JEE Main 2023 (10 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let the quadratic curve passing through the point $(- 1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f (x)$ . Then the $x$ -intercept of the normal to the curve at the point $(α, α + 1)$ in the first quadrant is

Q: Why do we reject $\alpha = -1$ when solving $(\alpha + 1)(\alpha - 3) = 0$?

The problem explicitly states that the point $(\alpha, \alpha + 1)$ is in the first quadrant. In the first quadrant, both the $x$-coordinate and the $y$-coordinate must be positive. If $\alpha = -1$, the point becomes $(-1, 0)$, which is on the $x$-axis, not in the first quadrant. Thus, we must choose $\alpha = 3$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. The curve is a parabola, and the condition that it 'touches' the line

y = x

(1, 1)

implies two things: the point

(1, 1)

lies on the curve, and the slope of the curve at that point is identical to the slope of the line

y = x

Hint 2

Hint 2 (The Arsenal): You have three unknowns (

a, b, c

) in the general quadratic equation

f (x) = a x^{2} + b x + c

. Use the three conditions: the curve passes through

(- 1, 0)

, it passes through

(1, 1)

, and its derivative

f^{'} (1)

equals the slope of the line

y = x

Hint 3

Hint 3 (The Execution): Once you have the equation

f (x) = \frac{1}{4} (x + 1)^{2}

, find the point

(α, α + 1)

by substituting

x = α

and

y = α + 1

into the curve equation. After finding

α = 3

(for the first quadrant), use the point-slope form

y - y_{1} = m_{n} (x - x_{1})

where

m_{n}

is the negative reciprocal of the tangent slope

f^{'} (3)

Visualized Solution

General Equation of Quadratic Curve

Let $f (x) = a x^{2} + b x + c$
We need to find the constants $a$ , $b$ , and $c$ using the given conditions.

Passing through $(- 1, 0)$

Curve passes through $(- 1, 0)$
Substitute $x = - 1, y = 0$ :
$a (- 1)^{2} + b (- 1) + c = 0 ⟹ a - b + c = 0$

Touching at $(1, 1)$

Curve passes through $(1, 1)$
Substitute $x = 1, y = 1$ :
$a (1)^{2} + b (1) + c = 1 ⟹ a + b + c = 1$

Solving for $b$

Subtracting the equations:
$(a + b + c) - (a - b + c) = 1 - 0$
$2 b = 1 ⟹ b = \frac{1}{2}$

Tangency Condition at $(1, 1)$

Curve touches $y = x$ at $(1, 1)$
Slope of $y = x$ is $m = 1$
Therefore, $f^{'} (1) = 1$
Differentiating $f (x)$ : $f^{'} (x) = 2 a x + b$

Solving for $a$ and $c$

Substitute $x = 1$ and $b = \frac{1}{2}$ into $f^{'} (x) = 1$ :
$2 a (1) + \frac{1}{2} = 1 ⟹ 2 a = \frac{1}{2} ⟹ a = \frac{1}{4}$
Using $a + c = \frac{1}{2}$ :
$\frac{1}{4} + c = \frac{1}{2} ⟹ c = \frac{1}{4}$

Equation of the Curve

The function is $f (x) = \frac{1}{4} x^{2} + \frac{1}{2} x + \frac{1}{4}$
This can be written as $f (x) = \frac{1}{4} (x + 1)^{2}$

Locating $(α, α + 1)$

Point $(α, α + 1)$ lies on $y = \frac{1}{4} (x + 1)^{2}$
Substitute $x = α, y = α + 1$ :
$α + 1 = \frac{1}{4} (α + 1)^{2}$

Solving for $α$

$4 (α + 1) = (α + 1)^{2}$
$(α + 1)^{2} - 4 (α + 1) = 0$
$(α + 1) (α + 1 - 4) = 0 ⟹ (α + 1) (α - 3) = 0$
$α = - 1 or α = 3$

First Quadrant Point

For the first quadrant, $α > 0$ , so $α = 3$
The point is $(3, 3 + 1) = (3, 4)$

Slope of Tangent at $(3, 4)$

Slope of tangent $m_{t} = f^{'} (3)$
$f^{'} (3) = 2 (\frac{1}{4}) (3) + \frac{1}{2}$
$m_{t} = \frac{3}{2} + \frac{1}{2} = 2$

Slope of Normal

Slope of normal $m_{n} = - \frac{1}{m _{t}}$
$m_{n} = - \frac{1}{2}$

Equation of the Normal

Equation of normal at $(3, 4)$ with slope $- \frac{1}{2}$ :
$y - 4 = - \frac{1}{2} (x - 3)$

Finding the $x$ -intercept

For $x$ -intercept, set $y = 0$ :
$- 4 = - \frac{1}{2} (x - 3)$
$8 = x - 3 ⟹ x = 11$

Final Answer

The $x$ -intercept of the normal is $11$ .
Key Takeaway: Tangency at a point implies both the function values and their derivatives are equal at that point.

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The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

We begin with the general form of a quadratic curve:

f (x) = a x^{2} + b x + c

. Our mission is to determine the constants

a, b,

and

c

using the provided geometric clues.

