JEE Main 2023 (06 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: The number of points, where the curve $y = x^{5} - 20 x^{3} + 50 x + 2$ crosses the $x$ -axis, is _____.

Enter Numerical Value:

Visualized Solution

Defining the Function

Let the given curve be represented by the function:
$f (x) = x^{5} - 20 x^{3} + 50 x + 2$
We need to find the number of points where it crosses the x-axis.
This is equivalent to finding the number of real roots of $f (x) = 0$ .

Finding the Derivative

To understand the shape of the curve, we need its turning points.
Differentiate $f (x)$ with respect to $x$ :
$f^{'} (x) = 5 x^{4} - 60 x^{2} + 50$
Factor out the constant $5$ :
$f^{'} (x) = 5 (x^{4} - 12 x^{2} + 10)$

Setting up for Critical Points

Critical points occur where the derivative is zero.
Set $f^{'} (x) = 0$ :
$5 (x^{4} - 12 x^{2} + 10) = 0$
$\Rightarrow x^{4} - 12 x^{2} + 10 = 0$

Solving the Bi-quadratic Equation

Let $u = x^{2}$ . The equation becomes $u^{2} - 12 u + 10 = 0$ .
Using the quadratic formula:
$x^{2} = \frac{12 \pm ( - 12 ) ^{2} - 4 ( 1 ) ( 10 )}{2}$
$x^{2} = \frac{12 \pm 144 - 40}{2}$
$x^{2} = \frac{12 \pm 104}{2} = 6 \pm 26$

Approximating Critical Values

Approximate $26 \approx 5.1$ .
$x^{2} \approx 6 \pm 5.1$
So, $x^{2} \approx 11.1$ or $x^{2} \approx 0.9$ .
Taking the square root gives four critical points:
$x \approx \pm 3.33$ and $x \approx \pm 0.95$

Visualizing Turning Points

The four critical points divide the x-axis into intervals.
These are the $x$ -coordinates where the curve will turn.
Four turning points mean the curve will oscillate up and down.

The Intermediate Value Theorem (IVT)

We use the Intermediate Value Theorem.
If $f (a)$ and $f (b)$ have opposite signs, there is at least one root in $(a, b)$ .
Let's start with an easy anchor point:
$f (0) = 0 - 0 + 0 + 2 = 2 > 0$

Testing Interval 1: Far Left

Check values on the extreme left:
$f (- 5) = - 3125 + 2500 - 250 + 2 = - 873 < 0$
$f (- 4) = - 1024 + 1280 - 200 + 2 = 58 > 0$
Sign change from negative to positive!
Root 1 exists in the interval $(- 5, - 4)$ .

Testing Interval 2: Moving Right

Check around the next turning point:
$f (- 2) = - 32 + 160 - 100 + 2 = 30 > 0$
$f (- 1) = - 1 + 20 - 50 + 2 = - 29 < 0$
Sign change from positive to negative!
Root 2 exists in the interval $(- 2, - 1)$ .

Testing Interval 3: Near Origin

We already know $f (- 1) = - 29 < 0$ .
And our anchor point $f (0) = 2 > 0$ .
Sign change from negative to positive!
Root 3 exists in the interval $(- 1, 0)$ .

Testing Interval 4: Right Side

Continuing past the origin:
$f (1) = 1 - 20 + 50 + 2 = 33 > 0$
$f (2) = 32 - 160 + 100 + 2 = - 26 < 0$
Sign change from positive to negative!
Root 4 exists in the interval $(1, 2)$ .

Testing Interval 5: Far Right

Check the final turning point region:
$f (4) = 1024 - 1280 + 200 + 2 = - 54 < 0$
$f (5) = 3125 - 2500 + 250 + 2 = 877 > 0$
Sign change from negative to positive!
Root 5 exists in the interval $(4, 5)$ .

Tracing the Curve

We have found 5 distinct intervals containing a root.
Since a 5th-degree polynomial can have at most 5 real roots, we have found them all.
Let's visualize the complete curve passing through these points.

Final Conclusion

The curve crosses the x-axis exactly 5 times.
Final Answer: The number of points is 5.
Key Takeaway: Use derivatives to find turning points, and IVT to confirm roots between them.

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Function

We are given the polynomial function:

f (x) = x^{5} - 20 x^{3} + 50 x + 2

To understand the topography of this curve and determine the number of real roots, we must identify its turning points. We begin by calculating the first derivative:

f^{'} (x) = 5 x^{4} - 60 x^{2} + 50

Finding Critical Points

Setting the derivative to zero allows us to find the stationary points where the function changes direction:

5 (x^{4} - 12 x^{2} + 10) = 0

By substituting

u = x^{2}

, we transform this into a quadratic equation:

u^{2} - 12 u + 10 = 0

Using the quadratic formula, we solve for

u

u = \frac{12 \pm 144 - 40}{2} = 6 \pm 26

Since

x^{2} = u

, we obtain four critical points:

x = \pm 6 + 26 \approx \pm 3.43

x = \pm 6 - 26 \approx \pm 0.94

Applying the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) guarantees that if a continuous function changes sign over an interval, it must cross the

x

-axis within that interval. We evaluate the function at strategic points to observe these sign changes:

$f (- 5) = - 3125 + 2500 - 250 + 2 = - 873$ (Negative)

f (- 4) = - 1024 + 1280 - 200 + 2 = 58

(Positive) $f (- 2) = - 32 + 160 - 100 + 2 = 30$ (Positive)

f (- 1) = - 1 + 20 - 50 + 2 = - 29

(Negative) $f (0) = 2$ (Positive)

f (1) = 1 - 20 + 50 + 2 = 33

(Positive) $f (2) = 32 - 160 + 100 + 2 = - 26$ (Negative)

f (4) = 1024 - 1280 + 200 + 2 = - 54

(Negative) *

f (5) = 3125 - 2500 + 250 + 2 = 877

(Positive)

Conclusion

By observing the sign changes, we identify the existence of roots in the following intervals: 1.

