JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding Points of a Curve with a Specified Normal Line Slope

Q: Why is dy/dx = 90 and not -1/90?

The normal line is parallel to the given line, meaning m_N = -1/90. Since normal and tangent are perpendicular, m_T · m_N = -1, which gives m_T = dy/dx = -1/(-1/90) = 90.

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 x+90 y+2=0 is:

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Step-by-step solutionView

Correct answer

The number of points on the curve where the normal lines are parallel to $x + 90 y + 2 = 0$ is $4$ , which corresponds to the number of real roots of the derivative equation $y^{'} = 90$ .

Option analysis

Why each option works or fails

A · 4

No misconception. This is the correct option. We correctly equated the normal slope $- 1 / y^{'}$ to the line's slope $- 1 / 90$ , yielding $y^{'} = 90$ , and found that the resulting equation $270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 120 = 0$ has $4$ distinct real roots.

B · 2

The student made an algebraic error when factoring or analyzing the roots of the derivative equation, concluding that only $2$ real roots exist. Ensure careful application of Descartes' Rule of Signs or interval analysis to determine the exact number of real roots of the polynomial equation.

C · 0

The student made an error in equating the slope of the normal, perhaps setting the tangent slope $y^{'} = - 1 / 90$ or making a sign error, leading to an equation with no real roots. Recall that the slope of the normal line is $- 1 / y^{'}$ . Since the given line $x + 90 y + 2 = 0$ has slope $- 1 / 90$ , we must solve $y^{'} = 90$ .

D · 3

The student miscounted the number of real roots of the simplified quartic equation, missing one root. Verify the signs of the function values at key test points (such as local extrema) to accurately determine the number of real roots.

Reviewed route

Solution

StepWorking

01Given

Curve equation $y = 54 x^{5} - 135 x^{4} - 70 x^{3} + 180 x^{2} + 210 x$ and line $x + 90 y + 2 = 0$ .

02Goal

Find the number of points on the curve where the normal lines are parallel to the given line.

03Approach

The slope of the line $x + 90 y + 2 = 0$ is $m_{L} = - 1 / 90$ . Since the normal is parallel to this line, the slope of the normal is $m_{N} = - 1 / 90$ . Consequently, the slope of the tangent is $m_{T} = - 1 / m_{N} = 90$ . We compute $d y / d x$ , set it equal to $90$ , and determine the number of distinct real roots.

04Execute

Differentiate $y$ with respect to $x$ : $\frac{d y}{d x} = 270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 210$ Equating $\frac{d y}{d x} = 90$ : $270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 210 = 90$ $270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 120 = 0$ Divide through by $30$ : $9 x^{4} - 18 x^{3} - 7 x^{2} + 12 x + 4 = 0$

05Execute

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

✓Verify

All 4 roots are distinct real numbers: $1, 2, - 1 / 3, - 2 / 3$ . For each $x$ , the polynomial $y (x)$ defines a unique single real value $y$ . Hence, there are exactly 4 distinct points on the curve.

Hints that build this answer step by step

What is the relation between the derivative of the curve $y^{'} = \frac{d y}{d x}$ and the slope of the given line $x + 90 y + 2 = 0$ for the normal line to be parallel to it?
$y^{'} = 90$
Differentiating $y = 54 x^{5} - 135 x^{4} - 70 x^{3} + 180 x^{2} + 210 x$ and setting $y^{'} = 90$ yields which simplified quartic equation?
$9 x^{4} - 18 x^{3} - 7 x^{2} + 12 x + 4 = 0$
How many real roots does the equation $9 x^{4} - 18 x^{3} - 7 x^{2} + 12 x + 4 = 0$ have?
$4$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is dy/dx = 90 and not -1/90?

The normal line is parallel to the given line, meaning $m_{N} = - 1 / 90$ . Since normal and tangent are perpendicular, $m_{T} \cdot m_{N} = - 1$ , which gives $m_{T} = d y / d x = - 1 / (- 1 / 90) = 90$ .

