Q6. The interval to which $b$ may belong so that the function

$$f(x) = \left(1 - \frac{\sqrt{21 - 4b - b^2}}{b + 1}\right)x^3 + 5x + \sqrt{6}$$

is increasing at every point of its domain, is (a) $(-7, -1)$ (b) $(-6, -2)$ (c) $(2, 2.5)$ (d) $(2, 3)$ (e) all the above

1. Ensuring the function is defined

The coefficient of $x^3$ involves a square–root:

$\sqrt{\,21 - 4b - b^2\,}$

For this to be real we need

$21 - 4b - b^2 \ge 0.$

Re–arranging:

$b^2 + 4b - 21 \le 0 \;\;\Longrightarrow\;\; (b + 7)(b - 3) \le 0.$

Hence

$-7 \le b \le 3.$

Also, the denominator $b + 1$ must not be zero:

$b \ne -1.$

So temporary domain for $b$

$b \in [-7,3]\setminus\{-1\}.$

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2. Making the cubic always increasing

For the cubic

$f(x) = A\,x^3 + 5x + \sqrt6,$

where

$A = 1 - \frac{\sqrt{21 - 4b - b^2}}{b + 1},$

the derivative (slope-checker) is

$f'(x) = 3A\,x^2 + 5.$

A function is strictly increasing everywhere if its derivative is strictly positive for all real $x$.

2.1 What kind of quadratic is $f'(x)$?

It has no $x$-term and its constant term is $5>0$. Therefore:

• If $A>0$, the quadratic opens upward, its minimum is $5>0$ at $x=0$, so $f'(x)>0$ for all $x$.
• If $A\le0$, the quadratic is never always positive (it eventually dips below zero).

So we really just need

$A>0.$

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3. Solving $A>0$

Write the inequality:

$1 - \frac{\sqrt{21 - 4b - b^2}}{b + 1} > 0.$

The sign of $b+1$ matters.

Case I: $b+1>0\;\;(b>-1)$

Multiply without flipping sign:

$1 > \frac{\sqrt{21 - 4b - b^2}}{b + 1}\;\;\Longrightarrow\;\; \sqrt{21 - 4b - b^2} < b+1.$

Squaring (allowed because both sides are \ge0):

$21 - 4b - b^2 < (b + 1)^2.$

Simplify:

$20 - 6b - 2b^2 < 0 \;\Longrightarrow\;\; b^2 + 3b - 10 > 0.$

Factor or use quadratic formula:

$(b+5)(b-2) > 0 \;\Longrightarrow\;\; b<-5 \;\text{or}\; b>2.$

Intersect with $b>-1$ and the earlier domain $[-7,3]$:

$b>2 \;\text{and}\; b\le3 \;\Rightarrow\; b \in (2,3].$

Case II: $b+1<0\;\;(b<-1)$

Here $\dfrac{\sqrt{\,\cdot\,}}{b+1}$ is negative, so

$1 - \text{(negative)} > 1 > 0,$

which automatically satisfies $A>0$. Combine with the earlier domain:

$b \in [-7,-1).$

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4. Final admissible $b$-intervals

$b \in \left[ -7, -1 \right) \;\cup\; \left( 2, 3 \right].$

Every choice of $b$ inside those two pieces makes the cubic strictly increasing.

What is the question?

We have a curvy line (a cubic function). We want to know for which numbers $b$ this line is always climbing upward (never comes down) everywhere you look.

How to think about it like a 10-year-old?

1. Upward slope → look at the "slope–checker" (the derivative). If that slope–checker is always positive, the curve is always going up.
2. The slope-checker of a cubic is usually a quadratic. A quadratic that opens upward and never crosses the ground (the $x$-axis) is always positive.
3. We just need to find those $b$’s that make the quadratic open upward and never touch the ground.
4. There is also a square-root in the original formula. Square-roots are picky: the stuff inside must stay non-negative. And, we must never divide by zero.
5. Put these rules together and the magic safe zones for $b$ appear!

Step 1 – Domain restrictions

$21-4b-b^2 \ge 0 \;\Longrightarrow\; (b+7)(b-3) \le 0 \;\Longrightarrow\; -7 \le b \le 3.$

Also $b\ne-1$ (denominator).

Step 2 – Derivative

$f'(x)=3A\,x^2+5,\;\; A=1-\frac{\sqrt{21-4b-b^2}}{b+1}.$

Step 3 – Condition for monotonic increase

Because $3A\,x^2+5$ has no $x$ term and $5>0$:

• Necessary and sufficient: $A>0$.

Step 4 – Solve $A>0$

Case I ($b>-1$):

$\sqrt{21-4b-b^2} < b+1 \;\Longrightarrow\; b^2+3b-10>0 \;\Longrightarrow\; b>2\;(\text{within }-1<b\le3).$

Thus $b\in(2,3]$.

Case II ($b<-1$): $A>0$ automatically. With $-7\le b<-1$ we get $b\in[-7,-1)$.

Step 5 – Combine

$b\in[-7,-1) \cup (2,3].$

Any $b$ in this set makes $f'(x)>0\;\forall x$, hence $f(x)$ is strictly increasing everywhere.

Matching with options

• (a) $(-7,-1)$ – inside valid set.
• (b) $(-6,-2)$ – inside valid set.
• (c) $(2,2.5)$ – inside valid set.
• (d) $(2,3)$ – inside valid set.

Therefore every listed interval is acceptable, so the best option is

$\boxed{\text{(e) all the above}}.$