JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Limits, Continuity and Differentiability: Mathematics | JEE Main

Q: Do multiple roots give the same point?

Since is a polynomial in , each unique gives exactly one unique , so distinct real -roots always correspond to distinct points on the curve.

The number of points on the curve

y=54 x^{5}-135 x^{4}-70 x^{3}+180 x^{2}+210 x

at which the normal lines are parallel to

\mathrm{x}+90 \mathrm{y}+2=0

is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

4

2

0

3

Step-by-step solutionView

Correct answer

Setting the derivative equal to 90 gives a degree-4 polynomial in $x$ that factors into $(3x^2 - x - 2)(9x^2 - 15x - 2) = 0$ , yielding 4 distinct real roots and thus 4 points.

Option analysis

Why each option works or fails

A ·

4

None. The student correctly relates the normal slope to the tangent slope, differentiates the curve, and solves for the number of real roots of the resulting quartic equation. This is correct because the slope of the given line is $-1/90$ , making the normal slope $-1/90$ and the tangent slope $dy/dx = 90$ . The quartic $270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$ reduces to $(3x^2 - x - 2)(9x^2 - 15x - 2) = 0$ , each quadratic having positive discriminant and no common roots, yielding 4 real points.

B ·

2

The student may assume only one of the quadratic factors has real roots, or mistakenly set the tangent slope equal to $-1/90$ instead of $90$ . Check the discriminant of both quadratic factors: $(-1)^2 - 4(3)(-2) = 25 > 0$ and $(-15)^2 - 4(9)(-2) = 297 > 0$ . Both factors have two distinct real roots.

C ·

0

The student may mistakenly believe the quartic has negative discriminants across all factors, leading to zero real solutions. Evaluate the constant and leading coefficients; since both quadratic factors have negative constant terms when the leading terms are positive, their discriminants are strictly positive, guaranteeing real roots.

D ·

3

The student may assume that one root is repeated between the two quadratic factors without verifying their resultant or common root condition. Solve the quadratics directly or test roots: $3x^2 - x - 2 = 0$ gives $x = 1, -2/3$ . Neither satisfies $9x^2 - 15x - 2 = 0$ , so all 4 roots are distinct.

Reviewed route

Solution

StepWorking

01given

Curve: $y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$ . Given line: $x + 90y + 2 = 0$ .

02goal

Find the number of points on the curve where the normal line is parallel to $x + 90y + 2 = 0$ .

03approach

Slope of the given line is $m = -1/90$ . Since the normal is parallel to this line, the slope of the normal is $-1/90$ , meaning the slope of the tangent $\frac{dy}{dx}$ must be $90$ . Differentiate $y$ with respect to $x$ , set $\frac{dy}{dx} = 90$ , and count the number of real roots of the resulting polynomial.

04execute

Compute the derivative: $\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$ Equating $\frac{dy}{dx} = 90$ : $270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$ Dividing through by $30$ : $9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$

05execute

Factor the quartic polynomial $P(x) = 9x^4 - 18x^3 - 7x^2 + 12x + 4$ . Testing rational roots: $P(1) = 9 - 18 - 7 + 12 + 4 = 0 \implies (x - 1)$ is a factor. $P(2) = 9(16) - 18(8) - 7(4) + 12(2) + 4 = 144 - 144 - 28 + 24 + 4 = 0 \implies (x - 2)$ is a factor. Dividing $P(x)$ by $(x - 1)(x - 2) = x^2 - 3x + 2$ yields $9x^2 + 9x + 2 = (3x + 1)(3x + 2)$ . Thus, $P(x) = (x - 1)(x - 2)(3x + 1)(3x + 2) = 0$ , giving 4 distinct real roots: $x = 1, 2, -1/3, -2/3$ .

✓verify

Since $y(x)$ is a single-valued polynomial function of $x$ , each distinct $x$ -value yields a unique point $(x, y(x))$ on the curve. All 4 roots are real and distinct, so there are 4 points.

Hints that build this answer step by step

What must the value of the derivative $\frac{dy}{dx}$ be at any point where the normal line is parallel to $x + 90y + 2 = 0$ ?
$\frac{dy}{dx} = 90$
Differentiating $y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$ and setting $\frac{dy}{dx} = 90$ , what simplified polynomial equation in $x$ is obtained after dividing by $30$ ?
$9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$
How many distinct real roots does the polynomial $9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$ have?
4 distinct real roots

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Do multiple roots give the same point?

Since $y$ is a polynomial in $x$ , each unique $x$ gives exactly one unique $y$ , so distinct real $x$ -roots always correspond to distinct points on the curve.

Answer

Setting the derivative equal to 90 gives a degree-4 polynomial in $x$ that factors into $(3x^2 - x - 2)(9x^2 - 15x - 2) = 0$ , yielding 4 distinct real roots and thus 4 points.

