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JEE • MATHEMATICS • MEAN VALUES THEOREM

Mean values theorem practice · Mixed difficulty

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61 Medium
61 of 86
Application +4 | -1
If Rolle's theorem holds for the function f(x)2x3+bx2f(x) 2 x^{3}+b x^{2} +cx,x∈[−1,1]+c x, x \in[-1,1], at the point x=12x=\frac{1}{2}, then 2b+c2 b+c equals :
View answer and explanation
Correct option: C
2
Option B (-1) is correct because the value of 2b+c2b+c is -1 after applying Rolle's theorem conditions to the given function.
62 Easy
62 of 86
Analyzing +4 | -1
If cc is a point at which Rolle’s Theorem holds for the function, f(x)=log⁡e(x2+α7x)f\left( x \right) = {\log _e}\left( {\frac{{{x^2} + \alpha }}{{7x}}} \right) in the interval [3, 4]\left[ {3,\,4} \right] where α ∈ R,\alpha \, \in \,R, then f′′(c)f''(c) is equal to
View answer and explanation
Correct option: C
112\frac{1}{{12}}
Option C (112\frac{1}{{12}}) is correct because after finding α\alpha using Rolle's Theorem, we find cc and then evaluate f′′(c)f''(c).
63 Easy
63 of 86
Application +4 | -1
The value of c prescribed by Lagrange's mean value theorem, when f(x)=x2−4,a=2f(x)=\sqrt{x^2-4}, a=2 and b=3,b=3, is
View answer and explanation
Correct option: B
5\sqrt{5}
Option B (5\sqrt{{5}}) is correct because it is the value of c that satisfies Lagrange's Mean Value Theorem for the given function and interval.
64 Easy
64 of 86
Understanding +4 | -1
If aa and bb are two distinct real roots of the polynomial f(x)f\left( x \right) such that a<ba < b, then there exists a real number cc lying between aa and bb, such that
View answer and explanation
Correct option: B
f′(c)=0f'\left( c \right) = 0
Option B (f′(c)=0f'\left( c \right) = 0) is correct because Rolle's Theorem states that between any two roots of a polynomial f(x)f(x), there is a root of f′(x)f'(x).
65 Easy
65 of 86
Analyzing +4 | -1
Let f(x)=ax5+bx4+cx3+dx2+ex,f(x)=a x^5+b x^4+c x^3+d x^2+e x, where a,b,c,a, b, c, d,e∈Rd, e \in R and f(x)=0f(x)=0 has a positive root α.\alpha. Then,
View answer and explanation
Correct option: D
All of the above
Option D is correct because Rolle's theorem and properties of polynomial roots ensure that all statements about the derivatives of f(x) are true.
66 Easy
66 of 86
Analyzing +4 | -1
If f(x)f\left( x \right) is a function given by
f(x)=∣sin⁡xsin⁡asin⁡bcos⁡xcos⁡acos⁡btan⁡xtan⁡atan⁡b∣,where  0<a<b<π2f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {\sin x}&{\sin a}&{\sin b}\\ {\cos x}&{\cos a}&{\cos b}\\ {\tan x}&{\tan a}&{\tan b} \end{array}} \right|,\mathrm{where\;}0 < a < b < \frac{\pi }{2}
Then the equation f′(x)=0f'\left( x \right) = 0
View answer and explanation
Correct option: A
Has at least one root in (a,b)\left( {a,b} \right)
Option A (Has at least one root in (a,b)) is correct because Rolle's theorem applies, as f(a) = f(b) = 0, implying the existence of at least one root for f'(x) in (a,b).
67 Easy
67 of 86
Application +4 | -1
Let f{f} be a continuous function in [0,3][0,3] and differentiable in (0,3)(0,3) such that f(3)=0{f}(3)=0. Then there exist some α∈(0,3)\alpha \in(0,3) such that
View answer and explanation
Correct option: D
αf1(α)+f(α)=0\alpha f^1(\alpha)+f(\alpha)=0
Option D (αf1(α)+f(α)=0) is correct because applying Rolle's theorem to a function g(x) = x^n f(x) leads to the given condition.
68 Easy
68 of 86
Application +4 | -1
The value of cc in the Lagrange’s mean value theorem for the function
f(x)= x3−4x2+8x+11,f(x) = \,{x^3} - 4{x^2} + 8x + 11, when x∈ [0, 1]x \in \,[0,\,1] is
View answer and explanation
Correct option: D
4−73\frac{{4 - \sqrt 7 }}{3}
Option D (4−73\frac{{4 - \sqrt 7 }}{3}) is correct because it is the value of cc obtained by applying Lagrange's Mean Value Theorem to the given function on the specified interval.
