Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Limits Continuity and Differentiability question

2022 · 25 Jul · Shift 2 · Q27
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Limits Continuity and Differentiability
  5. /2022 · 25 Jul · Shift 2 · Q27

Limits Continuity and Differentiability question

2022 · 25 Jul · Shift 2 · Q27

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→π482−(cos⁡x+sin⁡x)72−2sin⁡2x\lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}x→4π​lim​2​−2​sin2x82​−(cosx+sinx)7​ is equal to
  1. A
    14
  2. B
    7
  3. C
    14 2\sqrt22​
  4. D
    7 2\sqrt22​
View written solutionFree

Correct answer: A

  1. Rewrite the expression using an identity

We use sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos xsin2x=2sinxcosx and (sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+sin⁡2x.(\sin x+\cos x)^2 = \sin^2 x+\cos^2 x+2\sin x\cos x = 1+\sin 2x.(sinx+cosx)2=sin2x+cos2x+2sinxcosx=1+sin2x.

So, (sin⁡x+cos⁡x)2=1+sin⁡2x.(\sin x+\cos x)^2 = 1+\sin 2x.(sinx+cosx)2=1+sin2x. At the same time, 2−2sin⁡2x=2(1−sin⁡2x).\sqrt{2}-\sqrt{2}\sin 2x = \sqrt{2}(1-\sin 2x).2​−2​sin2x=2​(1−sin2x).

Also, 1−sin⁡2x=1−(sin⁡x+cos⁡x)2+1=?1-\sin 2x = 1-(\sin x+\cos x)^2+1 = ?1−sin2x=1−(sinx+cosx)2+1=? A cleaner factorization is: 1−sin⁡2x=(sin⁡x−cos⁡x)2.1-\sin 2x = (\sin x-\cos x)^2.1−sin2x=(sinx−cosx)2. Hence, 2−2sin⁡2x=2(sin⁡x−cos⁡x)2.\sqrt{2}-\sqrt{2}\sin 2x = \sqrt{2}(\sin x-\cos x)^2.2​−2​sin2x=2​(sinx−cosx)2.

  1. Substitute a simpler variable

Let t=sin⁡x+cos⁡x.t=\sin x+\cos x.t=sinx+cosx. Then as x→π4x\to \frac\pi4x→4π​, t→2.t\to \sqrt2.t→2​. Also, t2=1+sin⁡2x  ⟹  1−sin⁡2x=2−t2.t^2=1+\sin 2x \implies 1-\sin 2x = 2-t^2.t2=1+sin2x⟹1−sin2x=2−t2. Therefore the denominator becomes 2(2−t2)=2(2−t)(2+t).\sqrt2(2-t^2)=\sqrt2(\sqrt2-t)(\sqrt2+t).2​(2−t2)=2​(2​−t)(2​+t).

The limit becomes lim⁡t→282−t72(2−t2).\lim_{t\to \sqrt2}\frac{8\sqrt2-t^7}{\sqrt2(2-t^2)}.limt→2​​2​(2−t2)82​−t7​. Now factor the numerator: 82=(2)7,8\sqrt2=(\sqrt2)^7,82​=(2​)7, so 82−t7=(2)7−t7=(2−t)((2)6+(2)5t+⋯+t6).8\sqrt2-t^7=(\sqrt2)^7-t^7=(\sqrt2-t)\big((\sqrt2)^6+(\sqrt2)^5t+\cdots+t^6\big).82​−t7=(2​)7−t7=(2​−t)((2​)6+(2​)5t+⋯+t6).

Thus,

= \frac{(\sqrt2-t)\big((\sqrt2)^6+(\sqrt2)^5t+(\sqrt2)^4t^2+(\sqrt2)^3t^3+(\sqrt2)^2t^4+\sqrt2\,t^5+t^6\big)}{\sqrt2(\sqrt2-t)(\sqrt2+t)}.$$ Cancel $(\sqrt2-t)$: $$= \frac{(\sqrt2)^6+(\sqrt2)^5t+(\sqrt2)^4t^2+(\sqrt2)^3t^3+(\sqrt2)^2t^4+\sqrt2\,t^5+t^6}{\sqrt2(\sqrt2+t)}.$$ 3. **Now take the limit $t\to \sqrt2$** Each term in the numerator becomes $(\sqrt2)^6=8$, so there are $7$ equal terms: $$\text{numerator} \to 7\cdot 8=56.$$ And the denominator becomes $$\sqrt2(\sqrt2+\sqrt2)=\sqrt2(2\sqrt2)=4.$$ Hence, $$\lim_{x\to \pi/4}\frac{8\sqrt2-(\cos x+\sin x)^7}{\sqrt2-\sqrt2\sin 2x} = rac{56}{4}=14.$$ 4. **Check options** - A: $14$ ✅ - B: $7$ - C: $14\sqrt2$ - D: $7\sqrt2$ So the correct answer is **A**.
PreviousNext

More from Limits Continuity and Differentiability

  • Let f(x) be a polynomial function such that f(x)+f′(x)+f′′(x)=x5+64. Then, the value of x→1lim​x−1f(x)​ is equal to:2022 · MCQ
  • x→2π​lim​(tan2x((2sin2x+3sinx+4)21​−(sin2x+6sinx+2)21​)) is equal to2022 · MCQ
  • Let f(x)=[2x2+1] and g(x)={2x−3,2x+3,​x<0x≥0​, where [t] is the greatest integer ≤ t. Then, in the open interval (−…2022 · Numerical
  • Let f : R → R be a continuous function such that f(3x)−f(x)=x. If f(8)=7, then f(14) is equal to :2022 · MCQ
  • If the function f(x)={secx−cosxloge​(1−x+x2)+loge​(1+x+x2)​k​,,​x∈(2−π​,2π​)−{0}x=0​…2022 · MCQ
  • If f(x)={x+a∣x−4∣​,,​x≤0x>0​ and g(x)={x+1(x−4)2+b​,,​x<0x≥0​…2022 · MCQ
  • Let f(x)={x3−x2+10x−7,−2x+log2​(b2−4),​x≤1x>1​. Then the set of all values of b, for which f(x) has maximum value at x = 1,…2022 · MCQ
  • Let β=x→0lim​αx(e3x−1)αx−(e3x−1)​ for some α∈R. Then the value of α+β is :2022 · MCQ