If y = 4x – 5 is tangent to the curve y2 =px3 +q at (2, 3), then
If y = 4x – 5 is tangent to the curve y2 =px3 +q at (2, 3), then
- p = –2, q = 7
- p =2, q=–7
- p = 2, q = 7
- p =–2, q=–7
The Correct Option is B
Solution and Explanation
First, let's find the slope of the curve at the point (2, 3). We can do this by taking the derivative of the curve equation with respect to x:
2yy' = 3px2
Substituting the coordinates of the point (2, 3), we get:
2(3)y' = 3p(22)
6y' = 12p
y' = 2p
Since the line y = 4x - 5 is tangent to the curve, the slope of the line at the point (2, 3) must be equal to the slope of the curve at that point. Therefore, we have:
2p = 4
Simplifying, we find:
p = 2
Now, let's substitute the value of p into the curve equation to find q:
y2 = px3 + q
(3)2 = (2)(23) + q
9 = 16 + q
q = -7
Therefore, the values of p and q that satisfy the given conditions are p = 2 and q = -7.
The correct option is (B) p = 2, q = -7.
Top Questions on Application of derivatives
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:(i)} Express the distance \( y \) between the wall and foot of the ladder in terms of \( h \) and height \( x \) on the wall at a certain instant. Also, write an expression in terms of \( h \) and \( x \) for the area \( A \) of the right triangle, as seen from the side by an observer.
- CBSE CLASS XII - 2025
- Mathematics
- Application of derivatives
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?
- CBSE CLASS XII - 2025
- Mathematics
- Application of derivatives
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:(iii) (a) Show that the area \( A \) of the right triangle is maximum at the critical point.
- CBSE CLASS XII - 2025
- Mathematics
- Application of derivatives
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:(ii)} Find the derivative of the area \( A \) with respect to the height on the wall \( x \), and find its critical point.
- CBSE CLASS XII - 2025
- Mathematics
- Application of derivatives
- Determine the values of $x$ for which the function $f(x) = \frac{x-4}{x+1}$ is an increasing or a decreasing function.
- CBSE CLASS XII - 2025
- Mathematics
- Application of derivatives
Questions Asked in MHT CET exam
- Evaluate the integral: \[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]
- MHT CET - 2025
- Integration
- What is the volume of oxygen required for complete combustion of 0.25 mole of methane at S.T.P.?
- MHT CET - 2025
- Stoichiometry and Stoichiometric Calculations
- A particle is moving with a constant velocity of 5 m/s in a circular path of radius 2 m. What is the centripetal acceleration of the particle?
- MHT CET - 2025
- centripetal acceleration
- Population of Town A and B was 20,000 in 1985. In 1989, the population of Town A was 25,000, and Town B had 28,000. What will be the difference in population between the two towns in 1993?
- MHT CET - 2025
- Population Growth Calculation
- Given the equation: \[ 81 \sin^2 x + 81 \cos^2 x = 30 \] Find the value of \( x \).
- MHT CET - 2025
- Trigonometric Identities
Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
- Given a function y = f(x), the equation of the tangent for this curve at x = x0
- Slope of tangent (at x=x0) m=dy/dx||x=x0
- A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
- The normal to a given curve y = f(x) at a point x = x0
- The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
Diagram Explaining Tangents and Normal:
