JEE Main 2021 (20 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let 'a' be a real number such that the function $f (x) = a x^{2} + 6 x - 15, x \in R$ is increasing in $(- \infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$ . Then the function $g (x) = a x^{2} - 6 x + 15, x \in R$ has a:

Q: Wait, why do we know the vertex is at $x = \frac{3}{4}$?

Great question! The problem states that $f(x)$ is increasing on $(-\infty, \frac{3}{4})$ and decreasing on $(\frac{3}{4}, \infty)$. In a quadratic function, the transition point between increasing and decreasing behavior is always the vertex. Since the function switches behavior exactly at $x = \frac{3}{4}$, that must be the x-coordinate of the vertex.

Q: How do we know for sure that $a$ is negative?

Think about the shape of the parabola. If a function increases and then decreases, it must look like a mountain—a downward-opening parabola. For a quadratic $f(x) = ax^2 + bx + c$, the parabola opens downwards if and only if the leading coefficient $a$ is negative. Our calculation $a = -4$ confirms this perfectly!

Q: Can we just use the vertex formula for $g(x)$ instead of finding the derivative?

Absolutely! You could use $x = -\frac{b}{2a}$ for $g(x) = -4x^2 - 6x + 15$. Here, $b = -6$ and $a = -4$, so $x = -\frac{-6}{2(-4)} = -\frac{6}{8} = -\frac{3}{4}$. Both the derivative method and the vertex formula method are valid and will lead you to the same result. The derivative method is just a more general tool that works for any differentiable function, not just quadratics.

Select Answer:

Visualized Solution

Analyzing $f (x)$ behavior

Given function: $f (x) = a x^{2} + 6 x - 15$
Increasing in $(- \infty, \frac{3}{4})$
Decreasing in $(\frac{3}{4}, \infty)$

Locating the Vertex

The transition point from increasing to decreasing is the vertex.
Therefore, the vertex of $f (x)$ is at $x = \frac{3}{4}$ .
Since it opens downwards, $a < 0$ .

Vertex Formula

For a quadratic $a x^{2} + b x + c$ , the x-coordinate of the vertex is:
$x = - \frac{b}{2 a}$

Substituting Values

For $f (x) = a x^{2} + 6 x - 15$ , we have $b = 6$ .
Substitute into the vertex formula:
$\frac{3}{4} = - \frac{6}{2 a}$

Solving for $a$

Simplify the equation:
$\frac{3}{4} = - \frac{3}{a}$
Cross-multiply to solve for $a$ :
$3 a = - 12 ⟹ a = - 4$

Defining $g (x)$

The second function is $g (x) = a x^{2} - 6 x + 15$ .
Substitute $a = - 4$ :
$g (x) = - 4 x^{2} - 6 x + 15$

Finding Critical Points

To find local extrema, we need the first derivative $g^{'} (x)$ .
$g^{'} (x) = \frac{d}{d x} (- 4 x^{2} - 6 x + 15)$
$g^{'} (x) = - 8 x - 6$

Solving $g^{'} (x) = 0$

Set the derivative to zero:
$- 8 x - 6 = 0$
$- 8 x = 6$
$x = - \frac{6}{8} = - \frac{3}{4}$

Second Derivative Test

Find the second derivative $g^{''} (x)$ :
$g^{''} (x) = \frac{d}{d x} (- 8 x - 6) = - 8$
Since $g^{''} (x) < 0$ , the curve is concave downwards.

Final Conclusion

Because $g^{''} (x) = - 8 < 0$ , the critical point is a local maximum.
Therefore, $g (x)$ has a local maximum at $x = - \frac{3}{4}$ .

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

Imagine you are standing on a landscape defined by a mathematical function. You are walking along the curve of

f (x) = a x^{2} + 6 x - 15

The problem states that you are walking uphill until you reach

x = \frac{3}{4}

, and then you begin to descend. This behavior indicates that the vertex of the parabola is located at

x = \frac{3}{4}

Because the graph transitions from increasing to decreasing, it must be a downward-opening parabola. This implies that the leading coefficient

a

must be negative.

The Algebra of the Vertex

We translate this geometric intuition into an algebraic equation. For any quadratic function

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

, the

x

-coordinate of the vertex is given by:

x = - \frac{b}{2 a}

We know the vertex is at

x = \frac{3}{4}

and the coefficient

b = 6

. Substituting these values into the formula, we obtain:

\frac{3}{4} = - \frac{6}{2 a}

Simplifying this expression, we get

\frac{3}{4} = - \frac{3}{a}

. Cross-multiplying yields

3 a = - 12

, which reveals that $a = - 4$ . The negative sign confirms our earlier intuition that the parabola opens downwards.

The Transformation to $g (x)$

With the value of

a

determined, we define the second function as

g (x) = a x^{2} - 6 x + 15

. Substituting

a = - 4

, we get:

g (x) = - 4 x^{2} - 6 x + 15

To find the local extrema of this new landscape, we utilize calculus. We must find where the slope of the tangent line is zero by calculating the first derivative,

g^{'} (x)

g^{'} (x) = \frac{d}{d x} (- 4 x^{2} - 6 x + 15) = - 8 x - 6

Setting the derivative to zero, we solve

- 8 x - 6 = 0

. This leads to

- 8 x = 6

, or:

x = - \frac{6}{8} = - \frac{3}{4}

The Final Verdict

We have identified the critical point at

x = - \frac{3}{4}

. To determine if this is a maximum or a minimum, we apply the second derivative test.

