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JEE • MATHEMATICS • MEAN VALUES THEOREM

Mean values theorem practice · Mixed difficulty

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Showing 1-30 of 86 questions
1 Medium
1 of 86
Application +4 | -1
If f(x)f(x) and g(x)g(x) are differentiable functions in [0,1][0,1] such that f(0)=2,f(1)=6,g(0)=0,g(1)=f(0)=2, f(1)=6, g(0)=0, g(1)= 2 , then there exists 0<c<10 < \mathrm{c} < 1 such that
View answer and explanation
Correct option: C
f′(c)=2g′(c)f^{\prime}(c)=2 g^{\prime}(c)
Option C (f′(c)=2g′(c)f^{\prime}(c)=2 g^{\prime}(c)) is correct because it is derived by applying Rolle's theorem to the function F(x)=f(x)−2g(x)F(x) = f(x) - 2g(x).
2 Medium
2 of 86
Analyzing +4 | -1
Assertion (A): If Rolle's theorem be applied in f(x)f(x), then L.M.V. theorem is also applicable for f(x)f(x).
Reason (R): Both Rolle's theorem & L.M.V. theorem can not be applied in f(x)=∣sin⁡∣x∣∣f(x)=|\sin | x || in [−π3,π3]\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]
The codes are given below, select the correct codes.
View answer and explanation
Correct option: B
Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
Option B (Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).) is correct because while both the assertion and the reason are true statements, the reason does not directly explain the assertion.
3 Medium
3 of 86
Analyzing +4 | -1
Let f(x)=x−1+x+24−10x−1,  1≤x≤26f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26 be real valued function, then f′(x)f'\left( x \right) for 1<x<261 < x < 26
View answer and explanation
Correct option: A
0
Option A is correct because the function simplifies to a constant value of 5 within the given interval, making its derivative zero.
4 Easy
4 of 86
Analyzing +4 | -1
The Rolle’s theorem is applicable in the interval −1≤x≤1- 1 \le x \le 1 for the function
View answer and explanation
Correct option: B
f(x)=x2f\left( x \right) = {x^2}
Option B (f(x)=x2f(x) = x^2) is correct because it satisfies all conditions of Rolle's theorem: continuity, differentiability, and f(−1)=f(1)f(-1) = f(1).
5 Easy
5 of 86
Analyzing +4 | -1
If f(x)f(x) satisfies the conditions for Rolle's theorem on [3,5][3,5], then ∫35f(x) dx\int\limits_3^5 {f\left( x \right)} \,dx equals
View answer and explanation
Correct option: D
−4/3-4 / 3
Option D (-4/3) is correct because integrating the function f(x) that satisfies Rolle's theorem conditions on [3,5] gives -4/3.
6 Medium
6 of 86
Application +4 | -1
The value of c, in the Lagrange&rsquo;s Mean value theorem f(b)−f(a)b−a=f′(c)\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right) for the function f(x)=x(x−1)(x−2)f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right) in the interval [0, 1/2], is
View answer and explanation
Correct option: B
1−2161 - \frac{{\sqrt {21} }}{6}
Option B (1−2161 - \frac{{\sqrt {21} }}{6}) is correct because it satisfies Lagrange's Mean Value Theorem within the given interval [0, 1/2].
7 Easy
7 of 86
Analyzing +4 | -1
If the functions f(x)f(x) and g(x)g(x) are continuous on [a,b][a, b] and differentiable on (a,b),(a, b), then in the interval (a,b),(a, b), the equation ∣f′(x)f(a)g′(x)g(a)∣=1a−b∣f(a)f(b)g(a)g(b)∣\left|\begin{array}{ll}f^{\prime}(x) & f(a) \\ g^{\prime}(x) & g(a)\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{ll}f(a) & f(b) \\ g(a) & g(b)\end{array}\right|
View answer and explanation
Correct option: A
Has at least one root
Option A is correct because Lagrange's Mean Value Theorem applied to a function defined by the determinant guarantees at least one root for the given equation in the interval (a, b).
8 Medium
8 of 86
Analyzing +4 | -1
The number of real values of kk such that the equation x3−3x−k=0x^3-3 x-k=0 has two distinct real roots in the interval (0,1)(0,1) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because Rolle's theorem cannot be satisfied within the interval (0,1) for the given function, thus there are no such real values of k.