Why each option works or fails

A · correct: $4$ - None. The student correctly relates the normal slope to the tangent slope, differentiates the curve, and solves for the number of real roots of the resulting quartic equation. This is correct because the slope of the given line is $-1/90$ , making the normal slope $-1/90$ and the tangent slope $dy/dx = 90$ . The quartic $270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$ reduces to $(3x^2 - x - 2)(9x^2 - 15x - 2) = 0$ , each quadratic having positive discriminant and no common roots, yielding 4 real points.
B: $2$ - The student may assume only one of the quadratic factors has real roots, or mistakenly set the tangent slope equal to $-1/90$ instead of $90$ . Check the discriminant of both quadratic factors: $(-1)^2 - 4(3)(-2) = 25 > 0$ and $(-15)^2 - 4(9)(-2) = 297 > 0$ . Both factors have two distinct real roots.
C: $0$ - The student may mistakenly believe the quartic has negative discriminants across all factors, leading to zero real solutions. Evaluate the constant and leading coefficients; since both quadratic factors have negative constant terms when the leading terms are positive, their discriminants are strictly positive, guaranteeing real roots.
D: $3$ - The student may assume that one root is repeated between the two quadratic factors without verifying their resultant or common root condition. Solve the quadratics directly or test roots: $3x^2 - x - 2 = 0$ gives $x = 1, -2/3$ . Neither satisfies $4$ , so all 4 roots are distinct.

Step-by-step solution

given: Curve: $y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$ . Given line: $x + 90y + 2 = 0$ .
goal: Find the number of points on the curve where the normal line is parallel to $x + 90y + 2 = 0$ .
approach: Slope of the given line is $m = -1/90$ . Since the normal is parallel to this line, the slope of the normal is $-1/90$ , meaning the slope of the tangent $\frac{dy}{dx}$ must be $90$ . Differentiate $y$ with respect to $x$ , set $\frac{dy}{dx} = 90$ , and count the number of real roots of the resulting polynomial.
execute: Compute the derivative: $\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$ Equating $\frac{dy}{dx} = 90$ : $270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$ Dividing through by $30$ : $9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$
execute: Factor the quartic polynomial $P(x) = 9x^4 - 18x^3 - 7x^2 + 12x + 4$ . Testing rational roots: $P(1) = 9 - 18 - 7 + 12 + 4 = 0 \implies (x - 1)$ is a factor. $P(2) = 9(16) - 18(8) - 7(4) + 12(2) + 4 = 144 - 144 - 28 + 24 + 4 = 0 \implies (x - 2)$ is a factor. Dividing $P(x)$ by $(x - 1)(x - 2) = x^2 - 3x + 2$ yields $9x^2 + 9x + 2 = (3x + 1)(3x + 2)$ . Thus, $P(x) = (x - 1)(x - 2)(3x + 1)(3x + 2) = 0$ , giving 4 distinct real roots: $x = 1, 2, -1/3, -2/3$ .
verify: Since $y(x)$ is a single-valued polynomial function of $x$ , each distinct $x$ -value yields a unique point $(x, y(x))$ on the curve. All 4 roots are real and distinct, so there are 4 points.

Limits, Continuity and Differentiability: Mathematics | JEE Main

What feels right?

Setting the derivative equal to 90 gives a degree-4 polynomial in $x$ that factors into $(3x^2 - x - 2)(9x^2 - 15x - 2) = 0$ , yielding 4 distinct real roots and thus 4 points.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

We think you should solve this next

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $a_1, a_2, a_3, \dots$ be an A.P. If $a_7 = 3$ , the product $a_1 a_4$ is minimum and…

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix…

Students also ask

Limits, Continuity and Differentiability: Mathematics | JEE Main

Setting the derivative equal to 90 gives a degree-4 polynomial in xxx that factors into (3x2−x−2)(9x2−15x−2)=0(3x^2 - x - 2)(9x^2 - 15x - 2) = 0(3x2−x−2)(9x2−15x−2)=0, yielding 4 distinct real roots and thus 4 points.

01given

02goal

03approach

04execute

05execute

✓verify

Let g:R→Rg: R \rightarrow Rg:R→R be a non constant twice differentiable such that…

Let a1,a2,a3,…a_1, a_2, a_3, \dotsa1​,a2​,a3​,… be an A.P. If a7=3a_7 = 3a7​=3, the product a1a4a_1 a_4a1​a4​ is minimum and…

Let y=f(x)y=f(x)y=f(x) represent a parabola with focus (−12,0)\left(-\frac{1}{2}, 0\right)(−21​,0) and directrix…

Students also ask

Setting the derivative equal to 90 gives a degree-4 polynomial in $x$ that factors into $(3x^2 - x - 2)(9x^2 - 15x - 2) = 0$ , yielding 4 distinct real roots and thus 4 points.

Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

Let $a_1, a_2, a_3, \dots$ be an A.P. If $a_7 = 3$ , the product $a_1 a_4$ is minimum and…

Let $y=f(x)$ represent a parabola with focus $\left(-\frac{1}{2}, 0\right)$ and directrix…