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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Mixed difficulty

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Showing 61-90 of 174 questions

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61 Easy
61 of 174
Analyzing +4 | -1
Let f(x)=x4+ax3+3x2+bx+1,a,b∈Rf(x)=x^4+a x^3+3 x^2+b x+1, a, b \in R. If f(x)≥0,∀x∈Rf(x) \geq 0, \forall x \in R, then the maximum value of a2+b2a^2+b^2 is equal to
View answer and explanation
Correct option: B
12
Option B (12) is correct because the condition f(x)≥0f(x) \geq 0 implies that a2+b2a^2 + b^2 must be less than or equal to 12, making 12 the maximum value.
62 Medium
62 of 174
Application +4 | -1
The function f(x)=x2+2xf\left( x \right) = \frac{x}{2} + \frac{2}{x} has local minimum at
View answer and explanation
Correct option: D
x=2x = 2
Option D (x=2x = 2) is correct because the function's second derivative is positive at x=2x = 2, indicating a local minimum.
63 Medium
63 of 174
Analyzing +4 | -1
If 4x4+9y4=644 x^4+9 y^4=64 then the maximum value of x2+y2x^2+y^2 is (where x\mathrm{x} and y\mathrm{y} are real)
View answer and explanation
Correct option: B
4313\frac{4}{3} \sqrt{13}
Option B (4313\frac{4}{3} \sqrt{13}) is correct because it represents the maximum value of x2+y2x^2 + y^2 obtained by using trigonometric substitution and finding the maximum of the resulting expression.
64 Medium
64 of 174
Understanding +4 | -1
The function y=a(1−cos⁡x)y = a\left( {1 - \cos x} \right) is maximum when x is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because the function y = a(1 - cos x) is maximum when cos x is minimum, which occurs at x = π\pi.
65 Medium
65 of 174
Application +4 | -1
If a,b>0a, b>0, then the minimum value y=b2a−x+a2x,0<x<ay=\frac{b^2}{a-x}+\frac{a^2}{x}, 0 < x < a is
View answer and explanation
Correct option: D
(a+b)2a\frac{(a+b)^2}{a}
Option D ((a+b)2a\frac{(a+b)^2}{a}) is correct because by finding the critical points of the function and checking the second derivative, we can determine the minimum value.
66 Hard
66 of 174
Analyzing +4 | -1
The maximum value of f(x)=x4+x+x2f\left( x \right) = \frac{x}{{4 + x + {x^2}}} on [-1,1] is
View answer and explanation
Correct option: C
16\frac{1}{6}
Option C (16\frac{1}{6}) is correct because it represents the maximum value of the function on the given interval, found by evaluating the function at critical points and endpoints.
67 Medium
67 of 174
Analyzing +4 | -1
On the interval [0, 1] the function x25(1−x)75{x^{25}}{\left( {1 - x} \right)^{75}} takes its maximum value at the point
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because the function attains its maximum value at x = 1/4, as determined by finding the critical points and analyzing the sign change of the derivative.
68 Medium
68 of 174
Application +4 | -1
The function f(x)=x4−62  x2+ax+9f\left( x \right) = {x^4} - 62\;{x^2} + ax + 9 attains its maximum value on the interval [0, 2] at x=1x = 1. Then, the value of aa is
View answer and explanation
Correct option: A
120
Option A (120) is correct because the derivative of the function at x=1 must be zero since it attains a maximum value at that point.
69 Medium
69 of 174
Analyzing +4 | -1
The greatest value of the function f(x)=xe−xf\left( x \right) = x{e^{ - x}} in [0,  ∞]\left[ {0,\;\infty } \right], is
View answer and explanation
Correct option: B
1e\frac{1}{e}
Option B (1e\frac{1}{e}) is correct because it represents the maximum value of the function f(x)=xe−xf(x) = xe^{-x} on the interval [0,∞)[0, \infty) found by analyzing its derivative.
70 Medium
70 of 174
Analyzing +4 | -1
The value of a for which the function f(x)={tan⁡−1a−3x2,  0<x<1−6x,    x≥1f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{{\tan }^{ - 1}}a - 3{x^2},\;0 < x < 1}\\ { - 6x,\;\;x \ge 1} \end{array}} \right. has a maximum at x=1x = 1, is
View answer and explanation
Correct option: D
−1- 1
Option D (−1- 1) is correct. The function must be continuous and have a non-positive derivative at x=1 for a maximum to occur. This condition is satisfied when a = -1.
71 Medium
71 of 174
Application +4 | -1
Greatest and least value of (sin⁡−1x)2+(cos⁡−1x)2\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2 are respectively
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Correct option: D
5π24\frac{5 \pi^2}{4} and π28\frac{\pi^2}{8}
Option D (5π24\frac{5 \pi^2}{4} and π28\frac{\pi^2}{8}) is correct because by using the identity (sin⁡−1x)2+(cos⁡−1x)2=(sin⁡−1x+cos⁡−1x)2−2sin⁡−1xcos⁡−1x(\sin^{-1} x)^2 + (\cos^{-1} x)^2 = (\sin^{-1} x + \cos^{-1} x)^2 - 2 \sin^{-1} x \cos^{-1} x and completing the square, we can find the maximum and minimum values.
72 Medium
72 of 174
Analyzing +4 | -1
The value(s) of 'aa' for which f(x)=ax3+(a+6)x2+4x+7f(x)=a x^3+(a+6) x^2+4 x+7 is monotonic ∀ x<0\forall \, x < 0 is/are
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because for f(x) to be monotonic for all x < 0, f'(x) = 0 must have both roots non-negative, implying a > 0, but this condition is not sufficient to guarantee monotonicity.
73 Medium
73 of 174
Analyzing +4 | -1
The maximum value of f(x)=cos⁡xsin⁡xf(x)=\cos x \sqrt{\sin x} on the interval [0,π/2][0, \pi / 2] is
View answer and explanation
Correct option: A
21/2⋅3−3/42^{1 / 2} \cdot 3^{-3 / 4}
Option A (21/2⋅3−3/42^{1 / 2} \cdot 3^{-3 / 4}) is correct because it represents the maximum value of the function found by using calculus and trigonometric identities.
74 Easy
74 of 174
Application +4 | -1
If α=2β=3γ\alpha=2 \beta=3 \gamma, then find the value of log⁡(76)(α2+β2+γ2αβ+βγ+γα)\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
2
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.
75 Medium
75 of 174
Application +4 | -1
The marks obtained by a stu first and seound terminal examinamuns are 62 antu-n 48 , respectively. The minimum marks he should get in the annual examination to have an average of at least 60 marks, are
View answer and explanation
Correct option: A
70
The correct answer is A (70) because it represents the minimum marks needed to achieve an average of at least 60, calculated using the average formula.
76 Easy
76 of 174
Analyzing +4 | -1
For any real θ\theta, the maximum value of cos⁡2(cos⁡θ)+sin⁡2(sin⁡θ)\cos ^2(\cos \theta)+\sin ^2(\sin \theta) is
View answer and explanation
Correct option: B
1+sin⁡211+\sin ^2 1
Option B is correct because the maximum value of cos⁡2(cos⁡θ)\cos^2(\cos \theta) is 1, and the maximum value of sin⁡2(sin⁡θ)\sin^2(\sin \theta) is sin⁡2(1)\sin^2(1), both occurring at θ=π2\theta = \frac{\pi}{2}.
77 Easy
77 of 174
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
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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.
78 Easy
78 of 174
Analyzing +4 | -1
If mm and nn respectively are the number of local maximum and local minimum points of the function f(x)=∫0x2t2−5t+42+et dt,f\left( x \right) = \int\limits_0^{{x^2}} {\frac{{{t^2} - 5t + 4}}{{2 + {e^t}}}} \,dt, then the ordered pair (m,n)\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.
