JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let f and g be twice differentiable even functions on (-2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $g (\frac{3}{4}) = 0, g (1) = 2$ Then, the minimum number of solutions of $f (x) g^{''} (x) + f^{'} (x) g^{'} (x) = 0$ in (-2, 2) is equal to ______.

Q: Wait, why do we assume $g'(0) = 0$? Is it possible for $g'(0)$ to be something else?

Great question! Because $g(x)$ is an even function, we know $g(x) = g(-x)$. If we differentiate both sides using the chain rule, we get $g'(x) = -g'(-x)$. This is the definition of an odd function. For any odd function, $g'(0) = -g'(0)$, which forces $2g'(0) = 0$, meaning $g'(0)$ must be zero. It is a beautiful consequence of symmetry!

Enter Numerical Value:

Visualized Solution

Define Auxiliary Function $h (x)$

The given expression is $f (x) g^{''} (x) + f^{'} (x) g^{'} (x)$ .
By the product rule, this is exactly $\frac{d}{d x} [f (x) g^{'} (x)]$ .
Let's define an auxiliary function: $h (x) = f (x) g^{'} (x)$ .

Given Roots of $f (x)$

We are given that $f (\frac{1}{4}) = 0$ and $f (\frac{1}{2}) = 0$ .
These are two positive roots of the function $f (x)$ .

Symmetry of $f (x)$ Roots

$f (x)$ is an even function, which means $f (x) = f (- x)$ .
By symmetry, $f (- \frac{1}{4}) = 0$ and $f (- \frac{1}{2}) = 0$ .
Thus, $f (x)$ has at least $4$ roots in $(- 2, 2)$ .

Analyze the Derivative $g^{'} (x)$

$g (x)$ is also given as an even function: $g (x) = g (- x)$ .
Differentiating both sides with respect to $x$ : $g^{'} (x) = - g^{'} (- x)$ .
This proves that $g^{'} (x)$ is an odd function.

Root of $g^{'} (x)$ at Origin

For the odd function $g^{'} (x)$ , substitute $x = 0$ .
$g^{'} (0) = - g^{'} (0) ⟹ 2 g^{'} (0) = 0 ⟹ g^{'} (0) = 0$ .
Thus, $g^{'} (x)$ has at least one root at $x = 0$ .

Consolidating Roots of $h (x)$

Recall $h (x) = f (x) g^{'} (x)$ .
$h (x) = 0$ whenever $f (x) = 0$ or $g^{'} (x) = 0$ .
The combined roots are ${- \frac{1}{2}, - \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}}$ .

Rolle's Theorem Setup

Rolle's Theorem states: If $h (a) = h (b) = 0$ , then there exists $c \in (a, b)$ such that $h^{'} (c) = 0$ .
We have $5$ roots, creating $4$ distinct intervals.

Applying Rolle's Theorem

Interval 1: $(- \frac{1}{2}, - \frac{1}{4}) ⟹$ at least one root of $h^{'} (x)$ .
Interval 2: $(- \frac{1}{4}, 0) ⟹$ at least one root of $h^{'} (x)$ .
Interval 3: $(0, \frac{1}{4}) ⟹$ at least one root of $h^{'} (x)$ .
Interval 4: $(\frac{1}{4}, \frac{1}{2}) ⟹$ at least one root of $h^{'} (x)$ .

Final Conclusion

Summing up, $h^{'} (x) = 0$ has at least $1 + 1 + 1 + 1 = 4$ solutions.
Since $h^{'} (x) = f (x) g^{''} (x) + f^{'} (x) g^{'} (x)$ , the given expression has a minimum of $4$ solutions in $(- 2, 2)$ .

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The Sigma Insight: Mean Value Theorems

Solution Diagram

Analyzing the Setup

When you first look at the expression

f (x) g^{''} (x) + f^{'} (x) g^{'} (x) = 0

, it might seem like a chaotic mess of derivatives. However, this expression is a direct manifestation of the product rule.

The Auxiliary Revelation

In mathematics, the most effective way to solve a problem is often to redefine it. Let us define an auxiliary function

h (x) = f (x) g^{'} (x)

If we differentiate this with respect to

x

, we apply the product rule:

h^{'} (x) = f^{'} (x) g^{'} (x) + f (x) g^{''} (x)

Suddenly, the "terrifying" expression in our problem is revealed to be nothing more than

h^{'} (x) = 0

. We have transformed a second-order differential equation into a root-finding mission for the derivative of a function we just created.

The Power of Symmetry

We are given that

f (x)

and

g (x)

are even functions. An even function is symmetric about the

y

-axis, meaning

f (x) = f (- x)

Given that

f (1/4) = 0

and

f (1/2) = 0

, the symmetry of

f

implies that these roots must be mirrored on the negative side. Thus, we immediately identify four roots for

f (x)

x \in {- 1/2, - 1/4, 1/4, 1/2}

Now, consider

g^{'} (x)

. Since

g (x)

is an even function, its derivative

g^{'} (x)

must be an odd function, satisfying

g^{'} (x) = - g^{'} (- x)

Evaluating this at

x = 0

yields

g^{'} (0) = - g^{'} (0)

