The equation of tangent to the curve y (1 + x 2 ) = 2 – x, where it crosses x-axis is:
(A) x + 5y = 2 (B) x – 5y = 2
(C) 5x – y = 2 (D) 5x + y = 2




Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
NCERT Exemplar - MCQs
Question 2 Important
Question 3 Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 9 Important
Question 10 You are here
Question 11 Important
Question 12 Important
Question 13
Question 14
Question 15 Important
Question 16 Important
Question 17 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams You are here
Question 11 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Question 10 The equation of tangent to the curve y (1 + x2) = 2 β x, where it crosses x-axis is: (A) x + 5y = 2 (B) x β 5y = 2 (C) 5x β y = 2 (D) 5x + y = 2 First, let us find the point at which curve cuts π₯βaxis Since π=π on π₯βaxis β΄ Point =(π,π) For πβ coordinate Putting π=π in π¦(1+π₯^2 )=2βπ₯ 0(1+π₯^2 )=2βπ₯ 0=2βπ₯ π=π β΄ Point =(π,π) Now, to find equation of tangent we find slope of tangent at (π, π) Finding Slope of tangent Given curve π¦(1+π₯^2 )=2βπ₯ Differentiating w.r.t x ππ¦/ππ₯ (1+π₯^2 )+π¦ Γ (0+2π₯)=0β1 ππ¦/ππ₯ (1+π₯^2 )+π¦(2π₯)=β1 ππ¦/ππ₯ (1+π₯^2 )+2π₯π¦=β1 ππ¦/ππ₯ (1+π₯^2 )=β1β2π₯π¦ π π/π π=(βπ β πππ" " )/((π + π^π ) )Since we need Slope at (2, 0) Putting x = 2, y = 0 in ππ¦/ππ₯ π π/π π =(β1 β (2 Γ 2 Γ 0))/(1 + 2^2 ) =(β1 β 0)/(1 + 4) =(βπ)/π Finding equation of tangent Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) β΄ Equation of tangent at (2, 0) & Slope (β1)/5 is (πβπ)=(βπ)/π (πβπ) π¦=(βπ)/π (π₯β2) 5π¦=β1(π₯β2) 5π¦=βπ₯+2 5π¦+π₯=2 π+ππ=π Hence, equation of tangent is π+ππ=π So, the correct answer is (A)