9 Easy
9 of 86
Analyzing +4 | -1
If 0<a<b<π2  and  f(a,b)=tan⁡b−tan⁡ab−a,0 < a < b < \frac{\pi }{2}\,\,{\text{and}}\,\,f\left( {a,b} \right) = \frac{{\tan b - \tan a}}{{b - a}}, then
View answer and explanation
Correct option: B
f(a,b)>1f(a, b)>1
Option B (f(a, b)>1) is correct because by applying Lagrange's mean value theorem to f(x) = tan x, we find that f(a, b) = sec2(c) > 1.
10 Medium
10 of 86
Application +4 | -1
If the function f(x)=ax3+bx2+11x−6f\left( x \right) = a{x^3} + b{x^2} + 11x - 6 satisfies the condition of Rolle’s theorem in [1, 3] and f′(2+13)=0f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0, then the values of a,ba,b are respectively
View answer and explanation
Correct option: C
1,−61, - 6
Option C (1,−61, - 6) is correct because these values of a and b satisfy the conditions of Rolle's theorem and the given derivative condition.
11 Medium
11 of 86
Application +4 | -1
The function f(x)=(x−3)2f(x)=(x-3)^2 satisfies all the conditions of mean value theorem in {3,4}\{3,4\}. A point on y=(x−3)2y=(x-3)^2, where the tangent is parallel to the chord joining (3,0)(3,0) and (4,1)(4,1) is
View answer and explanation
Correct option: B
(72,14)\left(\frac{7}{2}, \frac{1}{4}\right)
Option B (72,14)\left(\frac{7}{2}, \frac{1}{4}\right) is correct because it satisfies the Mean Value Theorem, where the tangent's slope equals the chord's slope.
12 Easy
12 of 86
Application +4 | -1
If the function f(x)=x3−6kx2+5xf(x)=x^3-6 k x^2+5 x satisfies the conditions of Lagrange's mean value theorem for the interval [1,2][1,2] and the tangent to the curve y=f(x)y=f(x) at x=74x=\frac{7}{4} is parallel to the chord that joins the points of intersection of the curve with the ordinates x=1\mathrm{x}=1 and x=2\mathrm{x}=2. Then the value of kk is
View answer and explanation
Correct option: B
3548\frac{35}{48}
The correct answer is B (3548\frac{35}{48}) because it satisfies the conditions of Lagrange's mean value theorem and the given tangent condition.
13 Medium
13 of 86
Application +4 | -1
If f(x)=x3−6x2+ax+bf(x)=x^{3}-6 x^{2}+a x+b is defined on [1,3][1,3] satisfies Rolle's theorem for c=23+13c=\frac{2 \sqrt{3}+1}{\sqrt{3}} then
View answer and explanation
Correct option: C
a=11,b∈Ra=11, b \in R
Option C (a=11, b∈\inR) is correct because Rolle's theorem requires f(1)=f(3), which leads to a=11, and the value of 'b' does not affect this condition, so b can be any real number.
14 Medium
14 of 86
Application +4 | -1
The value of cc in Lagrange&rsquo;s theorem for the function f(x)=log⁡esin⁡xf\left( x \right) = {\log _e}\sin x in the interval [π/6,5  π/6]\left[ {\pi /6,5\;\pi /6} \right] is
View answer and explanation
Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi}{2}) is correct because it satisfies the condition of Lagrange's Mean Value Theorem within the given interval.
15 Easy
15 of 86
Application +4 | -1
Let a function ff be differentiable on R. If f(1)=−2f(1)=-2 and f(x)≥2,∀x∈[1,6]f(x) \geq 2, \forall x \in[1,6].
View answer and explanation
Correct option: B
f(6)≥8f(6) \geq 8
Option B (f(6)≥8f(6) \geq 8) is correct because by applying the Mean Value Theorem and using the given condition f′(x)≥2f'(x) \geq 2, we can deduce that f(6)≥8f(6) \geq 8.

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16 Hard
16 of 86
Analyzing +4 | -1
If f(x)=xαlog⁡xf\left( x \right) = {x^\alpha }\log x and f(0)=0f\left( 0 \right) = 0, then the value of &alpha; for which Rolle&rsquo;s theorem can be applied in [0, 1], is
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because it satisfies the conditions for Rolle's theorem to be applicable in the given interval.
17 Medium
17 of 86
Analyzing +4 | -1
If C01  +  C12  +  C23=0\frac{{{C_0}}}{1}\,\, + \,\,\frac{{{C_1}}}{2}\,\, + \,\,\frac{{{C_2}}}{3} = 0, where C0,  C1,  C2{C_0},\,\,{C_1},\,\,{C_2} are all real, then the quadratic equation C2 x2  +  C1 x  +  C0=0{C_2}\,{x^2}\,\, + \,\,{C_1}\,x\,\, + \,\,{C_0} = 0 has
View answer and explanation
Correct option: A
At least one root in (0,1)\left( {\mathrm{0,1}} \right)
The correct answer is A (At least one root in (0,1)) because Rolle's theorem guarantees a root within the interval (0,1) given the conditions.
18 Easy
18 of 86
Analyzing +4 | -1
Let ff be twice differentiable function satisfying f(1)=1,f(2)=4,f(3)=9,f(1)=1, f(2)=4, f(3)=9, then
View answer and explanation
Correct option: C
There exists at least one x∈(1,3)x \in(1,3) such that f′′(x)=2f^{\prime \prime}(x)=2
Option C is correct because by defining a function g(x) = f(x) - x2 and applying Rolle's theorem twice, we can show that there exists at least one x in (1,3) such that f''(x) = 2.
19 Medium
19 of 86
Understanding +4 | -1
Let A=R×R,RA = R \times R,R the real number system and R={((x,y),(a,b)∈A×A)}R = \left\{ {\left( {\left( {x,y} \right),\left( {a,b} \right) \in A \times A} \right)} \right\} either ((x,y),(a,b))∈R\left( {\left( {x,y} \right),\left( {a,b} \right)} \right) \in R and ((a,b),(p,q))∈R?\left( {\left( {a,b} \right),\left( {p,q} \right)} \right) \in R?
View answer and explanation
Correct option: A
((x,y),(p,q))∈R\left( {\left( {x,y} \right),\left( {p,q} \right)} \right) \in R
Option A (((x,y),(p,q)) ∈\in R) is correct because the relation R is transitive, meaning if (x,y) is related to (a,b) and (a,b) is related to (p,q), then (x,y) is related to (p,q).
20 Medium
20 of 86
Analyzing +4 | -1
If f(x)f\left( x \right) and g(x)g\left( x \right) are differentiable functions for 0≤x≤10 \le x \le 1 such that f(0)=2,g(0)=0,f(1)=6,g(1)=2,f\left( 0 \right) = 2,g\left( 0 \right) = 0,f\left( 1 \right) = 6,g\left( 1 \right) = 2, then in the interval (0,1)
View answer and explanation
Correct option: B
f′(x)=2  g′(x)f'\left( x \right) = 2\;g'\left( x \right) for at least one xx
Option B (f′(x)=2  g′(x)f'\left( x \right) = 2\;g'\left( x \right) for at least one xx) is correct because it follows from Rolle's theorem applied to the function ϕ(x)=f(x)−2g(x)\phi(x) = f(x) - 2g(x).
21 Easy
21 of 86
Analyzing +4 | -1
Let f(x)f(x) be twice differentiable continuous function defined on [a,b][a, b] such that f′(x)>0f^{\prime}(x)>0 and f′′(x)<0f^{\prime \prime}(x)<0 for all x∈(a,b).x \in(a, b). If c∈(a,b),c \in(a, b), then f(c)−f(a)f(b)−f(c)\frac{f(c)-f(a)}{f(b)-f(c)} is greater than
View answer and explanation
Correct option: D
c−ab−c\frac{c-a}{b-c}
Option D is correct because by applying Lagrange's mean value theorem and using the given conditions on the first and second derivatives, we can derive the inequality relating the function values.
22 Easy
22 of 86
Understanding +4 | -1
If f′′(x)<0f^{\prime \prime}(x)<0 for all x∈(a,b),x \in(a, b), then f′(x)=0f^{\prime}(x)=0
View answer and explanation
Correct option: B
At most once in (a,b)(a, b)
Option B (At most once in (a,b)(a, b)) is correct because if f′(x)=0f'(x) = 0 at two distinct points, Rolle's theorem would imply f′′(x)=0f''(x) = 0 at some point, contradicting f′′(x)<0f''(x) < 0.
