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JEE • MATHEMATICS • MAXIMA & MINIMA

Maxima & minima practice · Mixed difficulty

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91 Medium
91 of 174
Analyzing +4 | -1
The minimum value of x2+11+x2  {x^2} + \frac{1}{{1 + {x^2}}}\; is at
View answer and explanation
Correct option: A
x=0x = 0
Option A (x=0x = 0) is correct because the function's derivative is zero at x=0x=0, indicating a minimum value.
92 Easy
92 of 174
Analyzing +4 | -1
If the function y=sin⁡(f(x))y=\sin (f(x)) is monotonic for all values of xx [where f(x)f(x) is continuous], then the maximum value of the difference between the maximum and the minimum value of f(x)f(x), is
View answer and explanation
Correct option: A
π\pi
Option A is correct because the sine function is monotonic on intervals of length π\pi, so the maximum difference between the maximum and minimum values of f(x)f(x) is π\pi.
93 Medium
93 of 174
Analyzing +4 | -1
Let f(x)=2x+tan⁡−1xf(x)=2 x+\tan ^{-1} x and g(x)=log⁡e(1+x2+x),x∈[0,3]g(x)=\log _{e}\left(\sqrt{1+x^{2}}+x\right), x \in[0,3]. Then
View answer and explanation
Correct option: D
max⁡f(x)>max⁡g(x)\max f(x) > \max g(x)
Option D (max⁡f(x)>max⁡g(x)\max f(x) > \max g(x)) is correct because the maximum value of f(x) is greater than the maximum value of g(x) in the given interval.
94 Easy
94 of 174
Application +4 | -1
The point of parallel 2y=x22y = {x^2}, which is nearest to the point (0, 3) is
View answer and explanation
Correct option: C
(±2,  2)\left( { \pm 2,\;2} \right)
Option C ((±2,  2)\left( { \pm 2,\;2} \right)) is correct because these points on the parabola are nearest to the point (0, 3).
95 Easy
95 of 174
Analyzing +4 | -1
Consider a cuboid of sides 2x, 4x  and 5x2x,\,4x\,\,{\text{and}}\,{\text{5x}} and a closed hemisphere of radius r.r. If the sum of their surface areas is a constant k,k, then the ratio x:r,x:r, for which the sum of their volumes is maximum, is
View answer and explanation
Correct option: B
19:4519:45
Option B (19:4519:45) is correct because it represents the ratio x:rx:r that maximizes the sum of the volumes given the constraint on the surface areas.
96 Easy
96 of 174
Application +4 | -1
If xx is real, then the minimum value of x2−8x+17x^2-8 x+17 is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the minimum value of the quadratic function x2−8x+17x^2 - 8x + 17 occurs at its vertex, which can be found by completing the square or using calculus.
97 Medium
97 of 174
Application +4 | -1
The curve y=f(x)y=f(x) satisfies d2ydx2=6x−4\frac{d^{2} y}{d x^{2}}=6 x-4 and f(x)f(x) has a local minimum value 5 when x=1x=1. Then f(0)f(0) is equal to :
View answer and explanation
Correct option: C
5
Option C (5) is correct because integrating the second derivative twice and using the given conditions results in f(0) = 5.
98 Easy
98 of 174
Analyzing +4 | -1
Let f:N→Nf: N \rightarrow N be such that f(n+1)>f(f(n))f(n+1)>f(f(n)) for all n∈Nn \in N, then
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct. The function f(n+1) > f(f(n)) implies f(n) = n, but none of the given options satisfy this condition.
99 Easy
99 of 174
Application +4 | -1
Let f(x)f(x) be a differentiable function and G\mathrm{G} is the graph of f(x)f(x). If P(a,f(a))\mathrm{P}(a, f(a)) on G\mathrm{G} is the point nearest to (0,0)(0,0), then f(a)f′(a)=f(a) f^{\prime}(a)=
View answer and explanation
Correct option: B
−a-a
Option B (−a-a) is correct because the point nearest to (0,0) implies the distance function is minimized, leading to the given relation.
100 Hard
100 of 174
Analyzing +4 | -1
Let f(x)=1+2x2+22x4  +...+210x20f\left( x \right) = 1 + 2{x^2} + {2^2}{x^4}\; + ... + {2^{10}}{x^{20}}. Then, f(x)f\left( x \right) has
View answer and explanation
Correct option: A
More than one minimum
Option A (More than one minimum) is correct because the derivative of the function has multiple roots, indicating multiple minima.
