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JEE Main · MCQ · Quadratic Equations · Inequalities
MCQ · Mathematics · Quadratic Equations
Q. Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + 2ax + (3a+10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:
A$\left(-\infty,\,-\dfrac{11}{5}\right) \cup (5,\,\infty)$
B$\left(-\infty,\,-\dfrac{11}{5}\right)$ ✓
C$(-\infty,\,-3)$
D$(-\infty,\,-2)\cup(5,\,\infty)$
✅ Correct Answer: (B) $\left(-\infty,\,-\dfrac{11}{5}\right)$
Step-by-Step Solution
1
Key insight: condition for $\alpha < 1 < \beta$
Let $f(x) = x^2 + 2ax + (3a+10)$.
The leading coefficient is $+1 > 0$, so the parabola opens upward.
The leading coefficient is $+1 > 0$, so the parabola opens upward.
📈 For an upward parabola with roots $\alpha$ and $\beta$:
• $f(x) > 0$ when $x < \alpha$ or $x > \beta$
• $f(x) < 0$ when $\alpha < x < \beta$
So $\alpha < 1 < \beta$ ⟺ $f(1) < 0$
• $f(x) > 0$ when $x < \alpha$ or $x > \beta$
• $f(x) < 0$ when $\alpha < x < \beta$
So $\alpha < 1 < \beta$ ⟺ $f(1) < 0$
2
Compute $f(1)$
$$f(1) = (1)^2 + 2a(1) + (3a+10)$$
$$= 1 + 2a + 3a + 10$$
$$= 5a + 11$$
3
Apply $f(1) < 0$
$$5a + 11 < 0$$
$$5a < -11$$
$$a < -\frac{11}{5}$$
4
Check: Does $f(1) < 0$ automatically guarantee two real roots?
Compute discriminant $D$:
$$D = (2a)^2 - 4(3a+10) = 4a^2 - 12a - 40$$
$$= 4(a^2 - 3a - 10) = 4(a-5)(a+2)$$
$D > 0$ when $a < -2$ or $a > 5$.
Our condition: $a < -\dfrac{11}{5} = -2.2$
Since $-\dfrac{11}{5} < -2$, the set $a < -\dfrac{11}{5}$ is a subset of $a < -2$.
So $D > 0$ is automatically satisfied whenever $f(1) < 0$.
Since $-\dfrac{11}{5} < -2$, the set $a < -\dfrac{11}{5}$ is a subset of $a < -2$.
So $D > 0$ is automatically satisfied whenever $f(1) < 0$.
5
Number line verification
$\longleftarrow \quad \underbrace{\quad\quad\quad\quad}_{a<-\frac{11}{5}} \quad -\frac{11}{5} \quad\quad -2 \quad\quad 0 \quad\quad 5 \quad \longrightarrow$
✅ $f(1)<0$: $a < -11/5$ | ❌ $D<0$: $-2
✅ $f(1)<0$: $a < -11/5$ | ❌ $D<0$: $-2
Answer: $a \in \left(-\infty,\,-\dfrac{11}{5}\right)$ → Option (B) ✓
Related Theory
📌 Location of Roots — Complete Framework
For $f(x) = ax^2+bx+c$ with $a>0$ and roots $\alpha \leq \beta$, here are all standard conditions used in JEE:
The most important for JEE: $k$ between roots ⟺ $f(k)<0$ when leading coeff $>0$.
| Condition Required | Mathematical Condition |
|---|---|
| Both roots $> k$ | $D\geq0$, $f(k)>0$, $-b/2a > k$ |
| Both roots $< k$ | $D\geq0$, $f(k)>0$, $-b/2a < k$ |
| $k$ lies between roots ($\alpha < k < \beta$) | $f(k) < 0$ (for $a>0$) |
| Both roots in $(k_1, k_2)$ | $D\geq0$, $f(k_1)>0$, $f(k_2)>0$, $k_1<-b/2a |
| Exactly one root in $(k_1, k_2)$ | $f(k_1)\cdot f(k_2) < 0$ |
$\alpha0$)
$\alpha0$ (for $a<0$)
📌 Why $f(1) < 0$ is the Complete Condition
Many students worry: "Should I also check $D>0$ separately?"
Answer: NO, not separately. Here's why:
If $f(1) < 0$ and leading coefficient $>0$, the parabola is below the x-axis at $x=1$. This means:
• The parabola MUST cross the x-axis on both sides of $x=1$
• So it has two distinct real roots — one less than 1 and one greater than 1
• $D>0$ is automatically guaranteed
In this problem: $f(1)<0 \Rightarrow a < -11/5 = -2.2 \Rightarrow a < -2$, which is already inside the $D>0$ region.