69 Medium
69 of 86
Application +4 | -1
The value of c in (0, 2) satisfying the Mean value theorem for the function f(x)=x(x−1)2,  x∈[0,  2]f\left( x \right) = x{\left( {x - 1} \right)^2},\;x \in \left[ {0,\;2} \right] is equal to
View answer and explanation
Correct option: B
43\frac{4}{3}
Option B (43\frac{4}{3}) is correct because it is the value of 'c' within the interval (0, 2) that satisfies the Mean Value Theorem for the given function.
70 Medium
70 of 86
Application +4 | -1
Suppose ff is differentiable on R,f(0)=0\mathbb{R}, f(0)=0 and 1≤f′(x)≤21 \leq f^{\prime}(x) \leq 2 for all x∈Rx \in \mathbb{R}. Then
View answer and explanation
Correct option: A
x≤f(x)≤2xx \leq f(x) \leq 2 x for all x≥0x \geq 0
Option A (x≤f(x)≤2xx \leq f(x) \leq 2x for all x≥0x \geq 0) is correct because it uses the Lagrange's Mean Value Theorem and the given bounds on f′(x)f'(x) to establish the inequality.
71 Hard
71 of 86
Analyzing +4 | -1
If [0, 1], Lagrange’s mean value theorem is NOT applicable to
View answer and explanation
Correct option: D
f(x)=∣x∣f\left( x \right) = \left| x \right|
Option D (f(x)=∣x∣f(x) = |x|) is correct because it is not differentiable at x = 0, which lies within the interval [0, 1], thus violating the conditions for Lagrange's Mean Value Theorem.
72 Easy
72 of 86
Application +4 | -1
Value of cc in the Lagrange's mean value theorem for the function

f(x)=x+1x on [12,2}f(x)=x+\frac{1}{x} \text { on }\left[\frac{1}{2}, 2\right\}
is
View answer and explanation
Correct option: C
1
Option C (1) is correct because it is the only value within the interval (12,2)(\frac{1}{2}, 2) that satisfies the Lagrange's Mean Value Theorem for the given function.
73 Medium
73 of 86
Application +4 | -1
If f(x)=x2−2x+4f(x)=x^2-2 x+4 and f(5)−f(1)5−1=f′(c)\frac{f(5)-f(1)}{5-1}=f^{\prime}(c) then value of c\mathrm{c} will be
View answer and explanation
Correct option: D
3
Option D (3) is correct because it is the value of c that satisfies the Mean Value Theorem, where the average rate of change equals the instantaneous rate of change.
74 Easy
74 of 86
Analyzing +4 | -1
Consider a quadratic equation ax2+bx+c=0a x^{2}+b x+c=0, where 2a+3b+6c=02 a+3 b+6 c=0 and let g(x)=ax33+bx22+cxg(x)=a \frac{x^{3}}{3}+b \frac{x^{2}}{2}+c x.
Statement 1: The quadratic equation has at least one root in the interval (0,1)(0,1).
Statement 2: The Rolle's theorem is applicable to function g(x)g(x) on the interval [0,1][0,1].
View answer and explanation
Correct option: D
Statement 1 is true, Statement 2 is true,, Statement 2 is a correct explanation for Statement 1.
Option D is correct. Statement 1 is true because Rolle's theorem applies to g(x) on [0,1], implying the quadratic equation has a root in (0,1). Statement 2 correctly explains Statement 1.
75 Easy
75 of 86
Understanding +4 | -1
Let a,b be two distinct roots of a polynomial f(x)f\left( x \right). Then there exists at least one root lying between aa and bb of the polynomial
View answer and explanation
Correct option: B
f′(x)f'\left( x \right)
Option B (f′(x)f'(x)) is correct because, according to Rolle's theorem, between any two roots of a polynomial, there exists at least one root of its derivative.
76 Easy
76 of 86
Understanding +4 | -1
Let SS be the set of all functions f:[0, 1]→R,f:\left[ {0,\,1} \right] \to R, which are continuous on [0,1] and differentiable on (0,1). Then for every ff in SS, there exists a c ∈(0,1),c\, \in (0,1), depending on ff, such that
View answer and explanation
Correct option: A
f(1)−f(c)1−c=f′(c′)\frac{{f(1) - f(c)}}{{1 - c}} = f'(c')
Option A (f(1)−f(c)1−c=f′(c′)\frac{{f(1) - f(c)}}{{1 - c}} = f'(c')) is correct because it directly applies the Mean Value Theorem (MVT) or Lagrange's Mean Value Theorem (LMVT).
77 Easy
77 of 86
Application +4 | -1
If Rolle’s theorem hold for the function f(x)=x3−ax2+bx+4,  x∈[1,2]f\left( x \right) = {x^3} - a{x^2} + bx + 4,\,\,x \in \left[ {1,2} \right] with f′(43)=0f'\left( {\frac{4}{3}} \right) = 0, then ordered pair (a, b)\left( {a,\,b} \right) is equal to
View answer and explanation
Correct option: A
(5, 8)\left( {5,\,8} \right)
Option A (5,8) is correct because it satisfies the conditions of Rolle's theorem for the given function.
