For all 'x', x² + 2ax + 10 - 3a 0, then the interval in which 'a' lies is (A) a

Q: Why is the condition $D < 0$ and not $D \le 0$?

Great question! The problem states the expression is strictly greater than zero ($> 0$). If $D = 0$, the parabola would touch the x-axis at one point, making the expression equal to zero at that point. Since we need it to be strictly positive everywhere, we must exclude the case where it touches the axis.

Q: Do I need to check the condition $A > 0$?

Yes, always! For a quadratic to be positive for all $x$, it must open upwards. Here, $A = 1$, which is already positive. If $A$ were a variable, you would have to ensure it is positive. Never skip this check!

Q: How do I know the interval is between the roots and not outside?

Use the wavy curve method! For a quadratic $(a - r_1)(a - r_2) < 0$, the expression is negative between the roots $r_1$ and $r_2$. You can also test a value in the interval, like $a = 0$, to see if it satisfies the inequality.

Question

For all 'x', $x^2 + 2ax + 10 - 3a > 0$, then the interval in which 'a' lies is Options: (A) $a < -5$ (B) $-5 < a < 2$ (C) $a > 5$ (D) $2 < a < 5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing in front of a graph. You have a quadratic expression, $x^2 + 2ax + 10 - 3a$, and you are told that for every single real value of $x$, this expression is strictly greater than zero. Since the coefficient of $x^2$ is $1$, which is positive, we know we are dealing with an upward-opening parabola. If this parabola is always positive, it means it never dips below the $x$-axis. It is, quite literally, a floating parabola. ### The Gatekeeper: The Discriminant For this parabola to stay completely above the $x$-axis, it cannot have any real roots. If it had real roots, it would cross the $x$-axis, and at those points, the expression would be zero or negative. Algebraically, the condition for a quadratic $Ax^2 + Bx + C > 0$ to hold for all $x$ is twofold: first, $A > 0$ (which is already satisfied here), and second, the discriminant $D$ must be strictly less than zero. This is our gatekeeper. We need: $$D = B^2 - 4AC < 0$$ ### The Algebraic Grind Now, let us get our hands dirty with the algebra. We identify our coefficients: $A = 1$, $B = 2a$, and $C = 10 - 3a$. Substituting these into our discriminant formula, we get: $$(2a)^2 - 4(1)(10 - 3a) < 0$$ Expanding this, we have $4a^2 - 4(10 - 3a) < 0$, which simplifies to $4a^2 - 40 + 12a < 0$. Rearranging the terms, we arrive at: $$4a^2 + 12a - 40 < 0$$ Notice how all these terms are multiples of $4$? We can divide the entire inequality by $4$ without changing the inequality sign, leaving us with: $$a^2 + 3a - 10 < 0$$ ### The Final Verdict: The Wavy Curve We are almost there. We need to factorize $a^2 + 3a - 10 < 0$. We look for two numbers that multiply to $-10$ and add to $3$. Those numbers are $5$ and $-2$. So, the inequality becomes: $$(a + 5)(a - 2) < 0$$ To solve this, we use the wavy curve method. The critical points are $a = -5$ and $a = 2$. These points divide the number line into three regions. Testing the regions, we find that the expression is negative only between the roots. Thus, the valid interval for $a$ is: $$-5 < a < 2$$ You have successfully navigated the constraints of the quadratic and found the hidden interval. Keep this logic in your toolkit; it is the key to mastering many such problems in your **JEE** journey.

Animated Solution for Mathematics - Quadratic Equations: For all 'x', $x^{2} + 2 a x + 10 - 3 a > 0$ , then the interval in which 'a' lies is

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Visualized Solution

Visualizing the Quadratic Expression

Condition for $f (x) > 0$

Setting up the Discriminant

Expanding the Expression

Simplifying the Inequality

Factorizing the Quadratic

Applying the Wavy Curve Method

Finding the Valid Interval

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Gatekeeper

The Algebraic Grind

The Final Verdict

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For all 'x', x2+2ax+10−3a>0, then the interval in which 'a' lies is