Complete set of value(s) of x satisfying the inequality x - 3 < x 2 + 4 x - 5 is(are)

( - ∞ , - 5 ] ∪ [ 1 , ∞ )

Find the range of the function f x = x 1 + x 2 .

Let, y = x 1 + x 2 ⇒ y 1 + x 2 = x ⇒ y x 2 - x + y = 0 , which is a quadratic in x . y ≠ 0 for this equation to be quadratic but y = 0 ⇒ x = 0 which is a real solution in x and hence y = 0 will be included in Range. Now, since x ∈ R , D ≥ 0 i.e., b 2 - 4 a c ≥ 0 , where a = y , b = - 1 , c = y ⇒ - 1 2 - 4 y y ≥ 0 ⇒ 1 - 4 y 2 ≥ 0 ⇒ 4 y 2 - 1 ≤ 0 ⇒ 2 y - 1 2 y + 1 ≤ 0 (solve by wavy-curve method) y ∈ - 1 2 , 1 2 Therefore, the range of the function f x is - 1 2 , 1 2 .

Solve 2 x 2 - 5 x + 2 = 5 - 9 + 8 x 4 , where x < 5 4

Given, 2 x 2 - 5 x + 2 = 5 - 9 + 8 x 4 ; x < 5 4 ⇒ 2 x 2 - 5 x + 2 = 5 4 - 1 4 9 + 8 x ⇒ 2 x 2 - 5 x + 2 - 5 4 = - 1 4 9 + 8 x ⇒ 2 x 2 - 5 x + 3 4 = - 1 4 9 + 8 x ⇒ 8 x 2 - 20 x + 3 = - 9 + 8 x Now plotting both the graph separately, y = 8 x 2 - 20 x + 3 ⇒ y = 8 x 2 - 5 2 x + 3 8 ⇒ 1 8 y = x 2 - 5 2 x + 25 16 - 25 16 + 3 8 ⇒ y 8 = x - 5 4 2 - 19 16 ⇒ 1 8 y + 19 2 = x - 5 4 2 Vertex ≡ 5 4 , - 19 2 y = - 9 + 8 x ⇒ y 2 = 9 + 8 x ⇒ y 2 = 8 x + 9 8 We will draw only negative branch. Vertex ≡ - 9 8 , 0 It is clear both the graphs does not cut each other ⇒ No solution

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 19

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Number of positive integral value(s) of $x$ satisfying the inequality $\frac{{(x - 4)}^{2013} \cdot {(x + 8)}^{2014} \cdot (x + 1)}{x^{2016} \cdot {(x - 2)}^{3} \cdot {(x + 3)}^{5} \cdot (x - 6) \cdot {(x + 9)}^{2012}} \leq 0$ is/are

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Important Questions on Quadratic Equations

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 20

Number of non-negative integral values of

x

satisfying the inequality

\frac{2}{x^{2} - x + 1} - \frac{1}{x + 1} - \frac{2 x - 1}{x^{3} + 1} \geq 0

is/are

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 21

Complete set of value(s) of

x

satisfying the inequality

x - 3 < \sqrt{x^{2} + 4 x - 5}

is(are)

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 22

Set of all real values of

x

satisfying the in equation

\sqrt{4 - x^{2}} \geq \frac{1}{x}

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 23

Find the range of the function

f (x) = \frac{x}{1 + x^{2}}

.

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 24

Solve $2 x^{2} - 5 x + 2 = \frac{5 - \sqrt{9 + 8 x}}{4},$ where $x < \frac{5}{4}$

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 25

Range of the function

f (x) = \frac{{(x - 2)}^{2}}{(x - 1) (x - 3)}

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Alpha Question Bank for Engineering: Mathematics>Chapter 4 - Quadratic Equations>Exercise-1>Q 26

If $α, β$ are the roots of the quadratic equation $c x^{2} - 2 b x + 4 a = 0$ then find the quadratic equation whose roots are $\frac{α}{2}, \frac{β}{2}$ .