range for all

y

15 Easy

15 of 58

Analyzing +4 | -1

In interval [1,e], the greatest value of

{x^2}\log x

View answer and explanation

Correct option: A

{e^2}

The correct answer is A (

{e^2}

) because it is the greatest value of the function

{x^2}\log x

in the interval [1,e].

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16 Easy

16 of 58

Analyzing +4 | -1

Let

f:(-\infty, \infty) \rightarrow(-\infty, \infty)

be defined by

f(x)=x^3+1

Statement 1: The function

f

has a local extremum at

x=0

Statement 2: The function

f

is continuous and differentiable on

(-\infty, \infty)

and

f^{\prime}(0)=0

View answer and explanation

Correct option: D

Statement 1 is false, Statement 2 is true.

Statement 1 is false, and Statement 2 is true. The function f(x) = x³ + 1 does not have a local extremum at x=0, but it is continuous and differentiable with f'(0) = 0.

17 Easy

17 of 58

Analyzing +4 | -1

The maximum value of

{\left( {1/x} \right)^x}

View answer and explanation

Correct option: C

{e^{1/e}}

Option C (

{e^{1/e}}

) is correct because it represents the maximum value of the function found by analyzing its derivatives.

18 Easy

18 of 58

Analyzing +4 | -1

The maximum value of

\frac{{\log x}}{x}\,\mathrm{in}\,\left( {2,\infty } \right),\,\mathrm{is}

View answer and explanation

Correct option: D

\frac{1}{e}

Option D (

\frac{1}{e}

) is correct because it represents the maximum value of the function, found by setting the derivative to zero and confirming it's a maximum.

19 Easy

19 of 58

Analyzing +4 | -1

Let

f(x)=x^4+a x^3+3 x^2+b x+1, a, b \in R

. If

f(x) \geq 0, \forall x \in R

, then the maximum value of

a^2+b^2

is equal to

View answer and explanation

Correct option: B

Option B (12) is correct because the condition

f(x) \geq 0

implies that

a^2 + b^2

must be less than or equal to 12, making 12 the maximum value.

20 Easy

20 of 58

Application +4 | -1

\alpha=2 \beta=3 \gamma

, then find the value of

\log _{\left(\frac{7}{6}\right)}\left(\frac{\alpha^2+\beta^2+\gamma^2}{\alpha \beta+\beta \gamma+\gamma \alpha}\right)

View answer and explanation

Correct option: A

Option A (2) is correct because substituting the given relationships between α, β, and γ simplifies the expression inside the logarithm to a constant, allowing for direct evaluation.

21 Easy

21 of 58

Analyzing +4 | -1

For any real

\theta

, the maximum value of

\cos ^2(\cos \theta)+\sin ^2(\sin \theta)

View answer and explanation

Correct option: B

1+\sin ^2 1

Option B is correct because the maximum value of

\cos^2(\cos \theta)

is 1, and the maximum value of

\sin^2(\sin \theta)

\sin^2(1)

, both occurring at

\theta = \frac{\pi}{2}

22 Easy

22 of 58

Analyzing +4 | -1

Divide 12 into two parts such that the product of the square on one part and the fourth power of the second port is maximum, are

View answer and explanation

Correct option: C

4,8

Option C (4,8) is correct because dividing 12 into 4 and 8 maximizes the product of the square of one part and the fourth power of the other, as determined by calculus.

23 Easy

23 of 58

Analyzing +4 | -1

m

and

n

respectively are the number of local maximum and local minimum points of the function

f\left( x \right) = \int\limits_0^{{x^2}} {\frac{{{t^2} - 5t + 4}}{{2 + {e^t}}}} \,dt,

then the ordered pair

\left( {m,n} \right)

is equal to

View answer and explanation

Correct option: B

(2, 3)

Option B (2, 3) is correct because the function has 2 local maximum points and 3 local minimum points, found by analyzing the sign changes of the derivative.

24 Easy

24 of 58

Application +4 | -1

The maximum value

{x^3} - 3x

in the interval [0, 2] is

View answer and explanation

Correct option: C

Option C (2) is correct because it is the maximum value of the function in the given interval, found by evaluating the function at critical points and endpoints.

25 Easy

25 of 58

Application +4 | -1

On the curve

x^3=12 y

, the values of

x

for which the abscissa changes at a faster rate than the ordinate is

View answer and explanation

Correct option: A

(-2,2)-\{0\}

Option A ((-2,2)-\{0\}) is correct because it represents the interval where the absolute value of the rate of change of x with respect to y is greater than 1.

26 Easy

26 of 58

Application +4 | -1

The minimum value of

2x + 3y

, when

xy = 6

, is

View answer and explanation

Correct option: B

Option B (12) is correct because it is the minimum value obtained by using calculus to find the minimum of the function 2x + 3y given the constraint xy = 6.

27 Easy

27 of 58

Analyzing +4 | -1

For the function

f\left( x \right) = x{e^x}

the point

View answer and explanation

Correct option: D

x = - 1\;\mathrm{is}\;a\;

minimum

The correct answer is D (

x = - 1\;\mathrm{is}\;a\;

minimum) because the first derivative is zero at x = -1 and the second derivative is positive at that point, indicating a minimum.

28 Easy

28 of 58

Analyzing +4 | -1

Let

f(x)=\left\{\begin{array}{cc}x^3-x^2+10 x-7, & x \leq 1 \\ -2 x+\log _2\left(b^2-4\right), & x>1\end{array}\right.

Then the set of all values of

b

, for which

f(x)

has maximum value at

x=1

, is

View answer and explanation

Correct option: C

[-6,-2) \cup(2,6]

The correct answer is C ([-6,-2) ∪(2,6]) because the function has a maximum at x=1 when f(1+) <= f(1), leading to the given interval for b.

29 Easy

29 of 58

Analyzing +4 | -1

If the function

y=\sin (f(x))

is monotonic for all values of

x

[where

f(x)

is continuous], then the maximum value of the difference between the maximum and the minimum value of

f(x)

, is

View answer and explanation

Correct option: A

\pi

Option A is correct because the sine function is monotonic on intervals of length

\pi

, so the maximum difference between the maximum and minimum values of

f(x)

\pi

30 Easy

30 of 58

Application +4 | -1

The point of parallel

2y = {x^2}

, which is nearest to the point (0, 3) is

View answer and explanation

Correct option: C

\left( { \pm 2,\;2} \right)

Option C (

\left( { \pm 2,\;2} \right)

) is correct because these points on the parabola are nearest to the point (0, 3).

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Key Takeaways

✓ This set contains 58 single-correct-answer MCQs on MAXIMA & MINIMA.
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