If the function ƒ defined on , ( \pi \over 6,\pi \over 3 ) by f(x) = \ \beginmatrix \sqrt 2 \mathop\rm cosx olimits - 1 \over \cot x - 1, & x e \pi \over 4 k, & x = \pi \over 4 \endmatrix . is continuous, then k is equal to

Limits Continuity and Differentiability PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q33 with Solution

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Correct answer: C

For continuity at $x=\dfrac{\pi}{4}$ , we need $k=\lim_{x\to \pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1}.$
Substitute $x=\dfrac{\pi}{4}$ to check the form: $\sqrt{2}\cos\frac{\pi}{4}-1=\sqrt{2}\cdot \frac{1}{\sqrt{2}}-1=1-1=0,$ $\cot\frac{\pi}{4}-1=1-1=0.$ So the limit is of the indeterminate form $\dfrac{0}{0}$ .
Simplify the denominator: $\cot x-1=\frac{\cos x}{\sin x}-1=\frac{\cos x-\sin x}{\sin x}.$ Hence

=\frac{\sqrt{2}\cos x-1}{(\cos x-\sin x)/\sin x} =\frac{(\sqrt{2}\cos x-1)\sin x}{\cos x-\sin x}.$$ 4. Now factor the numerator using $$\sqrt{2}\cos x=\cos x+\sin x \quad \text{at } x=\frac{\pi}{4},$$ and more usefully, $$\sqrt{2}\cos x-1=(\cos x+\sin x)-1.$$ So the limit becomes $$\lim_{x\to \pi/4}\frac{(\cos x+\sin x-1)\sin x}{\cos x-\sin x}.$$ This is still not the easiest form, so apply L'Hospital's Rule to the original expression. 5. Using L'Hospital's Rule: $$\lim_{x\to \pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1} =\lim_{x\to \pi/4}\frac{-\sqrt{2}\sin x}{-\csc^2 x}.

Thus

=\lim_{x\to \pi/4}\sqrt{2}\sin^3 x.$$ Now $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$, so $$k=\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)^3 =\sqrt{2}\cdot \frac{1}{2\sqrt{2}} =\frac{1}{2}.$$ 6. Therefore, $$k=\frac{1}{2}.$$ So the correct option is **C**.

View written solutionFree

Correct answer: C

For continuity at $x=\dfrac{\pi}{4}$ , we need $k=\lim_{x\to \pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1}.$
Substitute $x=\dfrac{\pi}{4}$ to check the form: $\sqrt{2}\cos\frac{\pi}{4}-1=\sqrt{2}\cdot \frac{1}{\sqrt{2}}-1=1-1=0,$ $\cot\frac{\pi}{4}-1=1-1=0.$ So the limit is of the indeterminate form $\dfrac{0}{0}$ .
Simplify the denominator: $\cot x-1=\frac{\cos x}{\sin x}-1=\frac{\cos x-\sin x}{\sin x}.$ Hence

=\frac{\sqrt{2}\cos x-1}{(\cos x-\sin x)/\sin x} =\frac{(\sqrt{2}\cos x-1)\sin x}{\cos x-\sin x}.$$ 4. Now factor the numerator using $$\sqrt{2}\cos x=\cos x+\sin x \quad \text{at } x=\frac{\pi}{4},$$ and more usefully, $$\sqrt{2}\cos x-1=(\cos x+\sin x)-1.$$ So the limit becomes $$\lim_{x\to \pi/4}\frac{(\cos x+\sin x-1)\sin x}{\cos x-\sin x}.$$ This is still not the easiest form, so apply L'Hospital's Rule to the original expression. 5. Using L'Hospital's Rule: $$\lim_{x\to \pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1} =\lim_{x\to \pi/4}\frac{-\sqrt{2}\sin x}{-\csc^2 x}.

Thus

=\lim_{x\to \pi/4}\sqrt{2}\sin^3 x.$$ Now $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$, so $$k=\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)^3 =\sqrt{2}\cdot \frac{1}{2\sqrt{2}} =\frac{1}{2}.$$ 6. Therefore, $$k=\frac{1}{2}.$$ So the correct option is **C**.