KCET · Maths · Basic of Mathematics
If \( y=\left(\tan ^{-1} x\right)^{2} \), then \( \left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1} \) is equal to
- A \( 00 \)
- B \( 11 \)
- C \( 04 \)
- D \( 12 \)
Answer & Solution
Correct Answer
(D) \( 12 \)
Step-by-step Solution
Detailed explanation
Given that, \( y=\left(\tan ^{-1} x\right)^{2} \)
So, \( y_{1}=\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}} \)
\( \Rightarrow \frac{d y}{d x}\left(1+x^{2}\right)=2 \tan ^{-1} x \)
\( \Rightarrow\left(1+x^{2}\right) y_{1}=2 \tan ^{-1} x \)
Now, \( \left(1+x^{2}\right) y_{2}+y_{1}(2 x)=\frac{2}{1+x^{2}} \)
\( \Rightarrow(1+x)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2 \)
Since \( y_{2}=\frac{d^{2} y}{d x^{2}} \)
So, \( y_{1}=\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}} \)
\( \Rightarrow \frac{d y}{d x}\left(1+x^{2}\right)=2 \tan ^{-1} x \)
\( \Rightarrow\left(1+x^{2}\right) y_{1}=2 \tan ^{-1} x \)
Now, \( \left(1+x^{2}\right) y_{2}+y_{1}(2 x)=\frac{2}{1+x^{2}} \)
\( \Rightarrow(1+x)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2 \)
Since \( y_{2}=\frac{d^{2} y}{d x^{2}} \)
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