101 Medium
101 of 174
Analyzing +4 | -1
The number of critical points for the function
f(x)=13sin⁡θtan⁡3x+(sin⁡θ−1)tan⁡x+θ−28−θf(x)=\frac{1}{3} \sin \theta \tan ^3 x+(\sin \theta-1) \tan x+\sqrt{\frac{\theta-2}{8-\theta}}
where π<θ<2π\pi<\theta<2 \pi is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the function has no critical points, implying no extrema.
102 Hard
102 of 174
Analyzing +4 | -1
The greatest value of f(x)=(x+1)1/3−(x−1)1/3f\left( x \right) = {\left( {x + 1} \right)^{1/3}} - {\left( {x - 1} \right)^{1/3}} on [0,1]is
View answer and explanation
Correct option: C
2
Option C (2) is correct. By finding the critical point and evaluating the function at the endpoints and critical point, we find the greatest value is 2.
103 Easy
103 of 174
Analyzing +4 | -1
Let a1,a2,a3,…a_1, a_2, a_3, \ldots be an A. P. with a6=2a_6=2. Then the common difference of this A.P., which maximises the product a1a4a_1 a_4 a5a_5, is :
View answer and explanation
Correct option: B
85\frac{8}{5}
Option B (85\frac{8}{5}) is correct. By expressing the product in terms of the common difference and maximizing it using calculus, the common difference is found to be 85\frac{8}{5}.
104 Easy
104 of 174
Analyzing +4 | -1
Let f(x)=x2−2xf(x)=x^2-2 x and g(x)=f(f(x)−1)+f(5−f(x))g(x)=f(f(x)-1)+f(5-f(x)), then
View answer and explanation
Correct option: D
g(x)≥0,∀x∈Rg(x) \geq 0, \forall x \in R
Option D (g(x)≥0,∀x∈Rg(x) \geq 0, \forall x \in R) is correct. By finding the derivative of g(x)g(x) and analyzing its critical points, we can determine that g(x)g(x) is always non-negative.
105 Medium
105 of 174
Application +4 | -1
If y=alog⁡∣x∣+bx2+xy=a \log |x|+b x^2+x has its extremum values at x=−1x=-1 and x=2,x=2, then
View answer and explanation
Correct option: B
a=2,b=−12a=2, b=\frac{-1}{2}
Option B (a=2,b=−12a=2, b=\frac{{-1}}{{2}}) is correct because it satisfies the conditions for extremum values at x=-1 and x=2.

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106 Hard
106 of 174
Understanding +4 | -1
If f(x)=∫x2x2dtlog⁡t,x>0f(x)=\int\limits_{{x^2}}^{x^2} \frac{d t}{\log t}, x>0, then
View answer and explanation
Correct option: D
f(x)f(x) is neither maximum nor minimum at x=1x=1
Option D (f(x)f(x) is neither maximum nor minimum at x=1x=1) is correct because the first derivative does not change sign at x=1.
107 Medium
107 of 174
Analyzing +4 | -1
Let p(x)=a0+a1x2+a2x4+…+anx2np\left( x \right) = {a_0} + {a_1}{x^2} + {a_2}{x^4} + \ldots + {a_n}{x^2}^n be a polynomial in a real variable xx with 0<a0<a1<a2…<an.  0 < {a_0} < {a_1} < {a_2} \ldots < {a_n}.\; The function p(x)p\left( x \right) has
View answer and explanation
Correct option: C
Only one minimum
Option C (Only one minimum) is correct because the derivative changes sign only once, from negative to positive, at x=0.
108 Medium
108 of 174
Analyzing +4 | -1
The function f(x)=x−x,  (x∈R)f\left( x \right) = {x^{ - x}},\;\left( {x \in R} \right) attains a maximum, value at xx which is
View answer and explanation
Correct option: C
1e\frac{1}{e}
Option C (1e\frac{1}{e}) is correct because the function attains a maximum value at this point, as confirmed by the first and second derivative tests.
109 Medium
109 of 174
Analyzing +4 | -1
A closed circular cylinder of volume kk cubic units is to be formed with minimum amount of material. The ratio of its height to the radius of the base is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the ratio of height to radius that minimizes the material used for a closed cylinder of fixed volume is 2.
