CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

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 Sept. 8, 2021 by

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)

CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.