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Application of Derivatives question

2010 · Shift 0 · Q39
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  5. /2010 · Shift 0 · Q39

Application of Derivatives question

2010 · Shift 0 · Q39

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Let f:R→Rf:R \to Rf:R→R be a continuous function defined by f(x)=1ex+2e−xf\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}f(x)=ex+2e−x1​ Statement - 1 : f(c)=13,f\left( c \right) = {1 \over 3},f(c)=31​, for some c∈Rc \in Rc∈R. Statement - 2 : 0<f(x)≤122,0 \lt f\left( x \right) \le {1 \over {2\sqrt 2 }},0<f(x)≤22​1​, for all x∈Rx \in Rx∈R
  1. A
    Statement - 1 is true, Statement -2 is true; Statement - 2 is not a correct explanation for Statement - 1.
  2. B
    Statement - 1 is true, Statement - 2 is false.
  3. C
    Statement - 1 is false, Statement - 2 is true.
  4. D
    Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1.
View written solutionFree

Correct answer: D

  1. Given function

f(x)=1ex+2e−x,x∈Rf(x)=\frac{1}{e^x+2e^{-x}}, \qquad x\in \mathbb Rf(x)=ex+2e−x1​,x∈R

We must examine both statements.


  1. Analyze Statement-2 first

We need to check whether

0<f(x)≤122for all x∈R.0<f(x)\le \frac{1}{2\sqrt 2}\quad \text{for all }x\in \mathbb R.0<f(x)≤22​1​for all x∈R.

Since ex>0e^x>0ex>0 and e−x>0e^{-x}>0e−x>0 for all xxx, we have

ex+2e−x>0e^x+2e^{-x}>0ex+2e−x>0

so

f(x)=1ex+2e−x>0.f(x)=\frac{1}{e^x+2e^{-x}}>0.f(x)=ex+2e−x1​>0.

Now find the maximum value of f(x)f(x)f(x).

Since f(x)f(x)f(x) is the reciprocal of ex+2e−xe^x+2e^{-x}ex+2e−x, maximizing f(x)f(x)f(x) is equivalent to minimizing

g(x)=ex+2e−x.g(x)=e^x+2e^{-x}.g(x)=ex+2e−x.

Differentiate:

g′(x)=ex−2e−x.g'(x)=e^x-2e^{-x}.g′(x)=ex−2e−x.

Set g′(x)=0g'(x)=0g′(x)=0:

ex−2e−x=0e^x-2e^{-x}=0ex−2e−x=0 ex=2e−xe^x=2e^{-x}ex=2e−x e2x=2e^{2x}=2e2x=2 x=12ln⁡2.x=\frac{1}{2}\ln 2.x=21​ln2.

Now

g′′(x)=ex+2e−x>0,g''(x)=e^x+2e^{-x}>0,g′′(x)=ex+2e−x>0,

so this gives a minimum of g(x)g(x)g(x).

Evaluate ggg there:

Let ex=2e^x=\sqrt 2ex=2​, then e−x=12e^{-x}=\frac{1}{\sqrt 2}e−x=2​1​.

Hence

gmin⁡=2+2⋅12=2+2=22.g_{\min}=\sqrt 2+2\cdot \frac{1}{\sqrt 2}=\sqrt 2+\sqrt 2=2\sqrt 2.gmin​=2​+2⋅2​1​=2​+2​=22​.

Therefore,

f(x)≤122.f(x)\le \frac{1}{2\sqrt 2}.f(x)≤22​1​.

Combining with positivity,

0<f(x)≤122∀x∈R.0<f(x)\le \frac{1}{2\sqrt 2}\quad \forall x\in \mathbb R.0<f(x)≤22​1​∀x∈R.

So Statement-2 is true.


  1. Now check Statement-1

Statement-1 says:

f(c)=13for some c∈R.f(c)=\frac13 \quad \text{for some } c\in \mathbb R.f(c)=31​for some c∈R.

From Statement-2, we know

0<f(x)≤122.0<f(x)\le \frac{1}{2\sqrt 2}.0<f(x)≤22​1​.

Now compare:

122=12.828…>13=13=0.333…\frac{1}{2\sqrt 2}=\frac{1}{2.828\dots}>\frac13=\frac{1}{3}=0.333\dots22​1​=2.828…1​>31​=31​=0.333…

Indeed,

22<3  ⟹  122>13.2\sqrt 2<3 \implies \frac{1}{2\sqrt 2}>\frac13.22​<3⟹22​1​>31​.

Also, fff is continuous on R\mathbb RR.

Now check some value:

f(0)=11+2=13.f(0)=\frac{1}{1+2}=\frac13.f(0)=1+21​=31​.

So directly, for c=0c=0c=0,

f(c)=f(0)=13.f(c)=f(0)=\frac13.f(c)=f(0)=31​.

Hence Statement-1 is true.


  1. Is Statement-2 the correct explanation for Statement-1?

Statement-2 tells us that the range satisfies

0<f(x)≤122.0<f(x)\le \frac{1}{2\sqrt 2}.0<f(x)≤22​1​.

Since

13<122,\frac13 < \frac{1}{2\sqrt 2},31​<22​1​,

and fff is continuous, while also

f(x)→0as x→±∞,f(x)\to 0 \quad \text{as } x\to \pm \infty,f(x)→0as x→±∞,

the value 13\frac1331​ is indeed attained (in fact, directly at x=0x=0x=0).

More simply, Statement-2 establishes that the function takes positive values up to at least 13\frac1331​, and continuity ensures intermediate values are attained. So it serves as the intended explanation for Statement-1.

Thus both statements are true, and Statement-2 is the correct explanation of Statement-1.


  1. Option check
  • A: Both true, but not explanation — incorrect
  • B: Statement-2 false — incorrect
  • C: Statement-1 false — incorrect
  • D: Both true, and Statement-2 is correct explanation — correct

Final Answer

D\boxed{\text{D}}D​

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