110 Medium
110 of 174
Analyzing +4 | -1
Given a function y=xx,x>0y=x^x, x>0 and 0<x<10 < x < 1. The values of xx for which the function attain values exceeding the values of its inverse are
View answer and explanation
Correct option: A
0<x<10 < x < 1
Option A (0<x<10 < x < 1) is correct because the graph of y=xxy=x^x lies above the line y=xy=x for 0<x<10 < x < 1.
111 Medium
111 of 174
Application +4 | -1
The maximum distance from the origin of a point on the curve x=asin⁡t−sin⁡(atb),  y=acos⁡t−bcos⁡(atb)x = a\sin t - \sin \left( {\frac{{at}}{b}} \right),\;y = a\cos t - b\cos \left( {\frac{{at}}{b}} \right),both a,  b>0a,\;b > 0 is
View answer and explanation
Correct option: B
a+ba + b
Option B (a+ba + b) is correct because it represents the maximum distance from the origin of a point on the given curve.
112 Easy
112 of 174
Analyzing +4 | -1
The number of points of local maxima of the function f(x)=(x−2)2/3(2x+1)f(x)=(x-2)^{2 / 3}(2 x+1) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the function has one local maximum at x=1. The first derivative test confirms this.
113 Medium
113 of 174
Application +4 | -1
If xy=c2xy = {c^2}, then minimum value of ax+byax + by is
View answer and explanation
Correct option: B
2cab2c\sqrt {ab}
Option B (2cab2c\sqrt {ab}) is correct because it is the minimum value of ax+byax + by subject to the constraint xy=c2xy = c^2, found using calculus or AM-GM inequality.
114 Medium
114 of 174
Analyzing +4 | -1
The minimum value of log⁡xx\frac{\log x}{x} in the interval [2,∞)[2, \infty) is:
View answer and explanation
Correct option: D
Does not exist
The minimum value of log(x)/x in the interval [2, infinity) does not exist because the function has a maximum at x=e, and it decreases as x approaches infinity.
115 Medium
115 of 174
Analyzing +4 | -1
The maximum value of log⁡xx\frac{{\log x}}{x} is
View answer and explanation
Correct option: A
1e\frac{1}{e}
Option A (1e\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.
116 Medium
116 of 174
Analyzing +4 | -1
If the function f(x)=ax+b(x−1)(x−4)f\left( x \right) = \frac{{ax + b}}{{\left( {x - 1} \right)\left( {x - 4} \right)}} has an extremum at P(2,  −1)P\left( {2,\; - 1} \right) then
View answer and explanation
Correct option: C
a=1,  b=0a = 1,\;b = 0
Option C (a = 1, b = 0) is correct because substituting these values into the function and its derivative satisfies the given conditions of having an extremum at P(2, -1).
117 Medium
117 of 174
Analyzing +4 | -1
The absolute maximum of x40−x20{x^{40}} - {x^{20}} on the interval [0, 1] is
View answer and explanation
Correct option: B
0
Option B (0) is correct because it represents the absolute maximum value of the function on the given interval, considering both critical points and endpoints.
118 Easy
118 of 174
Analyzing +4 | -1
If, f(x)=2x3−21x2+36x−30f\left( x \right) = 2{x^3} - 21{x^2} + 36x - 30, then which one of the following is correct
View answer and explanation
Correct option: C
f(x)has  maximum  at  x=1f\left( x \right)\mathrm{has\;maximum}\;at\;x = 1
Option C (f(x) has maximum at x=1f(x) \text{ has maximum at } x = 1) is correct because the first derivative is zero and the second derivative is negative at x=1x=1.
119 Medium
119 of 174
Analyzing +4 | -1
If f′(x)=(x−a)2n(x−b)2m+1f'\left( x \right) = {\left( {x - a} \right)^{2n}}{\left( {x - b} \right)^{2m + 1}} where m,  n∈Nm,\;n \in N, then
View answer and explanation
Correct option: A
x=bx = b is a point of minimum
Option A is correct because the sign of the derivative changes from negative to positive at x = b, indicating a point of minimum.
120 Medium
120 of 174
Analyzing +4 | -1
The maximum value of function f(x)=sin⁡x(1+cos⁡x),  x∈Rf\left( x \right) = \sin x\left( {1 + \cos x} \right),\;x \in R is
View answer and explanation
Correct option: A
33/2    4\frac{{{3^{3/2}}\;\;}}{4}
Option A (33/2;4\frac{{{{3^{{3/2}}}}\\;}}{4}) is correct because it represents the maximum value of the function, found by differentiation and checking the second derivative.
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