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If the tangent to the curve, y = x3 + ax - b at the point (1, -5) is perpendicular to the line, - x + y + 4 = 0 , then which one of the following points lies on the curve?
- a)( 2, -1)
- b)( -2, 2)
- c)( 2, -2)
- d)( -2,1)
Correct answer is option 'C'. Can you explain this answer?
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If the tangent to the curve, y = x3 + ax - b at the point (1, -5) is p...
f (1) = -5 ⇒ 1 + a - b = -5 ⇒ a - b = -6 ......(1)
f1(x) = 3x2 +a f1(1) = 3+a = -1⇒ a = -4
b = a + 6 = 2
f ( x ) = x3 - 4 x - 2 check options (( 2, -2) satisfies )
f1(x) = 3x2 +a f1(1) = 3+a = -1⇒ a = -4
b = a + 6 = 2
f ( x ) = x3 - 4 x - 2 check options (( 2, -2) satisfies )
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If the tangent to the curve, y = x3 + ax - b at the point (1, -5) is p...
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If the tangent to the curve, y = x3 + ax - b at the point (1, -5) is p...
To find the point on the curve, we need to determine the values of a and b first.
Given curve: y = x^3 - ax - b
Step 1: Find the derivative of the curve
The derivative of y with respect to x gives us the slope of the curve at any point.
dy/dx = 3x^2 - a
Step 2: Find the slope of the tangent at point (1, -5)
To find the slope of the tangent at point (1, -5), substitute x = 1 into the derivative.
dy/dx = 3(1)^2 - a
dy/dx = 3 - a
Step 3: Find the slope of the given line
The given line is -x + y + 4 = 0. We can rewrite it in slope-intercept form y = mx + c, where m is the slope.
-x + y + 4 = 0
y = x + 4
Comparing with y = mx + c, we get m = 1.
Step 4: Determine the condition for the slopes to be perpendicular
For two lines to be perpendicular, the product of their slopes should be -1.
(m1)(m2) = -1
(3 - a)(1) = -1
3 - a = -1
a = 4
Step 5: Substitute the value of a into the equation of the curve
y = x^3 - ax - b
y = x^3 - 4x - b
Step 6: Substitute the coordinates of the given point into the equation
-5 = (1)^3 - 4(1) - b
-5 = 1 - 4 - b
-5 = -3 - b
b = 2
Step 7: Substitute the values of a and b back into the equation of the curve
y = x^3 - 4x - 2
Step 8: Check which point lies on the curve
Substitute the x and y values of each option into the equation of the curve to check which point lies on the curve.
a) (2, -1):
-1 = (2)^3 - 4(2) - 2
-1 = 8 - 8 - 2
-1 = -2 (not true)
b) (-2, 2):
2 = (-2)^3 - 4(-2) - 2
2 = -8 + 8 - 2
2 = -2 (not true)
c) (2, -2):
-2 = (2)^3 - 4(2) - 2
-2 = 8 - 8 - 2
-2 = -2 (true)
d) (-2, 1):
1 = (-2)^3 - 4(-2) - 2
1 = -8 + 8 - 2
1 = -2 (not true)
Therefore, the point (2, -2) lies on the curve y = x^3 - 4x - 2.
Given curve: y = x^3 - ax - b
Step 1: Find the derivative of the curve
The derivative of y with respect to x gives us the slope of the curve at any point.
dy/dx = 3x^2 - a
Step 2: Find the slope of the tangent at point (1, -5)
To find the slope of the tangent at point (1, -5), substitute x = 1 into the derivative.
dy/dx = 3(1)^2 - a
dy/dx = 3 - a
Step 3: Find the slope of the given line
The given line is -x + y + 4 = 0. We can rewrite it in slope-intercept form y = mx + c, where m is the slope.
-x + y + 4 = 0
y = x + 4
Comparing with y = mx + c, we get m = 1.
Step 4: Determine the condition for the slopes to be perpendicular
For two lines to be perpendicular, the product of their slopes should be -1.
(m1)(m2) = -1
(3 - a)(1) = -1
3 - a = -1
a = 4
Step 5: Substitute the value of a into the equation of the curve
y = x^3 - ax - b
y = x^3 - 4x - b
Step 6: Substitute the coordinates of the given point into the equation
-5 = (1)^3 - 4(1) - b
-5 = 1 - 4 - b
-5 = -3 - b
b = 2
Step 7: Substitute the values of a and b back into the equation of the curve
y = x^3 - 4x - 2
Step 8: Check which point lies on the curve
Substitute the x and y values of each option into the equation of the curve to check which point lies on the curve.
a) (2, -1):
-1 = (2)^3 - 4(2) - 2
-1 = 8 - 8 - 2
-1 = -2 (not true)
b) (-2, 2):
2 = (-2)^3 - 4(-2) - 2
2 = -8 + 8 - 2
2 = -2 (not true)
c) (2, -2):
-2 = (2)^3 - 4(2) - 2
-2 = 8 - 8 - 2
-2 = -2 (true)
d) (-2, 1):
1 = (-2)^3 - 4(-2) - 2
1 = -8 + 8 - 2
1 = -2 (not true)
Therefore, the point (2, -2) lies on the curve y = x^3 - 4x - 2.
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If the tangent to the curve, y = x3 + ax - b at the point (1, -5) is perpendicular to the line, - x + y + 4 = 0 , then which one of the following points lies on the curve?a)( 2, -1)b)( -2, 2)c)( 2, -2)d)( -2,1)Correct answer is option 'C'. Can you explain this answer?
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