Question

Find the domain of $$f\left( x \right) = \sqrt {{{\left( {0.625} \right)}^{4 - 3x}} - {{\left( {1.6} \right)}^{x\left( {x + 8} \right)}}} $$

A. $$\left[ { - 3,\,2} \right]$$
B. $$\left[ { 1,\,4} \right]$$
C. $$\left[ {2,\,5} \right]$$
D. $$\left[ { - 4,\, - 1} \right]$$  
Answer :   $$\left[ { - 4,\, - 1} \right]$$
Solution :
$$\eqalign{ & {\text{Clearly, }}{\left( {0.625} \right)^{4 - 3x}} \geqslant {\left( {1.6} \right)^{x\left( {x + 8} \right)}} \cr & {\text{or }}{\left( {\frac{5}{8}} \right)^{4 - 3x}} \geqslant {\left( {\frac{8}{5}} \right)^{x\left( {x + 8} \right)}} \cr & {\text{or }}{\left( {\frac{8}{5}} \right)^{3x - 4}} \geqslant {\left( {\frac{8}{5}} \right)^{x\left( {x + 8} \right)}} \cr & {\text{or }}3x - 4 \geqslant {x^2} + 8x \cr & {\text{or }}{x^2} + 5x + 4 \leqslant 0 \cr & {\text{or }} - 4 \leqslant x \leqslant - 1 \cr} $$
Hence, the domain of function $$f\left( x \right)$$  is $$x\, \in \left[ { - 4,\, - 1} \right]$$

Releted MCQ Question on
Calculus >> Function

Releted Question 1

Let $$R$$ be the set of real numbers. If $$f:R \to R$$   is a function defined by $$f\left( x \right) = {x^2},$$   then $$f$$ is:

A. Injective but not surjective
B. Surjective but not injective
C. Bijective
D. None of these.
Releted Question 2

The entire graphs of the equation $$y = {x^2} + kx - x + 9$$     is strictly above the $$x$$-axis if and only if

A. $$k < 7$$
B. $$ - 5 < k < 7$$
C. $$k > - 5$$
D. None of these.
Releted Question 3

Let $$f\left( x \right) = \left| {x - 1} \right|.$$    Then

A. $$f\left( {{x^2}} \right) = {\left( {f\left( x \right)} \right)^2}$$
B. $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$$
C. $$f\left( {\left| x \right|} \right) = \left| {f\left( x \right)} \right|$$
D. None of these
Releted Question 4

If $$x$$ satisfies $$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$       then

A. $$0 \leqslant x \leqslant 4$$
B. $$x \leqslant - 2\,{\text{or}}\,x \geqslant 4$$
C. $$x \leqslant 0\,{\text{or}}\,x \geqslant 4$$
D. None of these

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