Let x = 2 be a root of the equation x 2 + p x + q = 0 and f ( x ) = { 1 - cos ( x 2 - 4 p x + q 2 + 8 q + 16 ) ( x - 2 p ) 4 , x 2 p 0 , x = 2 p Then lim x 2 p + [ f ( x ) ] , where [ ? · ? ] denotes the greatest integer function, is

Question

$Let x = 2 be a root of the equation x^{2} + p x + q = 0 and$ $f (x) =$ ${\begin{matrix} \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{{(x - 2 p)}^{4}}, & x \neq 2 p \\ 0, & x = 2 p \end{matrix}$

Then $\lim_{x \to 2 p^{+}} [f (x)],$ where $[\cdot]$ denotes the greatest integer function, is [2023]

Smart Achievers · Accepted Answer

(4)

$Put x = 2 in x^{2} + p x + q = 0 \Rightarrow 4 + 2 p + q = 0 \Rightarrow - 2 p = (q + 4)$

Now, $\lim_{x \to 2 p^{+}} \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{{(x - 2 p)}^{4}}$

$= \lim_{x \to 2 p^{+}} \frac{1 - \cos [x^{2} - 2 x \cdot 2 p + q^{2} + 2 \cdot q \cdot 4 + 4^{2}]}{{(x - 2 p)}^{4}}$

$= \lim_{x \to 2 p^{+}} \frac{1 - \cos {(x - 2 p)}^{2}}{{(x - 2 p)}^{4}} = \lim_{x \to 2 p^{+}} \frac{2 \sin^{2} (\frac{{(x - 2 p)}^{2}}{2})}{\frac{{(x - 2 p)}^{4}}{4}} \cdot \frac{1}{4}$

$= \lim_{x \to 2 p^{+}} 2 \times 1 \times \frac{1}{4} = \frac{1}{2}$ $(∵ \lim_{θ \to 0} \frac{\sin θ}{θ} = 1)$

Now, $[1 / 2] = 0$

Solution

1 2

2 1

3 - 1

4 0

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Solution

Q. Let x=2 be a root of the equation x2+px+q=0 and f(x)={1-cos(x2-4px+q2+8q+16)(x-2p)4,x≠2p0,x=2p Then limx→2p+[f(x)], where [·] denotes the greatest integer function, is [2023]

1 2

2 1

3 - 1

4 0

Ans. (4) Put x=2 in x2+px+q=0 ⇒4+2p+q=0 ⇒-2p=(q+4) Now, limx→2p+1-cos(x2-4px+q2+8q+16)(x-2p)4 =limx→2p+1-cos[x2-2x·2p+q2+2·q·4+42](x-2p)4 =limx→2p+1-cos(x-2p)2(x-2p)4=limx→2p+2sin2((x-2p)22)(x-2p)44·14 =limx→2p+2×1×14=12 (∵limθ→0sinθθ=1) Now, [1/2]=0

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