JEE Main 2024 (09 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let the set of all values of $p$ , for which $f (x) = (p^{2} - 6 p + 8) (sin^{2} 2 x - cos^{2} 2 x) + 2 (2 - p) x + 7$ does not have any critical point, be the interval $(a, b)$ . Then $16 ab$ is equal to _______

Q: Wait, why do we reject the case where $p = 2$?

Great observation! If you substitute $p = 2$ into the derivative $f'(x)$, you get $f'(x) = 0$ for all $x$. This means every single point on the function is a critical point, which is the exact opposite of what we want. We need the derivative to never be zero, so $p = 2$ must be excluded.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure calculus problem disguised in function notation! While physics often uses energy conservation, here we are dealing with the fundamental definition of critical points in calculus. The 'energy' here is the function itself, and we are analyzing its rate of change, which is the derivative. Stick to the calculus toolkit for this one.

Q: How do we know the amplitude of the sine wave must be less than 1?

We have the condition $2(p - 4)\sin 4x = 1$. If the maximum value of the left side (which is $|2(p - 4)|$) is less than $1$, then the expression $2(p - 4)\sin 4x$ can never reach $1$. It's like trying to jump over a wall that is higher than your maximum jump height; you will never touch the top of the wall.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): To ensure the function

f (x)

has no critical points, we must ensure its derivative

f^{'} (x)

never equals zero. Visualize the derivative as a function that must stay strictly above or below the x-axis for all real

x

Hint 2

Hint 2 (The Arsenal): Before differentiating, simplify the function using the trigonometric identity

cos^{2} θ - sin^{2} θ = cos 2 θ

. This will transform the expression into a much more manageable form, allowing you to easily factor out common terms after differentiation.

Hint 3

Hint 3 (The Execution): After differentiating, you will arrive at an expression of the form

2 (p - 2) [2 (p - 4) sin 4 x - 1] = 0

. For this to have no solutions, the term inside the bracket must never be zero. This implies that the amplitude of the sine wave must be restricted such that it never reaches the value

1

Visualized Solution

$C o n d i t i o n f or N o C r i t i c a l P o in t s$

A function $f (x)$ has no critical points if its derivative never vanishes.
Mathematically, $f^{'} (x) \neq = 0$ for all $x \in R$ .

$F a c t or i z in g t h e Q u a d r a t i c C oe f f i c i e n t$

Given function: $f (x) = (p^{2} - 6 p + 8) (sin^{2} 2 x - cos^{2} 2 x) + 2 (2 - p) x + 7$
Factorize the quadratic term: $p^{2} - 6 p + 8 = (p - 2) (p - 4)$

$A ppl y in g T r i g o n o m e t r i c I d e n t i t y$

Recall the double angle identity: $cos^{2} θ - sin^{2} θ = cos 2 θ$
Apply it to our term: $sin^{2} 2 x - cos^{2} 2 x = - (cos^{2} 2 x - sin^{2} 2 x)$
Result: $- cos 4 x$

$R e w r i t in g t h e F u n c t i o n f (x)$

Substitute the simplified terms back into the original function.
$f (x) = - (p - 2) (p - 4) cos 4 x + 2 (2 - p) x + 7$

$D i f f er e n t ia t in g w i t h R es p ec tt o x$

Differentiate $f (x)$ with respect to $x$ :
$f^{'} (x) = - (p - 2) (p - 4) (- sin 4 x \cdot 4) + 2 (2 - p)$
$f^{'} (x) = 4 (p - 2) (p - 4) sin 4 x - 2 (p - 2)$

$F a c t or in g t h eD er i v a t i v e$

Notice that $2 (2 - p) = - 2 (p - 2)$ .
Factor out $2 (p - 2)$ from both terms:
$f^{'} (x) = 2 (p - 2) [2 (p - 4) sin 4 x - 1]$

$A ppl y in g t h e N o C r i t i c a l P o in t s C o n d i t i o n$

We require $f^{'} (x) \neq = 0$ for all $x$ .
$2 (p - 2) [2 (p - 4) sin 4 x - 1] \neq = 0$
If $p = 2$ , $f^{'} (x) = 0$ everywhere (infinite critical points). Thus, $p \neq = 2$ .
Therefore, we must have $2 (p - 4) sin 4 x - 1 \neq = 0$ .

$V i s u a l i z in g t h e S in e T er m C o n s t r ain t$

Rearranging gives: $2 (p - 4) sin 4 x \neq = 1$ for all $x$ .
Let's visualize the function $y = 2 (p - 4) sin 4 x$ .
If its amplitude is too large, it will intersect the line $y = 1$ .

$B o u n d in g t h e A m pl i t u d e$

To ensure the sine wave NEVER touches $y = 1$ or $y = - 1$ , its maximum height must be strictly less than $1$ .
Amplitude of $2 (p - 4) sin 4 x$ is $∣2 (p - 4) ∣$ .
Therefore, $∣2 (p - 4) ∣ < 1$ .

