JEE Main 2024 (30 Jan Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

Q: Why does $f''(x)=0$ for *at least* two values?

Rolle's Theorem guarantees that if a function $f'(x)$ has roots at $x_1$ and $x_2$, then there exists at least one point $c$ in $(x_1, x_2)$ where $f''(c)=0$. We found three roots for $f'(x)$: at $1/2$, $1$, and $3/2$. This gives us two intervals: $(1/2, 1)$ and $(1, 3/2)$. Applying Rolle's Theorem to each interval guarantees at least one root for $f''(x)$ in each, leading to at least two roots in total.

Select Answer:

Visualized Solution

Understanding the Given Function $f (x)$

Given: $g$ is non-constant and twice differentiable.
Condition: $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ .
Function: $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ .
Objective: Analyze the roots of $f^{''} (x) = 0$ in $(0, 2)$ .

Differentiating $f (x)$ using Chain Rule

Differentiating $f (x)$ with respect to $x$ :
$f^{'} (x) = \frac{d}{d x} (\frac{1}{2} [g (x) + g (2 - x)])$
Using Chain Rule: $\frac{d}{d x} g (2 - x) = g^{'} (2 - x) \cdot (- 1)$
$f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]$

Evaluating $f^{'} (\frac{1}{2})$

Substitute $x = \frac{1}{2}$ into $f^{'} (x)$ :
$f^{'} (\frac{1}{2}) = \frac{1}{2} [g^{'} (\frac{1}{2}) - g^{'} (2 - \frac{1}{2})]$
$f^{'} (\frac{1}{2}) = \frac{1}{2} [g^{'} (\frac{1}{2}) - g^{'} (\frac{3}{2})]$
Since $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ , then $f^{'} (\frac{1}{2}) = 0$ .

Evaluating $f^{'} (\frac{3}{2})$

Substitute $x = \frac{3}{2}$ into $f^{'} (x)$ :
$f^{'} (\frac{3}{2}) = \frac{1}{2} [g^{'} (\frac{3}{2}) - g^{'} (2 - \frac{3}{2})]$
$f^{'} (\frac{3}{2}) = \frac{1}{2} [g^{'} (\frac{3}{2}) - g^{'} (\frac{1}{2})]$
Since $g^{'} (\frac{3}{2}) = g^{'} (\frac{1}{2})$ , then $f^{'} (\frac{3}{2}) = 0$ .

Evaluating $f^{'} (1)$

Substitute $x = 1$ into $f^{'} (x)$ :
$f^{'} (1) = \frac{1}{2} [g^{'} (1) - g^{'} (2 - 1)]$
$f^{'} (1) = \frac{1}{2} [g^{'} (1) - g^{'} (1)]$
Therefore, $f^{'} (1) = 0$ .

Applying Rolle's Theorem on $[\frac{1}{2}, 1]$

Consider the interval $[\frac{1}{2}, 1]$ .
We have $f^{'} (\frac{1}{2}) = 0$ and $f^{'} (1) = 0$ .
By Rolle's Theorem, there exists at least one $c_{1} \in (\frac{1}{2}, 1)$ such that $f^{''} (c_{1}) = 0$ .

Applying Rolle's Theorem on $[1, \frac{3}{2}]$

Consider the interval $[1, \frac{3}{2}]$ .
We have $f^{'} (1) = 0$ and $f^{'} (\frac{3}{2}) = 0$ .
By Rolle's Theorem, there exists at least one $c_{2} \in (1, \frac{3}{2})$ such that $f^{''} (c_{2}) = 0$ .

Final Conclusion

Since $c_{1} \in (\frac{1}{2}, 1)$ and $c_{2} \in (1, \frac{3}{2})$ , both points lie within $(0, 2)$ .
Conclusion: The equation $f^{''} (x) = 0$ holds for at least two values of $x$ in $(0, 2)$ .
Key Takeaway: If a differentiable function has $n$ roots, its derivative has at least $n - 1$ roots between them.

00:00 / 00:00

The Sigma Insight: Mean Value Theorems

Solution Diagram

Analyzing the Setup

We are given a twice-differentiable function

g (x)

and a composite function defined as:

f (x) = \frac{1}{2} [g (x) + g (2 - x)]

Our objective is to analyze the behavior of the second derivative

f^{''} (x)

within the interval

(0, 2)

Unveiling the First Derivative

To reach the second derivative, we must first differentiate

f (x)

with respect to

x

. Applying the derivative operator and the chain rule to the second term, we obtain:

f^{'} (x) = \frac{1}{2} [g^{'} (x) + g^{'} (2 - x) \cdot (- 1)]

This simplifies to the following expression:

f^{'} (x) = \frac{1}{2} [g^{'} (x) - g^{'} (2 - x)]

The Discovery of Roots

We are given the condition

g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})

. Let us evaluate

f^{'} (x)

at specific points to identify its roots.