Calculating the second derivative, we find:

g^{''} (x) = \frac{d}{d x} (- 8 x - 6) = - 8

Since the second derivative is a negative constant, the function is concave downwards everywhere. Therefore, the critical point at $x = - \frac{3}{4}$ is a local maximum.

Common Traps & FAQs

Wait, why do we know the vertex is at

x = \frac{3}{4}

Great question! The problem states that

f (x)

is increasing on

(- \infty, \frac{3}{4})

and decreasing on

(\frac{3}{4}, \infty)

. In a quadratic function, the transition point between increasing and decreasing behavior is always the vertex. Since the function switches behavior exactly at

x = \frac{3}{4}

, that must be the x-coordinate of the vertex.

How do we know for sure that

a

is negative?

Think about the shape of the parabola. If a function increases and then decreases, it must look like a mountain—a downward-opening parabola. For a quadratic

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

, the parabola opens downwards if and only if the leading coefficient

a

is negative. Our calculation

a = - 4

confirms this perfectly!

Can we just use the vertex formula for

g (x)

instead of finding the derivative?

Absolutely! You could use

x = - \frac{b}{2 a}

for

g (x) = - 4 x^{2} - 6 x + 15

. Here,

b = - 6

and

a = - 4

, so

x = - \frac{- 6}{2 ( - 4 )} = - \frac{6}{8} = - \frac{3}{4}

. Both the derivative method and the vertex formula method are valid and will lead you to the same result. The derivative method is just a more general tool that works for any differentiable function, not just quadratics.

Animated Solution for Mathematics - Differentiation: Let 'a' be a real number such that the function $f (x) = a x^{2} + 6 x - 15, x \in R$ is increasing in $(- \infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$ . Then the function $g (x) = a x^{2} - 6 x + 15, x \in R$ has a:

Select Answer:

Visualized Solution

Analyzing $f (x)$ behavior

Locating the Vertex

Vertex Formula

Substituting Values

Solving for $a$

Defining $g (x)$

Finding Critical Points

Solving $g^{'} (x) = 0$

Second Derivative Test

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Algebra of the Vertex

The Transformation to $g (x)$

The Final Verdict

Similar Questions

Let $f : R \to R$ be a function defined by $f (x) = (x - 3)^{n_{1}} (x - 5)^{n_{2}}, n_{1}, n_{2} \in N$ . Then, which of the following is NOT true?

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2}$ , then $α$ and $α^{2}$ are the roots of the equation :

Let $f (x) = ⎩ ⎨ ⎧ e^{x}, 2 - e^{x - 1}, x - e, 0 \leq x \leq 1 1 < x \leq 2 2 < x \leq 3$ and $g (x) = \int_{0}^{x} f (t) d t, x \in [1, 3]$ then $g (x)$ has

Comprehension Passage

Which of the following is true?

Which of the following is true?

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

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 .........

Let $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 8 t + 15}{e ^{t}}$ dt, $x \in R$ Then the numbers of local maximum and local minimum points of f, respectively, are:

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

Let $a > 0$ . If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to :-

The function $f (x) = 2 x + 3 (x)^{\frac{2}{3}}, x \in R$ , has

Select Answer:

Visualized Solution

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

Locating the Vertex

Vertex Formula

Substituting Values

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

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

Finding Critical Points

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

Second Derivative Test

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Algebra of the Vertex

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

The Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function defined by f(x)=(x−3)n1​(x−5)n2​,n1​,n2​∈N. Then, which of the following is NOT true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=2x3−9ax2+12a2x+1,a>0 has a local maximum at x=α and a local minimum at x=α2, then α and α2 are the roots of the equation :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=⎩⎨⎧​ex,2−ex−1,x−e,​0≤x≤11<x≤22<x≤3​ and g(x)=∫0x​f(t)dt,x∈[1,3] then g(x) has

Comprehension Passage

Which of the following is true?

Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=∫0ex​1+t2f′(t)​dt. Which of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫0x2​et​(t−2)(t−3)dt for all x∈(0,∞), then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫0x2​ett2−8t+15​ dt, x∈R Then the numbers of local maximum and local minimum points of f, respectively, are:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) is cubic polynomial with f(2)=18 and f(1)=−1. Also f(x) has local maxima at x=−1 and f′(x) has local minima at x=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=2x+3(x)32​,x∈R, has

Analyzing $f (x)$ behavior

Solving for $a$

Defining $g (x)$

Solving $g^{'} (x) = 0$

The Transformation to $g (x)$

Let $f : R \to R$ be a function defined by $f (x) = (x - 3)^{n_{1}} (x - 5)^{n_{2}}, n_{1}, n_{2} \in N$ . Then, which of the following is NOT true?

If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2}$ , then $α$ and $α^{2}$ are the roots of the equation :

Let $f (x) = ⎩ ⎨ ⎧ e^{x}, 2 - e^{x - 1}, x - e, 0 \leq x \leq 1 1 < x \leq 2 2 < x \leq 3$ and $g (x) = \int_{0}^{x} f (t) d t, x \in [1, 3]$ then $g (x)$ has

Let $g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t$ . Which of the following is true?

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Let $f (x) = \int_{0}^{x^{2}} \frac{t ^{2} - 8 t + 15}{e ^{t}}$ dt, $x \in R$ Then the numbers of local maximum and local minimum points of f, respectively, are:

$f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

The function $f (x) = 2 x + 3 (x)^{\frac{2}{3}}, x \in R$ , has