JEE Main 2023 — Mathematics Calculus

Given,

$x=2$x=2 be the root of the given equation ${x}^{2}+px+q=0$x2+px+q=0,

Putting $x=2$x=2 in given equation we get,

$$\Rightarrow 4+2p+q=0\Rightarrow q+4=-2p$$⇒4+2p+q=0⇒q+4=−2p

$$\because {x}^{2}-4px+{q}^{2}+8q+16$$∵x2−4px+q2+8q+16

$$={x}^{2}-4px+{(q+4)}^{2}$$=x2−4px+(q+4)2

$$={x}^{2}-4px+4{p}^{2}(\because q+4=-2p)$$=x2−4px+4p2(∵q+4=−2p)

$$=(x-2p{)}^{2}$$=(x−2p)2

Now, solving the limit $\underset{x\rightarrow 2{p}^{+}}{\mathrm{lim}}[f(x)]=\underset{x\rightarrow 2{p}^{+}}{\mathrm{lim}}[\frac{1-\mathrm{cos}({(x-2p)}^{2})}{{(x-2p)}^{4}}]$x→2p+lim​[f(x)]=x→2p+lim​[(x−2p)41−cos((x−2p)2)​]

Let $x-2p=\theta$x−2p=θ

$$\Rightarrow \underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}[f(x)]=\underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}[\frac{1-\mathrm{cos}({\theta }^{2})}{{(\theta )}^{4}}]$$⇒θ→0+lim​[f(x)]=θ→0+lim​[(θ)41−cos(θ2)​]

$\because \underset{\theta \rightarrow 0}{\mathrm{lim}}(\frac{1-\mathrm{cos}({\theta }^{2})}{{(\theta )}^{4}})=\frac{1}{2}$∵θ→0lim​((θ)41−cos(θ2)​)=21​ (Using L'Hospital's)

$$\therefore \underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}[f(x)]=\underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}[\frac{1}{2}]$$∴θ→0+lim​[f(x)]=θ→0+lim​[21​]

$$\therefore \underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}[f(x)]=0$$∴θ→0+lim​[f(x)]=0