ISI2015-MMA-15

Question

The number of real solutions of the equations $(9/10)^x = -3+x-x^2$ is

• A. $2$
• B. $0$
• C. $1$
• D. none of the above

Comments

https://gateoverflow.in/211219/isi-sample-2014-15

Answer 1

Answer: $\mathbf B$

Explanation:

Algebraic solution:
$\Big({\underbrace{\frac{9}{10}}}_{\bf{+ve}}\Big)^x = -(x^2-x+3) = -\Big(x^2-2.\frac{1}{2}x+(\frac{1}{2})^2+3(\frac{1}{2})^2\Big) =\underbrace{ -\Big(\underbrace{(x-\frac{1}{2})^2 }_{+ve}+ \underbrace{\frac{11}{4}}_{+ve}\Big)}_{-ve}$

So, a positive value can never be equal to a negative value.

Hence, the equation does not contain any solution.

So, option $\mathbf B$ is the right answer.

GRAPHICAL SOLUTION:

Above curves represent the graph of $a^x$ where $0 \lt x \lt1$ and the quadratic equation.

The quadratic equation is downward open parabola since the coefficient of $x^2$ is negative.

Vertex of the parabola is located at $x = \frac{-b}{2a} = \frac{1}{2}$

Also note that it will never intersect the $x-axis$ as its Discriminant, D is $-ve$

$D = b^2-4ac = 1^2-4(-1)(-3) = 1-12 = -11 \lt0$

Hence, the two curves will never intersect each other, which means no solution is possible to the given inequality.

$\therefore \mathbf B$ is the correct option.