Question
Answers

The following question consists of two statements, one labeled as ‘Assertion (A)’ and the other labeled as ‘Reason (R)’. You have to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the reason (R) is a correct explanation for the given Assertion (A). Select your answer to these items using the codes given below and then select the correct option.
Codes:
(A) Both A and R are individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Derivative of ${{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right)$ with respect to ${{\cos }^{-1}}\theta \left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right)$ is 1 for $0Reason (R): \[{{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right)={{\cos }^{-1}}\left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right)\] for $-1\le x\le 1$
a) A
b) B
c) C
d) D

Answer Verified Verified
Hint: To verify the assertion use the formula given below.
 \[\dfrac{d}{dx}\left[ f(x), with \text{ } respect \text{ } to ,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]
And to verify the Reason use the formulae of inverse trigonometric functions with considering their domains.

Firstly we will check the Assertion (a) is correct or not,
Assertion (A):$\begin{align}
  & p={{\sin }^{-1}}\left( \dfrac{2x}{1+\mathop{x}^{2}} \right) and q={{\cos }^{-1}}\theta \left( \dfrac{1-\mathop{x}^{2}}{1+\mathop{x}^{2}} \right) \\
 & Domain(p)\equiv -1\le x\le 1 \\
\end{align}$
Consider,
$p={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and, $q={{\cos }^{-1}}\theta \left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ For $ 0\le x\le 1 $
Substitute,$x=\tan \theta $ and $\therefore \theta ={{\tan }^{-1}}x$…………………………………………. (1)
$p={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$ and $q={{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)$
Now, to proceed further we should know the Half Angle formula of ‘sin’ and ‘cosine’ in the form of ‘tan’
Formulae:
$\sin 2\theta =\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$
$\cos 2\theta =\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)$
By using above formulae we can write,
$\therefore p={{\sin }^{-1}}\left( \sin 2\theta \right)$ And $q={{\cos }^{-1}}\left( \cos 2\theta \right)$ ………………………………. (2)
As we know the domains of p and q,
$Domain(p)\equiv -1\le x\le 1$ and $Domain(q)=-1\le x\le 1$………………………… (A)
By observing both the domains carefully we can say that the required interval is the subset of given intervals i.e. $\left( 0Therefore now we can solve equation (2) with the use of formulae given below,
Formulae:
\[{{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta \] $-1\le x\le 1$
\[{{\cos }^{-1}}\theta \left( \cos 2\theta \right)=2\theta \] $-1\le x\le 1$
Therefore equation (2) will become,
\[\therefore p=2\theta \] And \[q=2\theta \]
Now put the value of \[\theta ={{\tan }^{-1}}x\] as evaluated in (1),
\[\therefore p=2{{\tan }^{-1}}x\] And \[q=2{{\tan }^{-1}}x\]
Differentiating p and q with respect to x using we will get,
\[\therefore \dfrac{dp}{dx}=2\dfrac{d}{dx}{{\tan }^{-1}}x\] And \[\dfrac{dq}{dx}=2\dfrac{d}{dx}{{\tan }^{-1}}x\]
Formula:
$\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$
By using above formula we can write p and q as,
$\therefore \dfrac{dp}{dx}=2\left( \dfrac{1}{1+{{x}^{2}}} \right)$ And $\dfrac{dq}{dx}=2\left( \dfrac{1}{1+{{x}^{2}}} \right)$ …………………………… (3)
As we have to calculate derivative of p with respect to q we should know the formula given below,
Formula:
If f(x) and g(x) are two functions then,
\[\dfrac{d}{dx}\left[ f(x),with\text{ }respect\text{ }to,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]
Therefore,
\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]={}^{\dfrac{dp}{dx}}/{}_{\dfrac{dq}{dx}}\]
From (3) we can write,
\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]=\dfrac{2\left( \dfrac{1}{1+{{x}^{2}}} \right)}{2\left( \dfrac{1}{1+{{x}^{2}}} \right)}\]
\[\dfrac{d}{dx}\left[ p,with\text{ }respect\text{ }to,q \right]=1\]
Therefore Assertion (a) is true.

Now, we will check whether the Reason is true or false,
Reason (R):
We will rewrite the equation given in the Reason,
${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$

Consider, Left Hand Side(L.H.S.)
$L.H.S.={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$
Put,$x=\tan \theta $ in above equation, $\therefore \theta ={{\tan }^{-1}}x$
$\therefore L.H.S.={{\sin }^{-1}}\left( \dfrac{2\tan x}{1+{{\tan }^{2}}x} \right)$
By using Half Angle formula we can write,
$\therefore L.H.S.={{\sin }^{-1}}\left( \sin 2\theta \right)$ $-1\le x\le 1$
To solve further we should know the formula given below,
Formula:
\[{{\sin }^{-1}}\left( \sin 2\theta \right)=2\theta \] $-1\le x\le 1$
By using this formula we can write,
\[\therefore L.H.S.=2\theta \]
Put, \[\theta ={{\tan }^{-1}}x\] as evaluated earlier,
\[\therefore L.H.S.=2{{\tan }^{-1}}x\]…………………………………………….(4)

Also Consider,
$R.H.S.={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ $-1\le x\le 1$
Put,$x=\tan \theta $ in above equation, $\therefore \theta ={{\tan }^{-1}}x$
$\therefore R.H.S.={{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)$
By using Half Angle formula we can write,
\[\therefore R.H.S.={{\cos }^{-1}}\left( \cos 2\theta \right)\] $-1\le x\le 1$
To solve further we should know the formula given below,
Formula:
\[{{\cos }^{-1}}\left( \cos 2\theta \right)=2\theta \] $-1\le x\le 1$
By using this formula we can write,
\[\therefore R.H.S.=2\theta \]
Put, \[\theta ={{\tan }^{-1}}x\] as evaluated earlier,
\[\therefore R.H.S.=2{{\tan }^{-1}}x\]…………………………………………………. (5)
From (4) and (5) we can write,
\[\therefore L.H.S.=R.H.S.\]
$\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$
Therefore, we can say that Reason (R) is true.
If we observe our solution carefully then we can say that the value obtained while solving could have covered nearly half the answer of Assertion and hence Reason is very much helpful in solving this example.
Therefore, both A and R is individually true and R is the correct explanation of A. Option (a) is the correct answer.


Note:
The formula given below is very much important to solve the problem,
\[\dfrac{d}{dx}\left[ f(x),with\text{ }respect\text{ }to,g(x) \right]=\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]
 In this problem Answer is very much dependent on Limits and you should be very much careful during considering them while solving.

Bookmark added to your notes.
View Notes
×