$S o l v in g t h e I n e q u a l i t y$

We have the inequality: $∣2 (p - 4) ∣ < 1$
Divide by 2: $∣ p - 4∣ < \frac{1}{2}$
Open the modulus: $- \frac{1}{2} < p - 4 < \frac{1}{2}$

$I d e n t i f y in g t h e I n t er v a l (a, b)$

Add $4$ to all parts of the inequality:
$4 - \frac{1}{2} < p < 4 + \frac{1}{2}$
$\frac{7}{2} < p < \frac{9}{2}$
The interval is $(a, b) = (\frac{7}{2}, \frac{9}{2})$ . So, $a = \frac{7}{2}$ and $b = \frac{9}{2}$ .

$F ina l C a l c u l a t i o n o f 16 ab$

We need to find the value of $16 ab$ .
Substitute $a$ and $b$ : $16 ab = 16 \cdot \frac{7}{2} \cdot \frac{9}{2}$
Simplify: $16 ab = 16 \cdot \frac{63}{4}$
Final Answer: $4 \cdot 63 = 252$

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

The function is defined as

f (x) = (p^{2} - 6 p + 8) (sin^{2} 2 x - cos^{2} 2 x) + 2 (2 - p) x + 7

. We seek the values of

p

such that

f (x)

has no critical points, meaning $f'(x) eq 0$ for all

x \in R

First, we factor the quadratic coefficient:

p^{2} - 6 p + 8 = (p - 2) (p - 4)

Next, we simplify the trigonometric component using the identity

cos 2 θ = cos^{2} θ - sin^{2} θ

. Since

sin^{2} 2 x - cos^{2} 2 x = - (cos^{2} 2 x - sin^{2} 2 x)

, this expression becomes

- cos 4 x

The function simplifies to:

f (x) = - (p - 2) (p - 4) cos 4 x + 2 (2 - p) x + 7

The Master Equation

To find the critical points, we differentiate

f (x)

with respect to

x

f^{'} (x) = 4 (p - 2) (p - 4) sin 4 x + 2 (2 - p)

We can rewrite

2 (2 - p)

- 2 (p - 2)

. Factoring out

2 (p - 2)

, we obtain:

f^{'} (x) = 2 (p - 2) [2 (p - 4) sin 4 x - 1]

For

f (x)

to have no critical points,

f^{'} (x)

must never be zero. Since $p eq 2$ (otherwise

f^{'} (x) = 0

identically), we require the bracketed term to never vanish:

2 (p - 4) sin 4 x e q 1 ⟹ sin 4 x e q \frac{1}{2 ( p - 4 )}

The No-Go Zone

The range of the sine function is

[- 1, 1]

. If the value

\frac{1}{2 ( p - 4 )}

lies within this interval, the derivative will necessarily equal zero for some

x

To ensure $f'(x) eq 0$, the value must lie strictly outside the interval

[- 1, 1]

. This implies:

\frac{1}{2 ( p - 4 )} > 1

This inequality is equivalent to:

∣2 (p - 4) ∣ < 1 ⟹ ∣ p - 4∣ < \frac{1}{2}

Solving the inequality

- \frac{1}{2} < p - 4 < \frac{1}{2}

by adding

4

to all sides yields:

\frac{7}{2} < p < \frac{9}{2}

Thus, the interval

(a, b)

(\frac{7}{2}, \frac{9}{2})

, where

a = \frac{7}{2}

and

b = \frac{9}{2}

Final Calculation

The problem asks for the value of

16 ab

. Substituting our values for

a

and

b

16 \cdot (\frac{7}{2}) \cdot (\frac{9}{2}) = 16 \cdot \frac{63}{4}

16 \cdot \frac{63}{4} = 4 \cdot 63 = 252

The final result is 252.

Common Traps & FAQs

Wait, why do we reject the case where

p = 2

Great observation! If you substitute

p = 2

into the derivative

f^{'} (x)

, you get

f^{'} (x) = 0

for all

x

. This means every single point on the function is a critical point, which is the exact opposite of what we want. We need the derivative to never be zero, so

p = 2

must be excluded.

Why can't we just use energy conservation or some other physics principle?

This is a pure calculus problem disguised in function notation! While physics often uses energy conservation, here we are dealing with the fundamental definition of critical points in calculus. The 'energy' here is the function itself, and we are analyzing its rate of change, which is the derivative. Stick to the calculus toolkit for this one.

How do we know the amplitude of the sine wave must be less than 1?

We have the condition

2 (p - 4) sin 4 x = 1

. If the maximum value of the left side (which is

∣2 (p - 4) ∣

) is less than

1

, then the expression

2 (p - 4) sin 4 x

can never reach

1

. It's like trying to jump over a wall that is higher than your maximum jump height; you will never touch the top of the wall.