(- 5, - 4)

(- 2, - 1)

(- 1, 0)

(1, 2)

(4, 5)

Because a 5th-degree polynomial can have at most 5 real roots, and we have identified 5 distinct intervals containing sign changes, we conclude that the function has exactly 5 real roots.

Common Traps & FAQs

Wait, why can't we just solve

f (x) = 0

directly using the formula?

Because this is a quintic equation! Unlike quadratic equations, there is no general algebraic formula to solve for the roots of a 5th-degree polynomial. We must rely on calculus and the Intermediate Value Theorem to 'trap' the roots. It is not about finding the exact value, but proving the existence of the crossing.

How do I know which values to test for the Intermediate Value Theorem?

Great question! You do not guess randomly. Use the critical points you found from the derivative. The function changes its direction only at these points. Therefore, the roots must be 'trapped' in the intervals defined by these critical points. By testing integers around these critical points, you are guaranteed to find the sign changes.

Is it possible for the curve to have fewer than 5 roots?

Absolutely. A 5th-degree polynomial can have 1, 3, or 5 real roots. However, by checking the sign changes across the critical intervals, we are systematically proving that in this specific case, the curve crosses the axis 5 times. We are not assuming; we are verifying.

Animated Solution for Mathematics - Differentiation: The number of points, where the curve $y = x^{5} - 20 x^{3} + 50 x + 2$ crosses the $x$ -axis, is _____.

Enter Numerical Value:

Visualized Solution

Defining the Function

Finding the Derivative

Setting up for Critical Points

Solving the Bi-quadratic Equation

Approximating Critical Values

Visualizing Turning Points

The Intermediate Value Theorem (IVT)

Testing Interval 1: Far Left

Testing Interval 2: Moving Right

Testing Interval 3: Near Origin

Testing Interval 4: Right Side

Testing Interval 5: Far Right

Tracing the Curve

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Function

Finding Critical Points

Applying the Intermediate Value Theorem

Conclusion

Similar Questions

The number of distinct real roots of $x^{4} - 4 x^{3} + 12 x^{2} + x - 1 = 0$ is .........

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

The number of distinct real roots of the equation $3 x^{4} + 4 x^{3} - 12 x^{2} + 4 = 0$ is .

The number of distinct real roots of $x^{4} - 4 x + 1 = 0$ is :

The curve $y (x) = a x^{3} + b x^{2} + c x + 5$ touches the x-axis at the point $P (- 2, 0)$ and cuts the y-axis at the point $Q$ , where $y^{'}$ is equal to 3. Then the local maximum value of $y (x)$ is :

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point:

If the set of all values of a, for which the equation $5 x^{3} - 15 x - a = 0$ has three distinct real roots, is the interval $(α, β)$ , then $β - 2 α$ is equal to

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f : R \to R$ be defined as $f (x) = ∣ x ∣ + ∣ x^{2} - 1∣$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of points, where the curve y=x5−20x3+50x+2 crosses the x-axis, is _____.

Enter Numerical Value:

Visualized Solution

Defining the Function

Finding the Derivative

Setting up for Critical Points

Solving the Bi-quadratic Equation

Approximating Critical Values

Visualizing Turning Points

The Intermediate Value Theorem (IVT)

Testing Interval 1: Far Left

Testing Interval 2: Moving Right

Testing Interval 3: Near Origin

Testing Interval 4: Right Side

Testing Interval 5: Far Right

Tracing the Curve

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Function

Finding Critical Points

Applying the Intermediate Value Theorem

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of distinct real roots of x4−4x3+12x2+x−1=0 is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of critical points of the function f(x)=(x−2)2/3(2x+1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of distinct real roots of the equation 3x4+4x3−12x2+4=0 is .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of distinct real roots of x4−4x+1=0 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The curve y(x)=ax3+bx2+cx+5 touches the x-axis at the point P(−2,0) and cuts the y-axis at the point Q, where y′ is equal to 3. Then the local maximum value of y(x) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum slope of the curve y=21​x4−5x3+18x2−19x occurs at the point:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the set of all values of a, for which the equation 5x3−15x−a=0 has three distinct real roots, is the interval (α,β), then β−2α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 5 such that x=±1 are its critical points. If limx→0​(2+x3f(x)​)=4, then which one of the following is not true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)=∣x∣+∣x2−1∣. The total number of points at which f attains either a local maximum or a local minimum is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 5 such that x=±1 are its critical points. If limx→0​(2+x3f(x)​)=4, then which one of the following is not true?

Animated Solution for Mathematics - Differentiation: The number of points, where the curve $y = x^{5} - 20 x^{3} + 50 x + 2$ crosses the $x$ -axis, is _____.

The number of distinct real roots of $x^{4} - 4 x^{3} + 12 x^{2} + x - 1 = 0$ is .........

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

The number of distinct real roots of the equation $3 x^{4} + 4 x^{3} - 12 x^{2} + 4 = 0$ is .

The number of distinct real roots of $x^{4} - 4 x + 1 = 0$ is :

The curve $y (x) = a x^{3} + b x^{2} + c x + 5$ touches the x-axis at the point $P (- 2, 0)$ and cuts the y-axis at the point $Q$ , where $y^{'}$ is equal to 3. Then the local maximum value of $y (x)$ is :

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point:

If the set of all values of a, for which the equation $5 x^{3} - 15 x - a = 0$ has three distinct real roots, is the interval $(α, β)$ , then $β - 2 α$ is equal to

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f : R \to R$ be defined as $f (x) = ∣ x ∣ + ∣ x^{2} - 1∣$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?