The curve passes through

(- 1, 0)

, which gives us the equation:

a (- 1)^{2} + b (- 1) + c = 0 \Rightarrow a - b + c = 0

It also passes through

(1, 1)

, leading to:

a (1)^{2} + b (1) + c = 1 \Rightarrow a + b + c = 1

Subtracting these two equations, we find that

2 b = 1

, so

b = \frac{1}{2}

The Tangency Trap

The curve touches the line

y = x

(1, 1)

, meaning the line is tangent to the parabola at that point. Since the slope of

y = x

1

, the derivative of our curve at

x = 1

must also be

1

We know

f^{'} (x) = 2 a x + b

. Substituting

x = 1

and

b = \frac{1}{2}

2 a (1) + \frac{1}{2} = 1 \Rightarrow 2 a = \frac{1}{2} \Rightarrow a = \frac{1}{4}

Returning to

a + b + c = 1

, we substitute

a = \frac{1}{4}

and

b = \frac{1}{2}

to find

c = \frac{1}{4}

. Our curve is:

f (x) = \frac{1}{4} x^{2} + \frac{1}{2} x + \frac{1}{4} = \frac{1}{4} (x + 1)^{2}

Locating the Point

The problem introduces a point

(α, α + 1)

on this curve. Since it lies on the parabola, it must satisfy

f (α) = α + 1

Substituting into our equation:

α + 1 = \frac{1}{4} (α + 1)^{2}

Multiplying by

4

and rearranging:

(α + 1)^{2} - 4 (α + 1) = 0 \Rightarrow (α + 1) (α - 3) = 0

This yields

α = - 1

α = 3

. Since the point is in the first quadrant, we must have

α = 3

, giving us the point

(3, 4)

Final Calculation

We now find the normal at

(3, 4)

. First, we calculate the tangent slope

m_{t}

m_{t} = f^{'} (3) = 2 (\frac{1}{4}) (3) + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = 2

The normal slope is the negative reciprocal,

m_{n} = - \frac{1}{2}

. The equation of the normal is:

y - 4 = - \frac{1}{2} (x - 3)

To find the

x

-intercept, we set

y = 0

- 4 = - \frac{1}{2} (x - 3) \Rightarrow 8 = x - 3 \Rightarrow x = 11

The $x$ -intercept is $11$ .

Common Traps & FAQs

Wait, why do we need the derivative condition

f^{'} (1) = 1

? Can't we just use the two points

(- 1, 0)

and

(1, 1)

Great question! Two points are not enough to define a quadratic curve

f (x) = a x^{2} + b x + c

, which has three constants (

a, b, c

). The word 'touches' is the key; it tells us the line

y = x

is tangent to the curve at

(1, 1)

. This tangency provides the third necessary condition: the slope of the curve at

x = 1

must be the same as the slope of the line

y = x

, which is

1

Why do we reject

α = - 1

when solving

(α + 1) (α - 3) = 0

The problem explicitly states that the point

(α, α + 1)

is in the first quadrant. In the first quadrant, both the

x

-coordinate and the

y

-coordinate must be positive. If

α = - 1

, the point becomes

(- 1, 0)

, which is on the

x

-axis, not in the first quadrant. Thus, we must choose

α = 3

How do we know the slope of the normal is

- 1/2

Recall that the normal to a curve at a point is the line perpendicular to the tangent at that same point. If the slope of the tangent is

m_{t}

, the slope of the normal

m_{n}

is given by

m_{n} = - \frac{1}{m _{t}}

. Since we calculated the tangent slope

f^{'} (3) = 2

, the normal slope is simply

- \frac{1}{2}

Animated Solution for Mathematics - Differentiation: Let the quadratic curve passing through the point $(- 1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f (x)$ . Then the $x$ -intercept of the normal to the curve at the point $(α, α + 1)$ in the first quadrant is

Enter Numerical Value:

Visualized Solution

General Equation of Quadratic Curve

Passing through $(- 1, 0)$

Touching at $(1, 1)$

Solving for $b$

Tangency Condition at $(1, 1)$

Solving for $a$ and $c$

Equation of the Curve

Locating $(α, α + 1)$

Solving for $α$

First Quadrant Point

Slope of Tangent at $(3, 4)$

Slope of Normal

Equation of the Normal

Finding the $x$ -intercept

Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Tangency Trap

Locating the Point

Final Calculation

Similar Questions

Find the equation of the normal to the curve $y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$ at $x = 0$

The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

The equation of the normal to the curve $y = (1 + x)^{2 y} + cos^{2} (sin^{- 1} x)$ at $x = 0$ is:

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Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point

Enter Numerical Value:

Visualized Solution

General Equation of Quadratic Curve

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

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

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

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

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Equation of the Curve

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

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

First Quadrant Point

Slope of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (3,4)

Slope of Normal

Equation of the Normal

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

Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Tangency Trap

Locating the Point

Final Calculation

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Passing through $(- 1, 0)$

Touching at $(1, 1)$

Solving for $b$

Tangency Condition at $(1, 1)$

Solving for $a$ and $c$

Locating $(α, α + 1)$

Solving for $α$

Slope of Tangent at $(3, 4)$

Finding the $x$ -intercept

Find the equation of the normal to the curve $y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$ at $x = 0$

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Let $P (h, k)$ be a point on the curve $y = x^{2} + 7 x + 2$ , nearest to the line, $y = 3 x - 3$ . Then the equation of the normal to the curve at $P$ is:

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If the line $a x + b y + c = 0$ is a normal to the curve $x y = 1$ , then

If the normal to the curve $y = f (x)$ at the point $(3, 4)$ makes an angle $\frac{3 π}{4}$ with the positive $x$ -axis, then $f^{'} (3) =$

Let the normal at a point $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y$ -axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to

Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point