Only check $D$ separately when the condition involves both roots on the same side of $k$ (e.g., both roots $>k$, or both in an interval).
Answer: NO, not separately. Here's why:
If $f(1) < 0$ and leading coefficient $>0$, the parabola is below the x-axis at $x=1$. This means:
• The parabola MUST cross the x-axis on both sides of $x=1$
• So it has two distinct real roots — one less than 1 and one greater than 1
• $D>0$ is automatically guaranteed
In this problem: $f(1)<0 \Rightarrow a < -11/5 = -2.2 \Rightarrow a < -2$, which is already inside the $D>0$ region.
Only check $D$ separately when the condition involves both roots on the same side of $k$ (e.g., both roots $>k$, or both in an interval).
📌 Vieta's Formulas
For $x^2+2ax+(3a+10)=0$ with roots $\alpha, \beta$:
$$\alpha + \beta = -2a \qquad \alpha\beta = 3a+10$$
These are useful for verification:
• If $a=-3$ (which satisfies $a<-11/5$): $\alpha+\beta=6$, $\alpha\beta=-9+10=1>0$ ✓ (both roots same sign)
• Check: $f(1)=5(-3)+11=-4<0$ ✓
Vieta's formulas: sum of roots $= -b/a$, product of roots $= c/a$ for $ax^2+bx+c=0$.
• If $a=-3$ (which satisfies $a<-11/5$): $\alpha+\beta=6$, $\alpha\beta=-9+10=1>0$ ✓ (both roots same sign)
• Check: $f(1)=5(-3)+11=-4<0$ ✓
Vieta's formulas: sum of roots $= -b/a$, product of roots $= c/a$ for $ax^2+bx+c=0$.
$\alpha+\beta = -2a$
$\alpha\beta = 3a+10$
$\alpha-\beta = \pm\sqrt{D}/a$
📌 Discriminant Analysis
$D = 4a^2 - 12a - 40 = 4(a-5)(a+2)$
Sign analysis of $D$:
• $D > 0$: $a < -2$ or $a > 5$ (two distinct real roots)
• $D = 0$: $a = -2$ or $a = 5$ (equal roots)
• $D < 0$: $-2 < a < 5$ (no real roots)
The answer option (A) $(-\infty,-11/5)\cup(5,\infty)$ is wrong because when $a>5$, $D>0$ but $f(1)=5a+11>0$, meaning 1 is NOT between the roots — both roots are less than 1 in that case. So $a>5$ does not satisfy $\alpha<1<\beta$.
Sign analysis of $D$:
• $D > 0$: $a < -2$ or $a > 5$ (two distinct real roots)
• $D = 0$: $a = -2$ or $a = 5$ (equal roots)
• $D < 0$: $-2 < a < 5$ (no real roots)
The answer option (A) $(-\infty,-11/5)\cup(5,\infty)$ is wrong because when $a>5$, $D>0$ but $f(1)=5a+11>0$, meaning 1 is NOT between the roots — both roots are less than 1 in that case. So $a>5$ does not satisfy $\alpha<1<\beta$.
$D=4(a-5)(a+2)$
$D>0$: $a<-2$ or $a>5$
$f(1)<0$ is the KEY condition
📌 Why Option (A) is a Common Wrong Answer
Many students compute $D>0$ which gives $a<-2$ or $a>5$, and think that's the answer — leading to option (A) or (D).
This is wrong because $D>0$ only ensures two distinct real roots exist, but does NOT ensure that 1 lies between them.
For $a>5$: $f(1)=5(5)+11=36>0$. So 1 is OUTSIDE both roots (either both $<1$ or both $>1$). Check vertex $x=-b/2a = -a$. For $a>5$, $-a<-5<1$, so vertex is to the left of 1, and $f(1)>0$ means 1 is to the right of both roots — so both $\alpha,\beta < 1$. This does NOT satisfy $\alpha<1<\beta$.
Moral: Always use $f(k)<0$ directly for "$k$ between roots" — never just $D>0$.
This is wrong because $D>0$ only ensures two distinct real roots exist, but does NOT ensure that 1 lies between them.
