Question 39 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

The point(s) on the curve y = x
^{
3
}
– 11x + 5 at which the tangent isy = x – 11 is/are:

Question 39 The point(s) on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11 is/are: (a) (−2, 19) (b) (2, −9) (c) (±2, 19) (d) (−2, 19) and (2, −9)
Equation of Curve is
𝑦=𝑥^3−11𝑥+5
We know that
Slope of tangent is 𝑑𝑦/𝑑𝑥
𝑑𝑦/𝑑𝑥=𝑑(𝑥^3 − 11𝑥 + 5)/𝑑𝑥
𝒅𝒚/𝒅𝒙=〖𝟑𝒙〗^𝟐−𝟏𝟏
Also,
Given tangent is 𝑦=𝑥−12
Comparing with 𝑦=𝑚𝑥+𝑐 , when m is the Slope
Slope of tangent =𝟏
From (1) and (2)
𝒅𝒚/𝒅𝒙=𝟏
3𝑥^2−11=1
3𝑥^2=1+11
3𝑥^2=12
𝑥^2=12/3
𝑥^2=4
𝒙=±𝟐
When 𝒙=𝟐
𝑦=(2)^3−11(2)+5
𝑦=8−22+5
𝑦=− 9
So, Point is (2, −9)
When 𝒙=−𝟐
𝑦=(−2)^3−11(−2)+5
𝑦=− 8+22+5
𝑦=19
So, Point is (−2, 19)
But (–2, 19) does not satisfy line y = x – 11
As 19 ≠ –2 – 11
∴ Only point is (2, –9)
So, the correct answer is (B)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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