Q.It is given that at $x = 1$, the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$. Find the value of $a$.

Concept & Intuition

We have a cubic whose roots are $\alpha, \beta, \gamma$ with $\beta < \gamma < \alpha$ and they form a geometric progression. That means there is a common ratio $r$ such that $\beta = a$, $\gamma = ar$, $\alpha = ar^2$ (or any cyclic ordering). The cubic’s coefficients give us symmetric sums of the roots, so we can solve for $a$ and $r$. Once we know the actual numbers, the quadratic $\gamma x^2 + 4\beta x + \alpha$ becomes a concrete expression whose extreme value (since the coefficient of $x^2$ is positive, it’s a minimum) can be found by completing the square or using the vertex formula.

Step-by-step solution

1. Set up the geometric progression

   Let the three roots be $\beta = t$, $\gamma = tr$, $\alpha = tr^2$, where $t > 0$ (since $\beta < \gamma < \alpha$ and all are real). The cubic is $8x^3 - 42x^2 + 63x - 27 = 0$.

2. Use Vieta’s formulas

   For a cubic $ax^3 + bx^2 + cx + d = 0$, the sum of roots $= -b/a$, sum of pairwise products $= c/a$, product $= -d/a$. Here:

3. Solve for $t$ and $r$ From the product: $t^3 r^3 = \frac{27}{8} \implies (tr)^3 = \frac{27}{8} \implies tr = \frac{3}{2}$. From the sum: $t(1 + r + r^2) = \frac{21}{4}$. Substitute $t = \frac{3}{2r}$ into the sum:

Multiply both sides by $4r$:

Divide by 3: $2r^2 - 5r + 2 = 0$.

Solve: $r = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$, so $r = 2$ or $r = \frac{1}{2}$.

4. Choose the correct ordering …

The key idea: $f(x) > 0$ for all real $x$ means the quadratic opens upward (which it does, since the coefficient of $x^2$ is $1 > 0$) and never touches the $x$-axis. That happens exactly when its discriminant is negative.

1. Write the discriminant condition For $f(x) = x^2 - 2(4K-1)x + g(K)$, the discriminant is

Factor the 4:

We need $\Delta < 0$, so

2. Substitute $g(K)$ Given $g(K) = 15K^2 - 2K - 7$, we have

Expand $(4K-1)^2 = 16K^2 - 8K + 1$, so

Simplify:

3. Solve the inequality for $K$ Factor:

This quadratic in $K$ opens upward and is negative between its roots. Hence

So the interval $(a,b) = (2,4)$.

4. Analyze $g(K)$ on $(2,4)$ $g(K) = 15K^2 - 2K - 7$ is a quadratic with leading coefficient $15 > 0$, so it is convex (opens upward). Its vertex occurs at

This vertex lies far left of the interval $(2,4)$.

On $(2,4)$, the function is strictly increasing (since the vertex is to the left). Therefore:

• The minimum on the open interval is approached as $K \to 2^+$ (but not attained, since $K=2$ is not included).
• The maximum is approached as $K \to 4^-$ (also not attained). …

The problem asks for the value of $x^2+2y^3$ when $x^2 y^3$ is maximised, given $x$ and $y$ are positive integers satisfying $x+2y=10$. The constraint is linear, so we can express $x$ in terms of $y$ (or vice versa) and turn the product into a function of one variable. Then we maximise that function — but because $x$ and $y$ are integers, we must check integer points near the continuous maximum.

1. Express $x$ in terms of $y$.

   From $x+2y=10$, we get $x = 10 - 2y$. Since $x>0$, we need $10-2y>0 \Rightarrow y<5$. Also $y>0$, so $y$ can be $1,2,3,4$.

2. Write the function to maximise.

   Let $f(y) = x^2 y^3 = (10-2y)^2 y^3$.

   We want the integer $y$ in $\{1,2,3,4\}$ that makes $f(y)$ largest.

3. Find the continuous maximum (to guide us).

   Differentiate $f(y)$ with respect to $y$ (treating $y$ as a real variable).

   $f(y) = (10-2y)^2 y^3 = (100 - 40y + 4y^2) y^3 = 100y^3 - 40y^4 + 4y^5$.

   $f'(y) = 300y^2 - 160y^3 + 20y^4 = 20y^2(15 - 8y + y^2)$.

   Set $f'(y)=0$: $y=0$ (not positive) or $y^2 - 8y + 15 = 0$.

   Solve: $y = \frac{8 \pm \sqrt{64-60}}{2} = \frac{8 \pm 2}{2}$, so $y=5$ or $y=3$.

   $y=5$ gives $x=0$, not positive. So the only positive critical point is $y=3$.

   Check the sign of $f'(y)$ around $y=3$: for $y<3$ (say $y=2$), $f'(2)=20\cdot4(15-16+4)=80\cdot3>0$; for $y>3$ (say $y=4$), $f'(4)=20\cdot16(15-32+16)=320\cdot(-1)<0$. So $y=3$ gives a local maximum for the continuous function.

4. Evaluate $f(y)$ at the integer candidates. …

Concept & Intuition

We want to maximize $x^2 y^3$ given $2x+3y=50$, with $x,y>0$ (integers, but we can first treat them as reals and then check integer feasibility). The exponents (2 and 3) suggest using the weighted AM–GM inequality: for positive numbers, the product is maximized when terms are proportional to their weights. Here, rewrite the constraint as a sum of five terms: two copies of $x$ and three copies of $y$, each with coefficients that match the exponents. Then AM–GM gives the condition for maximum directly.

Step-by-step solution

1. Set up the weighted AM–GM We want to maximize $x^2 y^3$. Notice that $2x+3y=50$ can be seen as the sum of five numbers: $x, x, y, y, y$. The AM–GM inequality says:

i.e.

Since $2x+3y=50$, we have

2. Equality condition …

We start with the inequality

Concept & Intuition

The product of three sines with angles spaced by $60^\circ$ is a classic identity: it simplifies to $\frac14\sin3\theta$. Why? Because $\sin3\theta = 3\sin\theta - 4\sin^3\theta$, but the triple-angle identity also emerges from combining these three factors. Once we rewrite the product, the inequality becomes a condition on $\sin3\theta$. The problem then asks: for which $\theta$ does $\sin3\theta$ reach its maximum value of $1$ while also satisfying that inequality? So we find all $\theta$ where $\sin3\theta=1$ and check if they meet $|\sin3\theta|\le\frac12$ — but wait, that seems contradictory. Let’s work carefully.

Step 1: Simplify the product

Use the identity

Let $A = \frac{\pi}{3}-\theta$ and $B = \frac{\pi}{3}+\theta$. Then

Since $\cos(-2\theta)=\cos2\theta$ and $\cos(2\pi/3)=-\frac12$, we get

Now multiply by $\sin\theta$:

Use $\sin\theta\cos2\theta = \frac12\bigl[\sin(3\theta) - \sin\theta\bigr]$:

Thus

Step 2: Rewrite the inequality

The given inequality becomes

Step 3: Interpret “attains its maximum”

The function $\sin3\theta$ has maximum value $1$. The problem asks for the number of $\theta$ in $[0,2\pi]$ such that $\sin3\theta = 1$ and the inequality holds.

But if $\sin3\theta = 1$, then $|\sin3\theta| = 1$, which is not $\le \frac12$. So no $\theta$ with $\sin3\theta=1$ satisfies the inequality? That would give 0 values, which isn’t among the options. …