Answer

The number of points on the curve where the normal lines are parallel to $x + 90 y + 2 = 0$ is $4$ , which corresponds to the number of real roots of the derivative equation $y^{'} = 90$ .

Why each option works or fails

A · correct: 4 - No misconception. This is the correct option. We correctly equated the normal slope $- 1 / y^{'}$ to the line's slope $- 1 / 90$ , yielding $y^{'} = 90$ , and found that the resulting equation $270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 120 = 0$ has $4$ distinct real roots.
B: 2 - The student made an algebraic error when factoring or analyzing the roots of the derivative equation, concluding that only $2$ real roots exist. Ensure careful application of Descartes' Rule of Signs or interval analysis to determine the exact number of real roots of the polynomial equation.
C: 0 - The student made an error in equating the slope of the normal, perhaps setting the tangent slope $y^{'} = - 1 / 90$ or making a sign error, leading to an equation with no real roots. Recall that the slope of the normal line is $- 1 / y^{'}$ . Since the given line $x + 90 y + 2 = 0$ has slope $- 1 / 90$ , we must solve $y^{'} = 90$ .
D: 3 - The student miscounted the number of real roots of the simplified quartic equation, missing one root. Verify the signs of the function values at key test points (such as local extrema) to accurately determine the number of real roots.

Step-by-step solution

Given: Curve equation $y = 54 x^{5} - 135 x^{4} - 70 x^{3} + 180 x^{2} + 210 x$ and line $x + 90 y + 2 = 0$ .
Goal: Find the number of points on the curve where the normal lines are parallel to the given line.
Approach: The slope of the line $x + 90 y + 2 = 0$ is $m_{L} = - 1 / 90$ . Since the normal is parallel to this line, the slope of the normal is $m_{N} = - 1 / 90$ . Consequently, the slope of the tangent is $m_{T} = - 1 / m_{N} = 90$ . We compute $d y / d x$ , set it equal to $90$ , and determine the number of distinct real roots.
Execute: Differentiate $y$ with respect to $x$ : $\frac{d y}{d x} = 270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 210$ Equating $\frac{d y}{d x} = 90$ : $270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 210 = 90$ $270 x^{4} - 540 x^{3} - 210 x^{2} + 360 x + 120 = 0$ Divide through by $30$ : $9 x^{4} - 18 x^{3} - 7 x^{2} + 12 x + 4 = 0$
Execute: Factor the quartic polynomial $P (x) = 9 x^{4} - 18 x^{3} - 7 x^{2} + 12 x + 4$ . Testing integer roots: $P (1) = 9 - 18 - 7 + 12 + 4 = 0 ⟹ (x - 1)$ is a factor. $P (2) = 9 (16) - 18 (8) - 7 (4) + 12 (2) + 4 = 144 - 144 - 28 + 24 + 4 = 0 ⟹ (x - 2)$ is a factor. Dividing $P (x)$ by $(x - 1) (x - 2) = x^{2} - 3 x + 2$ : $\frac{9 x^{4} - 18 x^{3} - 7 x^{2} + 12 x + 4}{x^{2} - 3 x + 2} = 9 x^{2} + 9 x + 2 = (3 x + 1) (3 x + 2)$ Thus, the factorization is $(x - 1) (x - 2) (3 x + 1) (3 x + 2) = 0$ , giving real roots $x = 1, 2, - 1 / 3, - 2 / 3$ .
Verify: All 4 roots are distinct real numbers: $1, 2, - 1 / 3, - 2 / 3$ . For each $x$ , the polynomial $y (x)$ defines a unique single real value $y$ . Hence, there are exactly 4 distinct points on the curve.

Finding Points of a Curve with a Specified Normal Line Slope

What feels right?

The number of points on the curve where the normal lines are parallel to $x + 90 y + 2 = 0$ is $4$ , which corresponds to the number of real roots of the derivative equation $y^{'} = 90$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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