Method of DifferentiationeasyFree

Method of Differentiation: Values

JEE Maths question with a full step-by-step solution.

Question
If y=1+x2+x41+x+x2y=\dfrac{1+x^{2}+x^{4}}{1+x+x^{2}} and dydx=ax+b\dfrac{dy}{dx}=ax+b, then the values of aa and bb are
Aa=2,  b=1a=2,\;b=1
Ba=−2,  b=1a=-2,\;b=1
Ca=2,  b=−1a=2,\;b=-1correct
Da=−2,  b=−1a=-2,\;b=-1
Solution
Step 1: Factor the numerator using the identity 1+x2+x4=(1+x+x2)(1−x+x2)1+x^{2}+x^{4}=(1+x+x^{2})(1-x+x^{2}). Verification: (1+x+x2)(1−x+x2)=1−x+x2+x−x2+x3+x2−x3+x4=1+x2+x4(1+x+x^{2})(1-x+x^{2})=1-x+x^{2}+x-x^{2}+x^{3}+x^{2}-x^{3}+x^{4}=1+x^{2}+x^{4}. Confirmed. Step 2: Simplify yy:
y=(1+x+x2)(1−x+x2)1+x+x2=1−x+x2y=\dfrac{(1+x+x^{2})(1-x+x^{2})}{1+x+x^{2}}=1-x+x^{2}
 Step 3: Differentiate:
dydx=2x−1=ax+b\dfrac{dy}{dx}=2x-1=ax+b
 Step 4: Comparing coefficients: a=2a=2 and b=−1b=-1. Correct answer: (3)
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