Answer

Verified

402.6k+ views

**Hint:**The given equation is a second-degree equation in ${\sin ^{ - 1}}x$. To find what all values $x$ can take (and thus the number of solutions), first we need to check what all values ${\sin ^{ - 1}}x$ can take. This can be done by solving the quadratic equation by taking ${\sin ^{ - 1}}x = y$(say). Then we can eliminate any values which does not belong to the range of ${\sin ^{ - 1}}x$.

**Formula used:**

A second-degree equation of the form $a{x^2} + bx + c = 0$ can be solved by

$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

**Complete step by step answer:**

The given quadratic equation is $2{({\sin ^{ - 1}}x)^2} - {\sin ^{ - 1}}x - 6 = 0$

We are asked to find the number of solutions.

Here the variable present is $x$. So, we need to find the number of values $x$ can take.

Let, ${\sin ^{ - 1}}x = y$

Then the equation becomes,

$\Rightarrow 2{y^2} - y - 6 = 0$, which is a quadratic equation in $y$.

A second-degree equation of the form $a{x^2} + bx + c = 0$ can be solved by

$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Here the variable is $y$.

Also $a = 2,b = - 1,c = - 6$

$\Rightarrow y = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times 2 \times - 6} }}{{2 \times 2}}$

Simplifying we get,

$ \Rightarrow y = \dfrac{{1 \pm \sqrt {1 + 48} }}{4} = \dfrac{{1 \pm \sqrt {49} }}{4}$

$ \Rightarrow y = \dfrac{{1 \pm 7}}{4}$

This gives $y = \dfrac{8}{4} = 2$ or $y = \dfrac{{ - 6}}{4} = \dfrac{{ - 3}}{2}$

Substituting for $y$ we get,

$\Rightarrow {\sin ^{ - 1}}x = 2{\text{ (or) }}{\sin ^{ - 1}}x = \dfrac{{ - 3}}{2}$

But ${\sin ^{ - 1}}x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2})$, which is the open interval with limits $ - \dfrac{\pi }{2},\dfrac{\pi }{2}$.

Here $\pi $ is in radians, which can be approximated to $3.14$.

This gives

$\Rightarrow \dfrac{\pi }{2} = \dfrac{{3.14}}{2} = 1.57$

So ${\sin ^{ - 1}}x$ cannot take the value $2$, since the value must be less than $1.57$.

Therefore ${\sin ^{ - 1}}x = \dfrac{{ - 3}}{2} \Rightarrow x = \sin (\dfrac{{ - 3}}{2})$, which is the only solution.

This gives the number of solutions is $1$.

**$\therefore $ The correct answer is option B.**

**Note:**

The alternative method to solve the quadratic equation without this formula.

$2{y^2} - y - 6 = 0$

By simple rearrangement we get,

$ \Rightarrow 2{y^2} - 4y + 3y - 6 = 0$

Taking common factors from first two terms and last two terms,

$ \Rightarrow 2y(y - 2) + 3(y - 2) = 0$

$ \Rightarrow (y - 2)(2y + 3) = 0$

The product of two terms is zero implies either one is zero.

$ \Rightarrow (y - 2) = 0{\text{ (or) }}(2y + 3) = 0$

$ \Rightarrow y = 2\left( {{\text{or}}} \right) = \dfrac{{ - 3}}{2}$

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail