Mean Value Theorems

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JEE Main 2014

LEVELJEE Main

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

Consider the function, $f (x) = ∣ x - 2∣ + ∣ x - 5∣, x \in R$ . Statement-1 : $f^{'} (4) = 0$ . Statement-2 : $f$ is continuous in $[2, 5]$ , differentiable in $(2, 5)$ and $f (2) = f (5)$ .

(A)

Statement-1 is false, Statement-2 is true.

(B)

Statement-1 is true, Statement-2 is true; statement-2 is a correct explanation for Statement-1.

(C)

Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.

(D)

Statement-1 is true, Statement-2 is false.

JEE Main 2007

LEVELJEE Main

A value of $c$ for which conclusion of Mean Value Theorem holds for the function $f (x) = lo g_{e} x$ on the interval $[1, 3]$ is

(A)

lo g_{3} e

(B)

lo g_{e} 3

(C)

2 \log_3 e

(D)

\frac{1}{2} lo g_{3} e

JEE Main 2005

LEVELJEE Main

Let $f$ be differentiable for all $x$ . If $f (1) = - 2$ and $f^{'} (x) \geq 2$ for $x \in [1, 6]$ , then

If the equation $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x = 0$ , $a_{1} \neq = 0, n \geq 2$ , has a positive root $x = α$ , then the equation $n a_{n} x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + \dots + a_{1} = 0$ has a positive root, which is

(A)

greater than

α

(B)

smaller than

α

(C)

greater than or equal to

α

(D)

equal to

α

JEE Main 2004

LEVELJEE Main

If $2 a + 3 b + 6 c = 0$ , then at least one root of the equation $a x^{2} + b x + c = 0$ lies in the interval

If $2 a + 3 b + 6 c = 0$ , then the quadratic equation $a x^{2} + b x + c = 0$ has

(A)

at least one root in

[0, 1]

(B)

at least one root in

[2, 3]

(C)

at least one root in

[4, 5]

Let $f, g : [- 1, 2] \to R$ be continuous functions which are twice differentiable on $(- 1, 2)$ . Let the values of $f$ and $g$ at the points $- 1, 0$ and $2$ be as given in the following table: \begin{center} \begin{array}{|c|c|c|c|} \hline & $x = - 1$ & $x = 0$ & $x = 2$ \\ \hline $f (x)$ & 3 & 6 & 0 \\ \hline $g (x)$ & 0 & 1 & -1 \\ \hline \end{array} \end{center} In each of the intervals $(- 1, 0)$ and $(0, 2)$ the function $(f - 3 g)^{''}$ never vanishes. Then the correct statement(s) is(are)

* Multiple Correct Options

(A)

f^{'} (x) - 3 g^{'} (x) = 0

has exactly three solutions in

(- 1, 0) \cup (0, 2)

(B)

f^{'} (x) - 3 g^{'} (x) = 0

has exactly one solution in

(- 1, 0)

(C)

f^{'} (x) - 3 g^{'} (x) = 0

has exactly one solution in

(0, 2)

(D)

f^{'} (x) - 3 g^{'} (x) = 0

has exactly two solutions in

(- 1, 0)

and exactly two solutions in

(0, 2)

JEE Advanced 2009

LEVELJEE Advanced

For the function $f (x) = x cos (1/ x), x \geq 1$ ,

* Multiple Correct Options

(A)

for at least one

x

in the interval

[1, \infty)

f (x + 2) - f (x) < 2

(B)

lim_{x \to \infty} f^{'} (x) = 1

(C)

for all

x

in the interval

[1, \infty)

f (x + 2) - f (x) > 2

(D)

f^{'} (x)

is strictly decreasing in the interval

[1, \infty)

JEE Advanced 2008

LEVELJEE Advanced

Let $f (x)$ be a non-constant twice differentiable function defined on $(- \infty, \infty)$ such that $f (x) = f (1 - x)$ and $f^{'} (\frac{1}{4}) = 0$ . Then,

* Multiple Correct Options

(A)

f^{''} (x)

vanishes at least twice on

[0, 1]

(B)

f^{'} (\frac{1}{2}) = 0

(C)

\int_{- 1/2}^{1/2} f (x + \frac{1}{2}) sin x d x = 0

(D)

\int_{0}^{1/2} f (t) e^{s i n π t} d t = \int_{1/2}^{1} f (1 - t) e^{s i n π t} d t

JEE Advanced 2007

LEVELJEE Main

Let $f (x) = 2 + cos x$ for all real $x$ . \\ STATEMENT-1: For each real $t$ , there exists a point $c$ in $[t, t + π]$ such that $f^{'} (c) = 0$ because \\ STATEMENT-2: $f (t) = f (t + 2 π)$ for each real $t$ .

