Question

How to solve above question

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Answer 1

Topic :-

Coordinate Geometry

Given :-

Area of the triangle formed by

$(0,0),(a^{x^2},0)\;and\;(0,a^{6x})\;is\;\dfrac{1}{2}a^5\;sq.\;units.$

To Find :-

Value of 'x'.

Concept Used :-

$Area\:of\:triangle\:formed\:by\:(x_1,y_1),(x_1,y_1)\:and\:(x_3,y_3)\:is$

$\dfrac{1}{2}\left | x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_1-x_1y_3 \right |$

Solution :-

Coordinates of given triangle are :-

$(x_1,y_1) \equiv (0,0)$

$(x_2,y_2) \equiv (a^{x^2},0)$

$(x_3,y_3) \equiv (0,a^{6x})$

Apply formula of Area of Triangle :-

$Area=\dfrac{1}{2}\left | (0)(0)-(a^{x^2})(0)+(a^{x^2})(a^{6x})-(0)(0)+(0)(0)-(0)(a^{6x}) \right |$

$Area=\dfrac{1}{2}\left | (a^{x^2})(a^{6x}) \right |$

$Area=\dfrac{1}{2}\left | (a^{x^2+6x}) \right |$

$(\because x^m \times x^n = x^{m+n} )$

$Area=\dfrac{1}{2}a^5$

$\dfrac{1}{2}\left | (a^{x^2+6x}) \right |=\dfrac{1}{2}a^5$

$\left | (a^{x^2+6x}) \right |=a^5$

Squaring both sides,

$\left | (a^{x^2+6x}) \right |^2=(a^5)^2$

$a^{2x^2+12x}=a^{10}$

$(\because (x^m)^n = x^{mn})$

On comparing powers,

$2x^2+12x=10$

$x^2+6x-5=0$

$x^2+6x+9-9-5=0$

$(x+3)^2-14=0$

$(x+3)^2=14$

Taking square root both sides,

$x+3=\pm \sqrt{14}$

$x=-3\pm \sqrt{14}$

Answer :-

So, possible value of x is :-

• -3 + √(14) or

• -3 - √(14)

Hence, none of the option is correct.

Answer 2

${\boxed{\underline{\tt{\orange{Required \: \: answer:-}}}}}$

1st PART :-

If the area taken as a^5/2

None of these options

★GIVEN:-

• $\displaystyle \sf{(0,0)}$

• $\displaystyle \sf{( {a}^{ {x}^{2},0} )}$

• $\displaystyle \sf{({0, {a}^{6x} } )}$

• Area = a^5/2

★TO FIND:-

• x?

★FORMULA USED:-

• $\displaystyle \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}$

★SOLUTION:-

$\displaystyle \sf{ \frac{{a}^{5}}{2} = \frac{1}{2} \times ( {a}^{ {x}^{2} } )( {a}^{6x})}$

$: \longrightarrow\displaystyle \sf{ {a}^{ {x}^{2} + 6x} = {a}^{5} }$

ON COMPARING:-

$: \longrightarrow\displaystyle \sf{ {x }^{2} + 6x = 5}$

$: \longrightarrow\displaystyle \sf{ {x }^{2} + 6x - 5 = 0}$

$\displaystyle \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}$

• a = 1

• b = 6

• c = -5

$: \longrightarrow \displaystyle \sf{x = \frac{ - 6 \pm \sqrt{ {6}^{2} - 4.1( - 5)} }{2.1} }$

$: \longrightarrow \displaystyle \sf{x = \frac{ - 6 \pm2 \sqrt{14} }{2} }$

$: \longrightarrow \displaystyle \sf{x = \frac{ - 6 + 2 \sqrt{14} }{2} } \sf{and} \: \displaystyle \sf{x = \frac{ - 6 - 2 \sqrt{14} }{2} }$

$: \longrightarrow \displaystyle { \boxed{\sf{x = - 3 + \sqrt{14}}} } \: \: \: \sf{and} \: \: \: \displaystyle{ \boxed{ \sf{x = - 3 - \sqrt{14} }}}$

NONE OF THESE OPTION IS THE ANSWER.

__________________________________________✿

2nd PART:-

IF THE AREA IS 1/2a^5

$\displaystyle \sf{d.) \: -1 \: or \: -5}$

★GIVEN:-

Three points:-

• $\displaystyle \sf{(0,0)}$
• $\displaystyle \sf{( {a}^{ {x}^{2},0} )}$
• $\displaystyle \sf{({0, {a}^{6x} } )}$

• Area = 1/2a^5

★TO FIND:-

• x?

★FORMULA USED:-

$\displaystyle \sf{Area = \frac{1}{2} [x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]}$

★SOLUTION:-

Put the values in the formula:-

$: \longrightarrow \displaystyle \sf{ \frac{1}{ {2a}^{5} } = \frac{1}{2} [0 + {a}^{ {x}^{2} }( {a}^{6x} - 0) + 0 ]}$

$: \longrightarrow \displaystyle \sf{ \frac{1}{ {a}^{5} } = {a}^{ {x}^{2} . } . {a}^{6x} = {a}^{ {x}^{2} + 6x} }$

$: \longrightarrow \displaystyle \sf{ {a}^{ - 5} = {a}^{ {x}^{2} + 6x } }$

ON COMPARING:-

$: \longrightarrow \displaystyle \sf{ - 5 = {x}^{2} + 6x }$

$: \longrightarrow \displaystyle \sf{ {x}^{2} + 6x + 5 = 0}$

$: \longrightarrow \displaystyle \sf{(x + 1)(x + 5) = 0}$

${ \boxed{ \underline{ \huge{ \pink{ \sf{ \bf{x \: = - 1 \: and \: - 5}}}}}}}$

Therefore, the answer is -1 , -5.