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Application of Derivatives question

2012 · Shift 2 · Q28
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  5. /2012 · Shift 2 · Q28

Application of Derivatives question

2012 · Shift 2 · Q28

JEE AdvancedMathematicsApplication of DerivativesMCQ+4 / −1
Let f(x)=(1−x)2  sin⁡2  x+x2f\left( x \right) = {\left( {1 - x} \right)^2}\,\,{\sin ^2}\,\,x + {x^2}f(x)=(1−x)2sin2x+x2 for all x∈IRx \in IRx∈IR and let g(x)=∫1x(2(t−1)t+1−In t)f(t)dtg\left( x \right) = \int\limits_1^x {\left( {{{2\left( {t - 1} \right)} \over {t + 1}} - In\,t} \right)f\left( t \right)dt}g(x)=1∫x​(t+12(t−1)​−Int)f(t)dt for all x∈(1, ∞)x \in \left( {1,\,\infty } \right)x∈(1,∞). Consider the statements: P:P:P: There exists some x∈Rx \in Rx∈R such that f(x)+2x=2(1+x2)f\left( x \right) + 2x = 2\left( {1 + {x^2}} \right)f(x)+2x=2(1+x2)Q:  Q:\,\,Q: There exists some x∈Rx \in Rx∈R such that 2 f(x)+1=2x(1+x)2\,f\left( x \right) + 1 = 2x\left( {1 + x} \right)2f(x)+1=2x(1+x) Then
  1. A
    both PPP and QQQ are true
  2. B
    PPP is true and QQQ is false
  3. C
    PPP is false and QQQ is true
  4. D
    both PPP and QQQ are false
View written solutionFree

Correct answer: C

  1. Given

We have f(x)=(1−x)2sin⁡2x+x2.f(x)=(1-x)^2\sin^2 x+x^2.f(x)=(1−x)2sin2x+x2.

Also, g(x)=∫1x(2(t−1)t+1−ln⁡t)f(t) dt,g(x)=\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right)f(t)\,dt,g(x)=∫1x​(t+12(t−1)​−lnt)f(t)dt, but note that the statements PPP and QQQ involve only f(x)f(x)f(x). So g(x)g(x)g(x) is not needed here.

  1. Analyze statement PPP

Statement PPP says: ∃x∈R such that f(x)+2x=2(1+x2).\exists x\in \mathbb R \text{ such that } f(x)+2x=2(1+x^2).∃x∈R such that f(x)+2x=2(1+x2).

Substitute f(x)f(x)f(x): (1−x)2sin⁡2x+x2+2x=2+2x2. (1-x)^2\sin^2 x+x^2+2x=2+2x^2.(1−x)2sin2x+x2+2x=2+2x2. Rearrange: (1−x)2sin⁡2x=2+2x2−x2−2x=x2−2x+2. (1-x)^2\sin^2 x=2+2x^2-x^2-2x=x^2-2x+2.(1−x)2sin2x=2+2x2−x2−2x=x2−2x+2. So (1−x)2sin⁡2x=(x−1)2+1. (1-x)^2\sin^2 x=(x-1)^2+1.(1−x)2sin2x=(x−1)2+1.

Now observe:

  • Left side satisfies (1−x)2sin⁡2x≤(1−x)2=(x−1)2, (1-x)^2\sin^2 x\le (1-x)^2=(x-1)^2,(1−x)2sin2x≤(1−x)2=(x−1)2, since sin⁡2x≤1\sin^2 x\le 1sin2x≤1.
  • Right side is (x−1)2+1>(x−1)2. (x-1)^2+1>(x-1)^2.(x−1)2+1>(x−1)2.

Hence (1−x)2sin⁡2x≤(x−1)2<(x−1)2+1, (1-x)^2\sin^2 x \le (x-1)^2 < (x-1)^2+1,(1−x)2sin2x≤(x−1)2<(x−1)2+1, so equality is impossible for any real xxx.

Therefore, PPP is false.

  1. Analyze statement QQQ

Statement QQQ says: ∃x∈R such that 2f(x)+1=2x(1+x).\exists x\in \mathbb R \text{ such that } 2f(x)+1=2x(1+x).∃x∈R such that 2f(x)+1=2x(1+x).

Substitute f(x)f(x)f(x): 2((1−x)2sin⁡2x+x2)+1=2x+2x2.2\big((1-x)^2\sin^2 x+x^2\big)+1=2x+2x^2.2((1−x)2sin2x+x2)+1=2x+2x2. Simplify: 2(1−x)2sin⁡2x+2x2+1=2x+2x2.2(1-x)^2\sin^2 x+2x^2+1=2x+2x^2.2(1−x)2sin2x+2x2+1=2x+2x2. So 2(1−x)2sin⁡2x=2x−1.2(1-x)^2\sin^2 x=2x-1.2(1−x)2sin2x=2x−1.

We need to check whether this has some real solution.

A very natural choice is x=12x=\frac12x=21​. Then right side becomes 2x−1=2⋅12−1=0.2x-1=2\cdot \frac12-1=0.2x−1=2⋅21​−1=0. Left side becomes

=2\cdot \frac14\sin^2\left(\frac12\right) =\frac12\sin^2\left(\frac12\right)>0,$$ so $x=\frac12$ is not a solution. Try instead to force the left side to be $0$. Since $$2(1-x)^2\sin^2 x=0$$ when either $x=1$ or $\sin x=0$. Take $x=1$. Then left side is $0$, and right side is $$2(1)-1=1,$$ so not a solution. Take $x=n\pi$ for integer $n$. Then left side is $0$. For equality we need $$2x-1=0 \implies x=\frac12,$$ which is not of the form $n\pi$. So those do not work directly. Let us instead define $$h(x)=2(1-x)^2\sin^2 x-(2x-1).$$ We seek some $x$ with $h(x)=0$. Check two values: - At $x=\frac12$, $$h\left(\frac12\right)=\frac12\sin^2\left(\frac12\right)>0.$$ - At $x=1$, $$h(1)=0-(2-1)=-1<0.$$ Since $h(x)$ is continuous, by the Intermediate Value Theorem there exists some $$x\in\left(\frac12,1\right)$$ such that $$h(x)=0.$$ Hence there exists a real $x$ satisfying $$2f(x)+1=2x(1+x).$$ Therefore, **$Q$ is true**. --- 4. **Conclusion** - $P$ is false - $Q$ is true Therefore the correct option is $$\boxed{\text{C}}$$ --- 5. **Comparison with stored answer** Stored correct answer: $C$ Our derived answer is also $C$, so they agree.
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