(A)

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B)

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C)

Statement-1 is True, Statement-2 is False

(D)

Statement-1 is False, Statement-2 is True

JEE Advanced 2005

LEVELJEE Main

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

(A)

f^{''} (x) = 2

for

\forall x \in (1, 3)

(B)

f^{''} (x) = f^{'} (x) = 5

for some

x \in (2, 3)

(C)

f^{''} (x) = 3

for

\forall x \in (2, 3)

(D)

f^{''} (x) = 2

for some

x \in (1, 3)

JEE Advanced 2004

LEVELJEE Main

If $f (x) = x^{α} lo g x$ and $f (0) = 0$ , then the value of $α$ for which Rolle's theorem can be applied in $[0, 1]$ is

(A)

- 2

(B)

- 1

(C)

0

(D)

1/2

JEE Advanced 2003

LEVELJEE Main

In $[0, 1]$ Lagranges Mean Value theorem is NOT applicable to

(A)

f (x) = {\frac{1}{2} - x, (\frac{1}{2} - x)^{2}, x < \frac{1}{2} x \geq \frac{1}{2}

(B)

f (x) = {\frac{s i n x}{x}, 1, x \neq = 0 x = 0

(C)

f (x) = x ∣ x ∣

(D)

f (x) = ∣ x ∣

JEE Advanced 1983

LEVELJEE Main

If $a + b + c = 0$ , then the quadratic equation $3 a x^{2} + 2 b x + c = 0$ has

(A)

at least one root in

[0, 1]

(B)

one root in

[2, 3]

and the other in

[- 2, - 1]

For a twice differentiable function $f (x), g (x)$ is defined as $g (x) = (f^{'} (x))^{2} + f^{''} (x) f (x)$ on $[a, e]$ . If for $a < b < c < d < e, f (a) = 0, f (b) = 2, f (c) = - 1, f (d) = 2, f (e) = 0$ then find the minimum number of zeros of $g (x)$ .

JEE Advanced 2005

LEVELJEE Advanced

$f (x)$ is a differentiable function and $g (x)$ is a double differentiable function such that $∣ f (x) ∣ \leq 1$ and $f^{'} (x) = g (x)$ . If $f^{2} (0) + g^{2} (0) = 9$ . Prove that there exists some $c \in (- 3, 3)$ such that $g (c) g^{''} (c) < 0$ .

JEE Advanced 2004

LEVELJEE Main

Using Rolle's theorem, prove that there is at least one root in $(4 5^{1/100}, 46)$ of the polynomial $P (x) = 51 x^{101} - 2323 x^{100} - 45 x + 1035$ .

JEE Advanced 2003

LEVELJEE Main

If the function $f : [0, 4] \to R$ is differentiable then show that (i) For $a, b \in (0, 4), (f (4))^{2} - (f (0))^{2} = 8 f^{'} (a) f (b)$

JEE Advanced 1982

LEVELJEE Main

If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

JEE Advanced 1981

LEVELJEE Main

For all $x$ in $[0, 1]$ , let the second derivative $f^{''} (x)$ of a function $f (x)$ exist and satisfy $∣ f^{''} (x) ∣ < 1$ . If $f (0) = f (1)$ , then show that $∣ f^{'} (x) ∣ < 1$ for all $x$ in $[0, 1]$ .

JEE Main 2020 (7 January Shift 1)

LEVELJEE Main

Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2$ , for all $x \in (- 7, 0)$ , then for all such functions $f$ , $f (- 1) + f (0)$ lies in the interval:

(A)

[- 6, 20]

(B)

(- \infty, 20]

(C)

(- \infty, 11]

(D)

[- 3, 11]

JEE Main 2020 (7 January Shift 2)

LEVELJEE Main

The value of $c$ in Lagrange’s mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ , where $x \in [0, 1]$ is :

(A)

\frac{4 - 7}{3}

(B)

\frac{2}{3}

(C)

\frac{7 - 2}{3}

(D)

\frac{4 - 5}{3}

JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

Let $S$ be the set of all functions $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ , depending on $f$ , such that:

(A)

\frac{f ( 1 ) - f ( c )}{1 - c} = f^{'} (c)

(B)

∣ f (c) - f (1) ∣ < ∣ f^{'} (c) ∣

(C)

∣ f (c) + f (1) ∣ < (1 + c) ∣ f^{'} (c) ∣

(D)

∣ f (c) - f (1) ∣ < (1 - c) ∣ f^{'} (c) ∣

JEE Main 2020 (9 January Shift 1)

