JEE Main 2024 — Mathematics Calculus

Let, $f(x)=\mathrm{log}(\frac{1-{x}^{2}}{1+{x}^{2}})$f(x)=log(1+x21−x2​)

$$\Rightarrow f(x)=\mathrm{log}(1-{x}^{2})-\mathrm{log}(1+{x}^{2})$$⇒f(x)=log(1−x2)−log(1+x2)

$$\Rightarrow {f}^{'}(x)=\frac{-2x}{(1-{x}^{2})}-\frac{2x}{(1+{x}^{2})}$$⇒f′(x)=(1−x2)−2x​−(1+x2)2x​

$$\Rightarrow {f}^{'}(x)=(-2x)[\frac{1+{x}^{2}+1-{x}^{2}}{1-{x}^{4}}]$$⇒f′(x)=(−2x)[1−x41+x2+1−x2​]

$$\Rightarrow {f}^{'}(x)=(\frac{4x}{{x}^{4}-1})$$⇒f′(x)=(x4−14x​)

$$\Rightarrow {f}^{"}(x)=\frac{({x}^{4}-1)(4)-(4x)(4{x}^{3})}{{({x}^{4}-1)}^{2}}$$⇒f"(x)=(x4−1)2(x4−1)(4)−(4x)(4x3)​

$$\Rightarrow {f}^{"}(x)=\frac{4(-3{x}^{4}-1)}{{({x}^{4}-1)}^{2}}$$⇒f"(x)=(x4−1)24(−3x4−1)​

$$\Rightarrow 225[{f}^{'}(x)-{f}^{"}(x)]=225[(\frac{4x}{{x}^{4}-1})-\frac{4(-3{x}^{4}-1)}{{({x}^{4}-1)}^{2}}]$$⇒225[f′(x)−f"(x)]=225[(x4−14x​)−(x4−1)24(−3x4−1)​]

Putting, $x=\frac{1}{2}$x=21​

$$\Rightarrow 225[{f}^{'}(\frac{1}{2})-{f}^{"}(\frac{1}{2})]=225[(\frac{2}{-\frac{15}{16}})-\frac{4(-\frac{3}{16}-1)}{{(\frac{1}{16}-1)}^{2}}]$$⇒225[f′(21​)−f"(21​)]=225[(−1615​2​)−(161​−1)24(−163​−1)​]

$$\Rightarrow 225[{f}^{'}(\frac{1}{2})-{f}^{"}(\frac{1}{2})]=225[(\frac{-32}{15})-\frac{4(\frac{-19}{16})}{{(\frac{-15}{16})}^{2}}]$$⇒225[f′(21​)−f"(21​)]=225[(15−32​)−(16−15​)24(16−19​)​]

$$\Rightarrow 225[{f}^{'}(\frac{1}{2})-{f}^{"}(\frac{1}{2})]=225[(\frac{-32}{15})+\frac{4\times 19\times 16}{225}]$$⇒225[f′(21​)−f"(21​)]=225[(15−32​)+2254×19×16​]

$$\Rightarrow 225[{f}^{'}(\frac{1}{2})-{f}^{"}(\frac{1}{2})]=[-480+1216]=736$$⇒225[f′(21​)−f"(21​)]=[−480+1216]=736

$$\Rightarrow 225[{f}^{'}(\frac{1}{2})-{f}^{"}(\frac{1}{2})]=736$$⇒225[f′(21​)−f"(21​)]=736

$$\Rightarrow 225[{y}^{'}(\frac{1}{2})-{y}^{"}(\frac{1}{2})]=736$$⇒225[y′(21​)−y"(21​)]=736