78 Medium
78 of 86
Understanding +4 | -1
f(x)f\left( x \right) satisfies the conditions of Rolle&rsquo;s theorem in [1, 2] and f(x)f\left( x \right) is continuous in [1,2]\left[ {1,2} \right], then ∫12f′(x)dx\int \nolimits_1^2 f'\left( x \right)dx is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because, according to Rolle's theorem, if f(x) satisfies the conditions in [1, 2], then f(1) = f(2), and the integral of f'(x) from 1 to 2 is f(2) - f(1) = 0.
79 Medium
79 of 86
Analyzing +4 | -1
If a,b,ca, b, c be non-zero real numbers such that ∫01(1+cos⁡8x)(ax2+bx+c)dx=∫02(1+cos⁡8x)(ax2+bx+c)dx=0,\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = 0,} } then the equation ax2+bx+c=0a x^2+b x+c=0 will have
View answer and explanation
Correct option: A
One root between 0 and 1 and other root between 1 and 2
Option A is correct because Rolle's theorem implies the derivative of the integral function has roots between the given intervals, corresponding to roots of the quadratic.
80 Easy
80 of 86
Analyzing +4 | -1
If f(x)=(x−p)(x−q)(x−r),f\left( x \right) = \left( {x - p} \right)\left( {x - q} \right)\left( {x - r} \right), where p<q<r,p < q < r, are real numbers, then the application of Rolle's theorem of f leads to
View answer and explanation
Correct option: C
(p+q+r)2>3(pq+qr+rp){\left( {p + q + r} \right)^2} > 3\left( {pq + qr + rp} \right)
Option C is correct because by applying Rolle's theorem to f(x) and analyzing the discriminant of f'(x), we can derive the inequality relating the sum and product of the roots.
81 Easy
81 of 86
Analyzing +4 | -1
If 2a+3b+6c=02a + 3b + 6c = 0, then at least one root of the equation ax2+bx+c=0a{x^2} + bx + c = 0 lies in the interval
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Correct option: A
(0, 1)
Option A is correct because Rolle's theorem guarantees at least one root of the derivative in (0, 1) when f(0) = f(1).
82 Easy
82 of 86
Application +4 | -1
If f(x)=4sin⁡xf(x)=4^{\sin x} satisfies the Rolle's theorem on [0,π][0, \pi], then the value of c∈(0,π)c \in(0, \pi) for which f′(c)=0f^{\prime}(c)=0 is
View answer and explanation
Correct option: C
c=π/2c=\pi / 2
Option C (c=π/2c=\pi / 2) is correct because it satisfies the condition f′(c)=0f'(c) = 0 within the interval (0,π)(0, \pi) after applying Rolle's theorem.
83 Medium
83 of 86
Analyzing +4 | -1
A function ff is defined by f(x)=exsin⁡xf\left( x \right) = {e^x}\sin x in [0,  π]\left[ {0,\;\pi } \right]. Which of the following is not correct?
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Correct option: D
Rolle’s theorem is not true in [0,  π]\left[ {0,\;\pi } \right]
Option B (a decreasing function) is correct because the derivative of the function is negative for all x>0x > 0.
84 Easy
84 of 86
Application +4 | -1
In the mean value theorem f(b)−f(a)=(b−a)f′(c), if a=4,  b=9f\left( b \right) - f\left( a \right) = \left( {b - a} \right)f'\left( c \right),\,if\,a = 4,\;b = 9 and f(x)=xf\left( x \right) = \sqrt x, then the value of cc is
View answer and explanation
Correct option: D
6.25
Option D (6.25) is correct because it is the value of 'c' that satisfies the Mean Value Theorem for the given function and interval.
85 Medium
85 of 86
Application +4 | -1
The position of a moving car at time tt is given by f(t)=at2+bt+c,t>0f(t)=a t^2+b t+c, t>0 where a,ba, b and cc are real numbers greater than 1. Then the average speed of the car over the interval [t1,t2]\left[t_1, t_2\right] is attained at the point
View answer and explanation
Correct option: A
t1+t22\frac{t_1+t_2}{2}
Option A (t1+t22\frac{{t_1+t_2}}{{2}}) is correct because the average speed is attained at the midpoint of the interval [t1, t2] according to Lagrange's Mean Value Theorem.
86 Medium
86 of 86
Application +4 | -1
If f(1)=3f(1)=3 and f′(x)≤1.4f^{\prime}({x}) \leq 1.4 for 1≤x≤81 \leq {x} \leq 8. The largest possible value which f(8){f} (8) can have, is
View answer and explanation
Correct option: A
12.8
Option C (12.8) is correct because it represents the maximum possible value of f(8) based on the given conditions and the Mean Value Theorem.
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