Exams › JEE Main › Maths
For the function y = m log x + n x² + x, if the extreme points occur at x = 1 and x = 2, then the value of 2m + 10n is:
- 1
- −4
- −2
- −3
Correct answer: −3
Solution
y'=m/x+2nx+1. Setting y'=0 at x=1 and x=2 gives m+2n+1=0 and m/2+4n+1=0, solving m=-2/3, n=-1/6. Then 2m+10n = -4/3 - 5/3 = -3.
Related JEE Main Maths questions
- For the curve defined by x² + y² = a², if k = 1/a, then k can be written as
- For the curve y² = 2x³, identify the point P at which the tangent is perpendicular to the line 4x - 3y + 2 = 0.
- For the function f(x) = 2x² - log|x|, with x ≠ 0, on which interval is the function increasing throughout?
- Two upright poles AP and BQ stand at points A and B respectively. If AP = 16 m, BQ = 22 m, and the distance AB = 20 m, then for a point R on AB, the value of AR for which RP² + RQ² is least is
- For the curve x⁴ + y⁴ = a⁴, a tangent drawn at an arbitrary point meets the coordinate axes at intercepts p and q. Then the quantity p^(-4/3) + q^(-4/3) equals
- A particle travels on the curve y = x³ + 2. Find the point(s) P on the curve where the y-coordinate changes at a rate 8 times the rate of change of the x-coordinate. The points are (4, 11) and (-4, -31/3).
⚔️ Practice JEE Main Maths free + battle 1v1 →