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Limits Continuity and Differentiability question

2017 · 9 Apr · Shift 1 · Q39
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Limits Continuity and Differentiability question

2017 · 9 Apr · Shift 1 · Q39

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
The value of k for which the function f(x)={(45)tan⁡ 4xtan⁡ 5x  ,0<x<π2k+25   ,x=π2f\left( x \right) = \left\{ {\begin{matrix} {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 \lt x \lt {\pi \over 2}} \\ {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \\ \end{matrix} } \right.f(x)={(54​)tan5xtan4x​,k+52​,​0<x<2π​x=2π​​ is continuous at x =π2,{\pi \over 2},2π​, is :
  1. A
    1720{{17} \over {20}}2017​
  2. B
    25{{2} \over {5}}52​
  3. C
    35{{3} \over {5}}53​
  4. D
    −25-{{2} \over {5}}−52​
View written solutionFree

Correct answer: C

  1. For continuity at x=π2x=\dfrac{\pi}{2}x=2π​, we need
lim⁡x→(π/2)−f(x)=f(π2).\lim_{x\to (\pi/2)^-} f(x)=f\left(\frac{\pi}{2}\right).x→(π/2)−lim​f(x)=f(2π​).

Here,

f(π2)=k+25.f\left(\frac{\pi}{2}\right)=k+\frac{2}{5}.f(2π​)=k+52​.

So we first compute

L=lim⁡x→(π/2)−(45)tan⁡4xtan⁡5x.L=\lim_{x\to (\pi/2)^-}\left(\frac45\right)^{\frac{\tan 4x}{\tan 5x}}.L=x→(π/2)−lim​(54​)tan5xtan4x​.
  1. As x→π2x\to \dfrac{\pi}{2}x→2π​,
tan⁡4x→tan⁡2π=0,tan⁡5x→tan⁡5π2 is also effectively 0\tan 4x\to \tan 2\pi=0, \qquad \tan 5x\to \tan \frac{5\pi}{2}\text{ is also effectively }0tan4x→tan2π=0,tan5x→tan25π​ is also effectively 0

so the exponent is of indeterminate form 00\dfrac0000​.

  1. Simplify the ratio using periodicity of tangent: Let
x=π2+h,x=\frac{\pi}{2}+h,x=2π​+h,

where h→0−h\to 0^-h→0−. Then

4x=2π+4h  ⟹  tan⁡4x=tan⁡4h,4x=2\pi+4h \implies \tan 4x=\tan 4h,4x=2π+4h⟹tan4x=tan4h,

and

5x=5π2+5h=π2+2π+5h  ⟹  tan⁡5x=tan⁡(π2+5h).5x=\frac{5\pi}{2}+5h=\frac{\pi}{2}+2\pi+5h \implies \tan 5x=\tan\left(\frac{\pi}{2}+5h\right).5x=25π​+5h=2π​+2π+5h⟹tan5x=tan(2π​+5h).

Now,

tan⁡(π2+θ)=−cot⁡θ.\tan\left(\frac{\pi}{2}+\theta\right)=-\cot\theta.tan(2π​+θ)=−cotθ.

So,

tan⁡5x=−cot⁡5h.\tan 5x=-\cot 5h.tan5x=−cot5h.

Hence

tan⁡4xtan⁡5x=tan⁡4h−cot⁡5h=−tan⁡4htan⁡5h.\frac{\tan 4x}{\tan 5x}=\frac{\tan 4h}{-\cot 5h}=-\tan 4h\tan 5h.tan5xtan4x​=−cot5htan4h​=−tan4htan5h.

As h→0h\to 0h→0,

tan⁡4h∼4h,tan⁡5h∼5h,\tan 4h\sim 4h, \qquad \tan 5h\sim 5h,tan4h∼4h,tan5h∼5h,

therefore

tan⁡4xtan⁡5x∼−20h2→0.\frac{\tan 4x}{\tan 5x}\sim -20h^2\to 0.tan5xtan4x​∼−20h2→0.
  1. Therefore,
L=(45)0=1.L=\left(\frac45\right)^0=1.L=(54​)0=1.
  1. Continuity requires
k+25=1.k+\frac25=1.k+52​=1.

So,

k=1−25=35.k=1-\frac25=\frac35.k=1−52​=53​.
  1. Thus the correct option is
35\boxed{\frac35}53​​

which is option C.

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