Consider the statements: P: There exists some x ∈ R, such that f (x) + 2x = 2(1+x²) Q: There exists some x ∈ R, such that 2 f (x) +1 = 2x(1+ x) Then [JEE Adv 2012 P2]

Correct Answer is : P is false and Q is true.

Correct Answer is : P is false and Q is true.

Test Index

Application of Derivatives

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Let f (x) =

(1-x)^2 \sin^2 x + x^2

for all x ∊ R, and let g (x) =

\int\limits_{1}^{x} \left( \frac{2(t-1)}{t+1} - \ln t \right)

f (t) dt for all x ∊ (1 , ∞).

Section: Mathematics

Question : 23 of 32

Marks: +1, -0

2(1+x^2)

Solution:

f (x) =

(1-x)^2 \sin^2 x + x^2

for all x ∊ R

g (x) =

\int\limits_{1}^{x}\left(\frac{2(t-1)}{t+1} - \ln t\right)

f (t) dt for all x ∊ (1 , ∞)

For statement P:

f (x) + 2x =

2(1+x^2)

91)

(-x)^2 \sin^2 x + x^2 + 2x

2+2x^2

(1-x)2 \mathrm{ssin}^2 x

x^2 - 2x + 2

(x-1)^2 + 1

(1-x)^2 (\sin^2 x - 1)

= 1

-(1-x)^2 \cos^2 x

= 1

(1-x)^2 \cos^2 x

= - 1

So, Eq. (1) will not have real solution. Thus, P is wrong.

For statement Q:

2(1-x)^2 \sin^2 x + 2x^2 + 1

2x+2x^2

(2)

2(1-x)^2 \sin^2 x

= 2x - 1

2\sin^2 x

\frac{2x-1}{(1-x)^2}

Let h (x) =

\frac{2x-1}{(1-x)^2} \cdot 2 \sin^2 x

Clearly, h (0) = negative,

\lim\limits_{x\to 1^{-}}

h (x) = + ∞.

So, by IVT, Eq. (2) will have solution.

So, statement Q is correct.

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