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Limits Continuity and Differentiability question

2022 · 26 Jul · Shift 1 · Q30
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  5. /2022 · 26 Jul · Shift 1 · Q30

Limits Continuity and Differentiability question

2022 · 26 Jul · Shift 1 · Q30

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If the function f(x)={log⁡e(1−x+x2)+log⁡e(1+x+x2)sec⁡x−cos⁡x,x∈(−π2,π2)−{0}k,x=0f(x) = \left\{ {\begin{matrix} {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \\ k & , & {x = 0} \\ \end{matrix} } \right.f(x)={secx−cosxloge​(1−x+x2)+loge​(1+x+x2)​k​,,​x∈(2−π​,2π​)−{0}x=0​ is continuous at x = 0, then k is equal to:
  1. A
    1
  2. B
    −-− 1
  3. C
    e
  4. D
    0
View written solutionFree

Correct answer: A

  1. For continuity at x=0x=0x=0, we need k=lim⁡x→0ln⁡(1−x+x2)+ln⁡(1+x+x2)sec⁡x−cos⁡x.k=\lim_{x\to 0}\frac{\ln(1-x+x^2)+\ln(1+x+x^2)}{\sec x-\cos x}.k=limx→0​secx−cosxln(1−x+x2)+ln(1+x+x2)​.

  2. Combine the logarithms: ln⁡(1−x+x2)+ln⁡(1+x+x2)=ln⁡((1−x+x2)(1+x+x2)).\ln(1-x+x^2)+\ln(1+x+x^2)=\ln\big((1-x+x^2)(1+x+x^2)\big).ln(1−x+x2)+ln(1+x+x2)=ln((1−x+x2)(1+x+x2)).

Now simplify the product: (1−x+x2)(1+x+x2)=(1+x2−x)(1+x2+x).(1-x+x^2)(1+x+x^2)=(1+x^2-x)(1+x^2+x).(1−x+x2)(1+x+x2)=(1+x2−x)(1+x2+x). This is of the form (A−x)(A+x)=A2−x2(A-x)(A+x)=A^2-x^2(A−x)(A+x)=A2−x2 with A=1+x2A=1+x^2A=1+x2, so (1−x+x2)(1+x+x2)=(1+x2)2−x2=1+x2+x4.(1-x+x^2)(1+x+x^2)=(1+x^2)^2-x^2=1+x^2+x^4.(1−x+x2)(1+x+x2)=(1+x2)2−x2=1+x2+x4.

Hence the numerator becomes ln⁡(1+x2+x4).\ln(1+x^2+x^4).ln(1+x2+x4).

  1. Simplify the denominator: sec⁡x−cos⁡x=1cos⁡x−cos⁡x=1−cos⁡2xcos⁡x=sin⁡2xcos⁡x.\sec x-\cos x=\frac{1}{\cos x}-\cos x=\frac{1-\cos^2 x}{\cos x}=\frac{\sin^2 x}{\cos x}.secx−cosx=cosx1​−cosx=cosx1−cos2x​=cosxsin2x​.

Therefore,

=\lim_{x\to 0}\frac{\cos x\,\ln(1+x^2+x^4)}{\sin^2 x}.$$ 4. Use standard limits near $x=0$: - $\cos x\to 1$ - $\ln(1+u)\sim u$ as $u\to 0$ - $\sin x\sim x$, so $\sin^2 x\sim x^2$ Let $u=x^2+x^4$. Then as $x\to 0$, $$\ln(1+x^2+x^4)\sim x^2+x^4.$$ So $$k=\lim_{x\to 0} \cos x\cdot \frac{x^2+x^4}{\sin^2 x}.$$ Using $\sin^2 x\sim x^2$, $$k=1\cdot \lim_{x\to 0}\frac{x^2(1+x^2)}{x^2}=1.$$ Thus, $$k=1.$$ 5. Checking options: - A: $1$ ✔ - B: $-1$ ✘ - C: $e$ ✘ - D: $0$ ✘ Therefore the correct option is **A**.
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