For

x = \frac{1}{2}

f^{'} (\frac{1}{2}) = \frac{1}{2} [g^{'} (\frac{1}{2}) - g^{'} (2 - \frac{1}{2})] = \frac{1}{2} [g^{'} (\frac{1}{2}) - g^{'} (\frac{3}{2})] = 0

For

x = \frac{3}{2}

f^{'} (\frac{3}{2}) = \frac{1}{2} [g^{'} (\frac{3}{2}) - g^{'} (2 - \frac{3}{2})] = \frac{1}{2} [g^{'} (\frac{3}{2}) - g^{'} (\frac{1}{2})] = 0

Finally, at the point of symmetry

x = 1

f^{'} (1) = \frac{1}{2} [g^{'} (1) - g^{'} (2 - 1)] = \frac{1}{2} [g^{'} (1) - g^{'} (1)] = 0

Thus, we have identified three distinct roots for

f^{'} (x)

x = \frac{1}{2}

x = 1

, and

x = \frac{3}{2}

Applying Rolle's Theorem

We now invoke Rolle's Theorem, which states that if a function is zero at two points, its derivative must vanish at some point in between.

Consider the interval

[\frac{1}{2}, 1]

. Since

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

and

f^{'} (1) = 0

, there must exist at least one

c_{1} \in (\frac{1}{2}, 1)

such that:

f^{''} (c_{1}) = 0

Next, consider the interval

[1, \frac{3}{2}]

. Since

f^{'} (1) = 0

and

f^{'} (\frac{3}{2}) = 0

, there must exist at least one

c_{2} \in (1, \frac{3}{2})

such that:

f^{''} (c_{2}) = 0

Conclusion

By applying the fundamental theorems of calculus to the inherent symmetry of the function, we have proven that $f^{''} (x) = 0$ at at least two distinct points in the interval

(0, 2)

Common Traps & FAQs

Wait, why do we check

x = 1

specifically? Is it just a guess?

That is a fantastic question! It is not a guess at all. Look at the symmetry of the function

f (x) = \frac{1}{2} [g (x) + g (2 - x)]

. The expression

2 - x

is the reflection of

x

about the point

x = 1

. When we evaluate

f^{'} (x)

x = 1

, we get

\frac{1}{2} [g^{'} (1) - g^{'} (2 - 1)]

, which is

\frac{1}{2} [g^{'} (1) - g^{'} (1)] = 0

. This point is the center of symmetry for our interval, and it naturally emerges as a root of the derivative.

Can we assume

g (x)

is a polynomial to solve this?

You could, but you do not need to! The problem only states that

g

is twice differentiable. The beauty of Rolle's Theorem is that it applies to any function that is continuous on

[a, b]

and differentiable on

(a, b)

. By using the theorem, we prove the result for *any* such function

g

, not just polynomials. This is the power of general calculus!

Why does

f^{''} (x) = 0

for *at least* two values?

Rolle's Theorem guarantees that if a function

f^{'} (x)

has roots at

x_{1}

and

x_{2}

, then there exists at least one point

c

(x_{1}, x_{2})

where

f^{''} (c) = 0

. We found three roots for

f^{'} (x)

: at

1/2

1

, and

3/2

. This gives us two intervals:

(1/2, 1)

and

(1, 3/2)

. Applying Rolle's Theorem to each interval guarantees at least one root for

f^{''} (x)

in each, leading to at least two roots in total.

Animated Solution for Mathematics - Differentiation: Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

Select Answer:

Visualized Solution

Understanding the Given Function $f (x)$

Differentiating $f (x)$ using Chain Rule

Evaluating $f^{'} (\frac{1}{2})$

Evaluating $f^{'} (\frac{3}{2})$

Evaluating $f^{'} (1)$

Applying Rolle's Theorem on $[\frac{1}{2}, 1]$

Applying Rolle's Theorem on $[1, \frac{3}{2}]$

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

Unveiling the First Derivative

The Discovery of Roots

Applying Rolle's Theorem

Conclusion

Similar Questions

Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

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,

If $f : R \to R$ is a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ , and $f (\frac{1}{2}) = f (1) = 1$ , then

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:

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

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

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 $[0, 1]$ .

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 $f^{'} (c) = 2 g^{'} (c)$ .

Select Answer:

Visualized Solution

Understanding the Given Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) using Chain Rule

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(21​)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(23​)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(1)

Applying Rolle's Theorem on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [21​,1]

Applying Rolle's Theorem on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1,23​]

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

Unveiling the First Derivative

The Discovery of Roots

Applying Rolle's Theorem

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be any continuous function on [0,2] and twice differentiable on (0,2). If f(0)=0,f(1)=1 and f(2)=2, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For all twice differentiable functions f:R→R, with f(0)=f(1)=f′(0)=0

.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; } If f:R→R is a twice differentiable function such that f′′(x)>0 for all x∈R, and f(21​)=f(1)=1, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is twice differentiable and continuous function in x∈[a,b] also f′(x)>0 and f′′(x)<0 and c∈(a,b) then f(b)−f(c)f(c)−f(a)​ is greater than

.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) 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).

Understanding the Given Function $f (x)$

Differentiating $f (x)$ using Chain Rule

Evaluating $f^{'} (\frac{1}{2})$

Evaluating $f^{'} (\frac{3}{2})$

Evaluating $f^{'} (1)$

Applying Rolle's Theorem on $[\frac{1}{2}, 1]$

Applying Rolle's Theorem on $[1, \frac{3}{2}]$

Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

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,

If $f : R \to R$ is a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ , and $f (\frac{1}{2}) = f (1) = 1$ , then

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

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

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 $f^{'} (c) = 2 g^{'} (c)$ .