For $a>5$: $f(1)=5(5)+11=36>0$. So 1 is OUTSIDE both roots (either both $<1$ or both $>1$). Check vertex $x=-b/2a = -a$. For $a>5$, $-a<-5<1$, so vertex is to the left of 1, and $f(1)>0$ means 1 is to the right of both roots — so both $\alpha,\beta < 1$. This does NOT satisfy $\alpha<1<\beta$.
Moral: Always use $f(k)<0$ directly for "$k$ between roots" — never just $D>0$.
📌 Graphical Understanding
Think of $f(x) = x^2+2ax+(3a+10)$ as an upward parabola:
Case $f(1) < 0$: The point $(1, f(1))$ is below the x-axis. Since the parabola opens upward, it must cross the x-axis once before $x=1$ and once after $x=1$. Therefore $\alpha < 1 < \beta$. ✅
Case $f(1) > 0$: The point $(1, f(1))$ is above the x-axis. Either both roots are between the vertex and 1 (both $<1$), or 1 is between vertex and both roots (both $>1$), or no real roots at all. In none of these cases does 1 lie between the two roots. ❌
Case $f(1) = 0$: $x=1$ is itself a root, so $\alpha=1$ or $\beta=1$ — strict inequalities not satisfied. ❌
Case $f(1) < 0$: The point $(1, f(1))$ is below the x-axis. Since the parabola opens upward, it must cross the x-axis once before $x=1$ and once after $x=1$. Therefore $\alpha < 1 < \beta$. ✅
Case $f(1) > 0$: The point $(1, f(1))$ is above the x-axis. Either both roots are between the vertex and 1 (both $<1$), or 1 is between vertex and both roots (both $>1$), or no real roots at all. In none of these cases does 1 lie between the two roots. ❌
Case $f(1) = 0$: $x=1$ is itself a root, so $\alpha=1$ or $\beta=1$ — strict inequalities not satisfied. ❌
📌 Common Mistakes to Avoid
❌ Mistake 1: Using $D>0$ alone as the condition. $D>0$ gives real roots but doesn't specify their location relative to 1.
❌ Mistake 2: Choosing option (A) by combining $D>0$ conditions — forgetting to apply $f(1)<0$.
❌ Mistake 3: Computing $f(1)$ as $1+2a+3a+10 = 5a+10$ — arithmetic error. Correct: $1+2a+3a+10 = 5a+11$.
❌ Mistake 4: Writing $5a+11<0 \Rightarrow a < 11/5$ (forgetting to flip sign when dividing by positive 5... or confusing sign). Correct: $a < -11/5$.
❌ Mistake 2: Choosing option (A) by combining $D>0$ conditions — forgetting to apply $f(1)<0$.
❌ Mistake 3: Computing $f(1)$ as $1+2a+3a+10 = 5a+10$ — arithmetic error. Correct: $1+2a+3a+10 = 5a+11$.
❌ Mistake 4: Writing $5a+11<0 \Rightarrow a < 11/5$ (forgetting to flip sign when dividing by positive 5... or confusing sign). Correct: $a < -11/5$.
📌 Key Results Summary
$f(x)=x^2+2ax+(3a+10)$
$\alpha<1<\beta \Leftrightarrow f(1)<0$
$f(1)=5a+11$
$5a+11<0 \Rightarrow a<-11/5$
$D>0$ auto-satisfied ✓
Answer: $(-\infty,-11/5)$
Frequently Asked Questions
1. What is the key condition for $\alpha<1<\beta$?
$f(1)<0$ when leading coefficient is positive.
2. What is $f(1)$?
$f(1)=1+2a+3a+10=5a+11$.
3. What inequality does $f(1)<0$ give?
$5a+11<0 \Rightarrow a<-11/5$.
4. What is the discriminant?
$D=4(a-5)(a+2)$. $D>0$ when $a<-2$ or $a>5$.
5. Do we need to check $D>0$ separately?
No. $a<-11/5$ is a subset of $a<-2$, so $D>0$ is automatically satisfied.
6. Why is option (A) wrong?
For $a>5$, $f(1)=5a+11>0$, so 1 is NOT between roots. Only $a<-11/5$ works.
7. What is $-11/5$ as decimal?
$-11/5=-2.2$.
8. What is $\alpha+\beta$ by Vieta's?
$\alpha+\beta=-2a$.
9. What is $\alpha\beta$ by Vieta's?
$\alpha\beta=3a+10$.
10. What if $f(1)=0$?
Then $x=1$ is a root itself, so strict inequality $\alpha<1<\beta$ is not satisfied.
11. What is the correct answer set?
$\left(-\infty, -\dfrac{11}{5}\right)$ — Option (B).
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