23 Easy
23 of 86
Analyzing +4 | -1
For f(x)=ln⁡(x2+α7x)f(x)=\ln \left(\frac{x^2+\alpha}{7 x}\right), Roll's theorem is applicable on [3,4][3,4] with c∈(3,4)c \in(3,4) such that f′(c)=0.f^{\prime}(c)=0. The value of f′′(c)f^{\prime \prime}(c) is
View answer and explanation
Correct option: A
112\frac{1}{12}
Option A is correct because by applying Rolle's theorem to the given function and solving for the constant α, we can find the value of c and then evaluate f''(c).
24 Medium
24 of 86
Understanding +4 | -1
If f(x)f\left( x \right) satisfies the requirements of Rolle&rsquo;s Theorem in [1,2]\left[ {\mathrm{1, 2}} \right] and f(x)f\left( x \right) is continuous in [1,2]\left[ {\mathrm{1, 2}} \right], then ∫12  f′(x)  dx\int \limits_1^2 \,\,f'\left( x \right)\,\,dx is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because Rolle's Theorem implies f(1) = f(2), and the integral of f'(x) from 1 to 2 is simply f(2) - f(1), which equals zero.
25 Easy
25 of 86
Application +4 | -1
If the function f(x)=x3−6  x2+ax+bf\left( x \right) = {x^3} - 6\;{x^2} + ax + b defined on [1,3], satisfies the Rolle’s theorem for c=23+13,c = \frac{{2\sqrt 3 + 1}}{{\sqrt 3 }},, then
View answer and explanation
Correct option: C
a=11,b∈Ra = 11,b \in R
Option C (a=11,b∈Ra = 11,b \in R) is correct because Rolle's theorem requires f(1) = f(3), which leads to a = 11, and b can be any real number.
26 Easy
26 of 86
Application +4 | -1
If 2a+3b+6c=0,2 a+3 b+6 c=0, then the equation ax2+bx+c=0a x^2+b x+c=0 has in the interval (0,1)(0,1)
View answer and explanation
Correct option: A
At least one root
The correct answer is A (At least one root) because applying Rolle's theorem to the function f(x)f(x) guarantees the existence of at least one root in the interval (0, 1).
27 Easy
27 of 86
Application +4 | -1
Let ff be continuous on [1, 5] and differentiable in (1,5)\left( {1,5} \right). If f(1)=−3f\left( 1 \right) = - 3 and f′(x)≥9f'\left( x \right) \ge 9 for all x∈(1,  5)x \in \left( {1,\;5} \right), then
View answer and explanation
Correct option: A
f(5)≥33f\left( 5 \right) \ge 33
Option A (f(5)≥33f\left( 5 \right) \ge 33) is correct because it represents the minimum possible value of f(5) based on the Mean Value Theorem and the given conditions.
28 Medium
28 of 86
Analyzing +4 | -1
If a+b+c=0,a+b+c=0, then the equation 3ax2+2bx+c=03 a x^2+2 b x+c=0 has, in the interval (0,1).(0,1).
View answer and explanation
Correct option: A
At least one root
Option A (At least one root) is correct because applying Rolle's theorem to the polynomial f(x) = ax3 + bx2 + cx shows that its derivative has at least one root in (0,1).
29 Medium
29 of 86
Application +4 | -1
If f(x)=(1−x)3/2f(x)=(1-x)^{3 / 2} satisfies the relation

f(x)=f(0)+xf′(0)+x22f′′(θx)f(x)=f(0)+x f^{\prime}(0)+\frac{x^2}{2} f^{\prime \prime}(\theta x)
then as x→1x \rightarrow 1, the value of θ\theta is
View answer and explanation
Correct option: D
925\frac{9}{25}
Option D (9/25) is correct because it is the value of θ that satisfies the given relation as x approaches 1, found by using Taylor's theorem.
30 Medium
30 of 86
Analyzing +4 | -1
If f(x)f\left( x \right) satisfies the conditions for Rolle’s theorem in [3, 5], then ∫35f(x)  dx\int \nolimits_3^5 f\left( x \right)\;dx equals
View answer and explanation
Correct option: D
−43- \frac{4}{3}
Option D (-4/3) is correct because it is the definite integral of a function that satisfies Rolle's theorem over the given interval, calculated using a specific example.
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