LEVELJEE Advanced

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$ . If for all $x \in (a, b)$ , $f^{'} (x) > 0$ and $f^{''} (x) < 0$ , then for any $c \in (a, b)$ , $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than:

(A)

\frac{b - c}{c - a}

(B)

1

(C)

\frac{c - a}{b - c}

(D)

\frac{b + a}{b - a}

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f and g are differentiable functions in [0,1] satisfying f(0)=2=g(1),g(0)=0 and f(1)=6, then for some c∈]0,1[

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the function, f(x)=∣x−2∣+∣x−5∣,x∈R. Statement-1 : f′(4)=0. Statement-2 : f is continuous in [2,5], differentiable in (2,5) and f(2)=f(5).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A value of c for which conclusion of Mean Value Theorem holds for the function f(x)=loge​x on the interval [1,3] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be differentiable for all x. If f(1)=−2 and f′(x)≥2 for x∈[1,6], then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2a+3b+6c=0, then at least one root of the equation ax2+bx+c=0 lies in the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2a+3b+6c=0, then the quadratic equation ax2+bx+c=0 has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the function f(x)=xcos(1/x),x≥1,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a non-constant twice differentiable function defined on (−∞,∞) such that f(x)=f(1−x) and f′(41​)=0. Then,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=2+cosx for all real x. \\ STATEMENT-1: For each real t, there exists a point c in [t,t+π] such that f′(c)=0 because \\ STATEMENT-2: f(t)=f(t+2π) for each real t.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is a twice differentiable function and given that f(1)=1;f(2)=4,f(3)=9, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=xαlogx and f(0)=0, then the value of α for which Rolle's theorem can be applied in [0,1] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In [0,1] Lagranges Mean Value theorem is NOT applicable to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a+b+c=0, then the quadratic equation 3ax2+2bx+c=0 has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using Rolle's theorem, prove that there is at least one root in (451/100,46) of the polynomial P(x)=51x101−2323x100−45x+1035.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f:[0,4]→R is differentiable then show that (i) For a,b∈(0,4),(f(4))2−(f(0))2=8f′(a)f(b)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) and g(x) are differentiable function for 0≤x≤1 such that f(0)=2,g(0)=0,f(1)=6,g(1)=2, then show that there exist c satisfying 0<c<1 and f′(c)=2g′(c).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of c in Lagrange’s mean value theorem for the function f(x)=x3−4x2+8x+11, where x∈[0,1] is :

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

Consider the function, $f (x) = ∣ x - 2∣ + ∣ x - 5∣, x \in R$ . Statement-1 : $f^{'} (4) = 0$ . Statement-2 : $f$ is continuous in $[2, 5]$ , differentiable in $(2, 5)$ and $f (2) = f (5)$ .

A value of $c$ for which conclusion of Mean Value Theorem holds for the function $f (x) = lo g_{e} x$ on the interval $[1, 3]$ is

Let $f$ be differentiable for all $x$ . If $f (1) = - 2$ and $f^{'} (x) \geq 2$ for $x \in [1, 6]$ , then

If $2 a + 3 b + 6 c = 0$ , then at least one root of the equation $a x^{2} + b x + c = 0$ lies in the interval

If $2 a + 3 b + 6 c = 0$ , then the quadratic equation $a x^{2} + b x + c = 0$ has

For the function $f (x) = x cos (1/ x), x \geq 1$ ,

Let $f (x)$ be a non-constant twice differentiable function defined on $(- \infty, \infty)$ such that $f (x) = f (1 - x)$ and $f^{'} (\frac{1}{4}) = 0$ . Then,

Let $f (x) = 2 + cos x$ for all real $x$ . \\ STATEMENT-1: For each real $t$ , there exists a point $c$ in $[t, t + π]$ such that $f^{'} (c) = 0$ because \\ STATEMENT-2: $f (t) = f (t + 2 π)$ for each real $t$ .

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

If $f (x) = x^{α} lo g x$ and $f (0) = 0$ , then the value of $α$ for which Rolle's theorem can be applied in $[0, 1]$ is

In $[0, 1]$ Lagranges Mean Value theorem is NOT applicable to

If $a + b + c = 0$ , then the quadratic equation $3 a x^{2} + 2 b x + c = 0$ has

Using Rolle's theorem, prove that there is at least one root in $(4 5^{1/100}, 46)$ of the polynomial $P (x) = 51 x^{101} - 2323 x^{100} - 45 x + 1035$ .

If the function $f : [0, 4] \to R$ is differentiable then show that (i) For $a, b \in (0, 4), (f (4))^{2} - (f (0))^{2} = 8 f^{'} (a) f (b)$

If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

The value of $c$ in Lagrange’s mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ , where $x \in [0, 1]$ is :