, which forces

g^{'} (0) = 0

. Consequently,

h (x) = f (x) g^{'} (x)

has a root at

x = 0

as well. We have now identified five distinct roots for

h (x)

x \in {- 1/2, - 1/4, 0, 1/4, 1/2}

The Dance of Rolle's Theorem

Rolle's Theorem states that if a function

h (x)

is differentiable and has roots at

a

and

b

, there must be at least one point

c

in the interval

(a, b)

where

h^{'} (c) = 0

Applying this to our identified roots, we map the following intervals: 1. In

(- 1/2, - 1/4)

, there is at least one root of

h^{'} (x)

. 2. In

(- 1/4, 0)

, there is at least one root of

h^{'} (x)

. 3. In

(0, 1/4)

, there is at least one root of

h^{'} (x)

. 4. In

(1/4, 1/2)

, there is at least one root of

h^{'} (x)

The Final Conclusion

By counting these intervals, we see that

h^{'} (x)

must vanish at least four times. Since

h^{'} (x)

is exactly the expression

f (x) g^{''} (x) + f^{'} (x) g^{'} (x)

, we have proven that the minimum number of solutions is 4.

By using the product rule to simplify the expression and leveraging the symmetry of even and odd functions, the path to the solution becomes clear. You do not need the exact form of

f (x)

g (x)

; you only need to understand their fundamental behavior.

Common Traps & FAQs

Wait, why do we assume

g^{'} (0) = 0

? Is it possible for

g^{'} (0)

to be something else?

Great question! Because

g (x)

is an even function, we know

g (x) = g (- x)

. If we differentiate both sides using the chain rule, we get

g^{'} (x) = - g^{'} (- x)

. This is the definition of an odd function. For any odd function,

g^{'} (0) = - g^{'} (0)

, which forces

2 g^{'} (0) = 0

, meaning

g^{'} (0)

must be zero. It is a beautiful consequence of symmetry!

Why can't we just use the given values to find the functions

f (x)

and

g (x)

The problem doesn't ask for the specific functions, and in fact, there are infinitely many functions that satisfy these conditions. We are looking for the *minimum* number of solutions, which is a classic property of Rolle's Theorem. We don't need the exact functions; we only need to know their behavior at specific points.

Does the fact that

f (1) = 1

and

g (1) = 2

matter for the final count of roots?

In this specific problem, those values confirm that the functions are not identically zero, which ensures that our roots are distinct and valid. However, for the purpose of counting roots within the interval

(- 2, 2)

, the values at

x = 1

are just boundary conditions that don't change the number of zeros we've already identified.

How do I know there aren't more roots than the ones we found?

We are asked for the *minimum* number of solutions. Rolle's Theorem guarantees at least one root between any two zeros of a differentiable function. While there could be more roots, we can only guarantee the ones that arise from the zeros we have already identified.

Why is the interval

(- 2, 2)

important?

The interval defines the domain where our functions are differentiable. By staying within

(- 2, 2)

, we ensure that all our roots are well-defined and that Rolle's Theorem is applicable throughout the entire range we are analyzing.

Enter Numerical Value:

Visualized Solution

Define Auxiliary Function $h (x)$

Given Roots of $f (x)$

Symmetry of $f (x)$ Roots

Analyze the Derivative $g^{'} (x)$

Root of $g^{'} (x)$ at Origin

Consolidating Roots of $h (x)$

Rolle's Theorem Setup

Applying Rolle's Theorem

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Auxiliary Revelation

The Power of Symmetry

The Dance of Rolle's Theorem

The Final Conclusion

Similar Questions

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 $g (x)$ .

Let $f : R \to R$ be a thrice differentiable function such that $f (0) = 0, f (1) = 1, f (2) = - 1, f (3) = 2$ and $f (4) = - 2$ . Then, the minimum number of zeros of $(3 f^{'} f^{''} + f f^{'''}) (x)$ is _______

Let $f : [0, 1] \to R$ be a twice differentiable function in $(0, 1)$ such that $f (0) = 3$ and $f (1) = 5$ . If the line $y = 2 x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$ , then the least number of points $x \in (0, 1)$ , at which $f^{''} (x) = 0$ , is ________

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

Let $f : (a, b) \to R$ be twice differentiable function such that $f (x) = \int_{a}^{x} g (t) d t$ for a differentiable function $g (x)$ . If $f (x) = 0$ has exactly five distinct roots in $(a, b)$ , then $g (x) g^{'} (x) = 0$ has at least :

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

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 $c \in] 0, 1 [$

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8$ , $f^{'} (2) = 5$ , $f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4$ for all $x \in (1, 6)$ , then :

Enter Numerical Value:

Visualized Solution

Define Auxiliary Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

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

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

Analyze the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)

Root of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x) at Origin

Consolidating Roots of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Rolle's Theorem Setup

Applying Rolle's Theorem

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Auxiliary Revelation

The Power of Symmetry

The Dance of Rolle's Theorem

The Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g:R→R be a non constant twice differentiable such that g′(21​)=g′(23​). If a real valued function f is defined as f(x)=21​[g(x)+g(2−x)], 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).

.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; } Let f be a twice differentiable function on (1,6). If f(2)=8, f′(2)=5, f′(x)≥1 and f′′(x)≥4 for all x∈(1,6), then :

Define Auxiliary Function $h (x)$

Given Roots of $f (x)$

Symmetry of $f (x)$ Roots

Analyze the Derivative $g^{'} (x)$

Root of $g^{'} (x)$ at Origin

Consolidating Roots of $h (x)$

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

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

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 $c \in] 0, 1 [$

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8$ , $f^{'} (2) = 5$ , $f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4$ for all $x \in (1, 6)$ , then :