Animated Solution for Mathematics - Differentiation: Let the set of all values of $p$ , for which $f (x) = (p^{2} - 6 p + 8) (sin^{2} 2 x - cos^{2} 2 x) + 2 (2 - p) x + 7$ does not have any critical point, be the interval $(a, b)$ . Then $16 ab$ is equal to _______

Enter Numerical Value:

Visualized Solution

$C o n d i t i o n f or N o C r i t i c a l P o in t s$

$F a c t or i z in g t h e Q u a d r a t i c C oe f f i c i e n t$

$A ppl y in g T r i g o n o m e t r i c I d e n t i t y$

$R e w r i t in g t h e F u n c t i o n f (x)$

$D i f f er e n t ia t in g w i t h R es p ec tt o x$

$F a c t or in g t h eD er i v a t i v e$

$A ppl y in g t h e N o C r i t i c a l P o in t s C o n d i t i o n$

$V i s u a l i z in g t h e S in e T er m C o n s t r ain t$

$B o u n d in g t h e A m pl i t u d e$

$S o l v in g t h e I n e q u a l i t y$

$I d e n t i f y in g t h e I n t er v a l (a, b)$

$F ina l C a l c u l a t i o n o f 16 ab$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Master Equation

The No-Go Zone

Final Calculation

Similar Questions

Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$ have a maxima for some value of $x < 0$ and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

The range of $a \in R$ for which the function $f (x) = (4 a - 3) (x + lo g_{e} 5) + 2 (a - 7) cot (\frac{x}{2}) sin^{2} (\frac{x}{2})$ , $x \neq = 2 nπ, n \in N$ , has critical points, is

Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x ^{2} + x + 2}{x ^{2} + 5 x + 6} < 0$ , be $(α, β)$ . Then $α^{2} + β^{2}$ is equal to _________

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{∣ t + 1∣}{t ^{2}}, t < 0$ , is strictly decreasing. Then the local maximum value of the function $g (x) = 2 lo g_{e} (x - 2) + α x^{2} + 4 x - α, x > 2$ , is .........

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where $a > 0$ , attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2} = q$ , then $a$ equals

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If the total maximum value of the function $f (x) = (\frac{3 e}{2 s i n x})^{s i n^{2} x}, x \in (0, \frac{π}{2})$ , is $k$ , then $\left( rac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8$ is equal to

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where $a > 0$ , attains its local maximum and local minimum values at $p$ and $q$ , respectively, such that $p^{2} = q$ , then $f (3)$ is equal to:

The maximum value of the function $f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48$ on the set $A = {x ∣ x^{2} + 20 \leq 9 x}$ is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ConditionforNoCriticalPoints

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FactorizingtheQuadraticCoefficient

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApplyingTrigonometricIdentity

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DifferentiatingwithRespecttox

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FactoringtheDerivative

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApplyingtheNoCriticalPointsCondition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VisualizingtheSineTermConstraint

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BoundingtheAmplitude

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SolvingtheInequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IdentifyingtheInterval(a,b)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCalculationof16ab

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Master Equation

The No-Go Zone

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum value of the function f(x)=2x3−15x2+36x−48 on the set A={x∣x2+20≤9x} is

$C o n d i t i o n f or N o C r i t i c a l P o in t s$

$F a c t or i z in g t h e Q u a d r a t i c C oe f f i c i e n t$

$A ppl y in g T r i g o n o m e t r i c I d e n t i t y$

$R e w r i t in g t h e F u n c t i o n f (x)$

$D i f f er e n t ia t in g w i t h R es p ec tt o x$

$F a c t or in g t h eD er i v a t i v e$

$A ppl y in g t h e N o C r i t i c a l P o in t s C o n d i t i o n$

$V i s u a l i z in g t h e S in e T er m C o n s t r ain t$

$B o u n d in g t h e A m pl i t u d e$

$S o l v in g t h e I n e q u a l i t y$

$I d e n t i f y in g t h e I n t er v a l (a, b)$

$F ina l C a l c u l a t i o n o f 16 ab$

Let the function $f (x) = 2 x^{3} + (2 p - 7) x^{2} + 3 (2 p - 9) x - 6$ have a maxima for some value of $x < 0$ and a minima for some value of $x > 0$ . Then, the set of all values of $p$ is

The range of $a \in R$ for which the function $f (x) = (4 a - 3) (x + lo g_{e} 5) + 2 (a - 7) cot (\frac{x}{2}) sin^{2} (\frac{x}{2})$ , $x \neq = 2 nπ, n \in N$ , has critical points, is

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where $a > 0$ , attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2} = q$ , then $a$ equals

The number of critical points of the function $f (x) = (x - 2)^{2/3} (2 x + 1)$ is

If the total maximum value of the function $f (x) = (\frac{3 e}{2 s i n x})^{s i n^{2} x}, x \in (0, \frac{π}{2})$ , is $k$ , then $\left( rac{k}{e}\right)^8 + \frac{k^8}{e^5} + k^8$ is equal to

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$ , where $a > 0$ , attains its local maximum and local minimum values at $p$ and $q$ , respectively, such that $p^{2} = q$ , then $f (3)$ is equal to:

The maximum value of the function $f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48$ on the set $A = {x ∣ x^{2} + 20 \leq 9 x}$ is