79 Medium
79 of 174
Analyzing +4 | -1
If f(x)=x2−1x2+1f\left( x \right) = \frac{{{x^2} - 1}}{{{x^2} + 1}}, for every real number xx, then minimum value of f(x)f\left( x \right)
View answer and explanation
Correct option: D
II equal to −1- 1
Option D (II equal to −1- 1) is correct because it is the minimum value of the function, found by analyzing its derivative.
80 Medium
80 of 174
Analyzing +4 | -1
The denominator of a fraction is greatest then 16 of the square of numerator, then least value of fraction is
View answer and explanation
Correct option: B
−18\frac{{ - 1}}{8}
Option B (−18\frac{{ - 1}}{8}) is correct because it represents the minimum value of the fraction, found by analyzing its derivative and critical points.
81 Easy
81 of 174
Application +4 | -1
The maximum value x3−3x{x^3} - 3x in the interval [0, 2] is
View answer and explanation
Correct option: C
2
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.
82 Hard
82 of 174
Analyzing +4 | -1
The maximum value of (1x)2x2{\left( {\frac{1}{x}} \right)^{2{x^2}}}, is
View answer and explanation
Correct option: B
ee\sqrt[e]{e}
Option B (ee\sqrt[e]{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.
83 Easy
83 of 174
Application +4 | -1
On the curve x3=12yx^3=12 y, the values of xx for which the abscissa changes at a faster rate than the ordinate is
View answer and explanation
Correct option: A
(−2,2)−{0}(-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.
84 Easy
84 of 174
Application +4 | -1
The minimum value of 2x+3y2x + 3y, when xy=6xy = 6, is
View answer and explanation
Correct option: B
12
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.
85 Medium
85 of 174
Analyzing +4 | -1
The function f(x)=∫−1xt(et−1)(t−1)(t−2)3(t−5)5dtf\left( x \right) = \int \limits_{ - 1}^x t\left( {{e^t} - 1} \right)\left( {t - 1} \right){\left( {t - 2} \right)^3}{\left( {t - 5} \right)^5}dt has local minimum at x=x =
View answer and explanation
Correct option: D
3
Option D (3) is correct because the derivative changes sign from negative to positive at x=3, indicating a local minimum.
86 Medium
86 of 174
Analyzing +4 | -1
In (−4,  4)\left( { - 4,\;4} \right) the function f(x)=∫−10x(t4−4)e−4t  dtf\left( x \right) = \int \nolimits_{ - 10}^x \left( {{t^4} - 4} \right){e^{ - 4t}}\;dt has
View answer and explanation
Correct option: C
Two extrema
Option C (Two extrema) is correct because the derivative of the function has two distinct real roots within the given interval.
87 Easy
87 of 174
Analyzing +4 | -1
For the function f(x)=xexf\left( x \right) = x{e^x} the point
View answer and explanation
Correct option: D
x=−1  is  a  x = - 1\;\mathrm{is}\;a\; minimum
The correct answer is D (x=−1  is  a  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.
88 Medium
88 of 174
Application +4 | -1
The real number xx when added to its inverse gives the minimum value of the sum at xx equals to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the function x+1xx + \frac{1}{x} has a minimum value at x=1x = 1, as determined by finding the critical points and using the second derivative test.
89 Easy
89 of 174
Analyzing +4 | -1
Let f(x)={x3−x2+10x−7,x≤1−2x+log⁡2(b2−4),x>1f(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 bb, for which f(x)f(x) has maximum value at x=1x=1, is
View answer and explanation
Correct option: C
[−6,−2)∪(2,6][-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.
90 Medium
90 of 174
Analyzing +4 | -1
Let f(x)=sin⁡xcos⁡2x,x∈[0,π]f(x)=\sin x \cos ^2 x, x \in[0, \pi]. Let mm be the number of values of xx at which f(x)f(x) is minimum and nn be the number of values of xx at which f(x)f(x) is maximum. Then
View answer and explanation
Correct option: D
m=1,n=2m=1, n=2
Option D (m=1,n=2m=1, n=2) is correct because the function has one minimum and two maxima in the given interval.
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