The sum of all real values of $x$ satisfying the equation $(x^{2} - 5 x + 5)^{(x^{2} + 4 x - 60)} = 1$ is

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3$
CASE $1 :$
$x^{2} + 4 x - 60$ can be any real number and $x^{2} - 5 x + 5 = 1$
i.e. $x^{2} - 5 x + 4 = 0$
$\Rightarrow x = 1, 4$

CASE $2 :$
$x^{2} + 4 x - 60 = 0$ and $x^{2} - 5 x + 5$ can be any real number except $0$ as $0^{0}$ is undefined.
i.e. $(x + 10) (x - 6) = 0$
$\Rightarrow x = 6, - 10$
and $x^{2} - 5 x + 5 \neq 0$
$\Rightarrow x \neq \frac{5 \pm \sqrt{5}}{2}$

CASE $3 :$
$(- 1)^{even} = 1$
i.e. $x^{2} - 5 x + 5 = - 1$ and $x^{2} + 4 x - 60$ is an even number

$\Rightarrow x^{2} - 5 x + 6 = 0$
$\Rightarrow (x - 2) (x - 3) = 0$
$\Rightarrow x = 2, 3$

At $x = 2,$ the value of $x^{2} + 4 x - 60$ is
$(2)^{2} + 4 (2) - 60 = - 48$
Which is even
At $x = 3,$ the value of $x^{2} + 4 x - 60$ is
$(3)^{2} + 4 (3) - 60 = - 39$
Which is odd
$∴ x = 3$ is rejected

Sum of all the real values of $x = 1 + 4 + 6 - 10 + 2 = 3$

Suggest Corrections

Similar questions

The sum of all real values of x satisfying the equation ${(x^{2} - 5 x + 5)}^{x^{2} + 4 x - 60} = 1$ is:

x

(x^{2} - 5 x + 5)^{(x^{2} + 4 x - 60)} = 1

\sqrt{5 x^{2} - 6 x + 8} - \sqrt{5 x^{2} - 6 x - 7} = 1

x

(x^{2} + 7 x + 11)^{(x^{2} - 4 x - 21)} = 1

x

\sqrt{3 x^{2} + 6 x + 7} + \sqrt{5 x^{2} + 10 x + 14} = 4 - 2 x - x^{2},

Join BYJU'S Learning Program