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and the other labelled as “Reason (R)”. You are 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 the 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 and 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): Let \[f:[0,\infty )\to [0,\infty ],\]be a function defined by \[y=f\left( x \right)={{x}^{2}}\], then \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=1\]

Reason (R): \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]

(a) A

(b) B

(c) C

(d) D

Answer

Verified

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Hint: Differentiate the given function with respect to \[x\] and with respect to \[y\] twice.

A: Given \[f:[0,\infty )\to [0,\infty ]\]be a function defined by \[y=f\left( x \right)={{x}^{2}}\], then we have to check the value of \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)\].

Now, we take the given function,

\[y={{x}^{2}}....\left( i \right)\]

Now, we differentiate it with respect to \[x\].

\[\left[ \text{Also}\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]\]

Therefore, \[\dfrac{dy}{dx}=2x.....\left( ii \right)\]

Again differentiating with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2....\left( iii \right)\]

As we have found that, \[\dfrac{dy}{dx}=2x\]

By taking reciprocal on both sides,

We get, \[\dfrac{dx}{dy}=\dfrac{1}{2x}....\left( iv \right)\]

Now, we differentiate with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{d}{dy}\left( \dfrac{1}{2x} \right)\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}\dfrac{dx}{dy}\]

Now, we put the value of \[\dfrac{dx}{dy}\].

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}.\dfrac{1}{2x}\]

Hence, \[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{4{{x}^{3}}}....\left( v \right)\]

Multiplying equation \[\left( iii \right)\]and \[\left( v \right)\],

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=2.\left( \dfrac{-1}{4{{x}^{3}}} \right)\]

Therefore, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}}

\right)=\dfrac{-1}{2{{x}^{3}}}\].

Hence, given Assertion (A) is wrong.

R: Here we have to check whether \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]or not.

We know that any quantity when multiplied by its reciprocal gives a result as \[1\].

That is \[a\times \dfrac{1}{a}=1\].

Now, we put \[a=\dfrac{dy}{dx}\].

We get, \[\dfrac{dy}{dx}\times \dfrac{1}{\dfrac{dy}{dx}}=1\]

Or, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]

To verify it further, we multiply the equation \[\left( ii \right)\]and \[\left( iv \right)\].

\[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=2x.\dfrac{1}{2x}\]

Therefore, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\] [Hence Proved]

Hence, Reason (R) is correct.

Therefore, option (d) is correct that is A is false and R is true

Note: Some students misunderstand that \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}\]and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] are reciprocal of each other like \[\dfrac{dx}{dy}\]and \[\dfrac{dy}{dx}\], but they are not as proved by above result also.

A: Given \[f:[0,\infty )\to [0,\infty ]\]be a function defined by \[y=f\left( x \right)={{x}^{2}}\], then we have to check the value of \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)\].

Now, we take the given function,

\[y={{x}^{2}}....\left( i \right)\]

Now, we differentiate it with respect to \[x\].

\[\left[ \text{Also}\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]\]

Therefore, \[\dfrac{dy}{dx}=2x.....\left( ii \right)\]

Again differentiating with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2....\left( iii \right)\]

As we have found that, \[\dfrac{dy}{dx}=2x\]

By taking reciprocal on both sides,

We get, \[\dfrac{dx}{dy}=\dfrac{1}{2x}....\left( iv \right)\]

Now, we differentiate with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{d}{dy}\left( \dfrac{1}{2x} \right)\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}\dfrac{dx}{dy}\]

Now, we put the value of \[\dfrac{dx}{dy}\].

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}.\dfrac{1}{2x}\]

Hence, \[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{4{{x}^{3}}}....\left( v \right)\]

Multiplying equation \[\left( iii \right)\]and \[\left( v \right)\],

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=2.\left( \dfrac{-1}{4{{x}^{3}}} \right)\]

Therefore, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}}

\right)=\dfrac{-1}{2{{x}^{3}}}\].

Hence, given Assertion (A) is wrong.

R: Here we have to check whether \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]or not.

We know that any quantity when multiplied by its reciprocal gives a result as \[1\].

That is \[a\times \dfrac{1}{a}=1\].

Now, we put \[a=\dfrac{dy}{dx}\].

We get, \[\dfrac{dy}{dx}\times \dfrac{1}{\dfrac{dy}{dx}}=1\]

Or, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]

To verify it further, we multiply the equation \[\left( ii \right)\]and \[\left( iv \right)\].

\[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=2x.\dfrac{1}{2x}\]

Therefore, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\] [Hence Proved]

Hence, Reason (R) is correct.

Therefore, option (d) is correct that is A is false and R is true

Note: Some students misunderstand that \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}\]and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] are reciprocal of each other like \[\dfrac{dx}{dy}\]and \[\dfrac{dy}{dx}\], but they are not as proved by above result also.