Tangents & normals practice · Mixed difficulty

Option B (3) is correct because the slope of the tangent at the point of inflection is found by taking the second derivative, setting it to zero to find the x-coordinate of the inflection point, and then substituting that x-value into the first derivative.

Option A is correct because it represents the differential equation derived from the given geometric conditions involving tangents and perpendicular lines.

Option A (on the left of $x=c$) is correct because the x-coordinate of the intersection point is less than c, indicating it lies to the left.

Option C (1/2 sq. unit) is correct because the area bounded by the coordinate axes and the normal is calculated to be 1/2 square unit.

Option C $\left( {a,\;b} \right)$ is the correct answer because it satisfies both the line and curve equations at the point of tangency.

Option B ($\frac{6}{7}$) is correct because it is the slope of the tangent calculated at the given point by finding dy/dx using parametric differentiation.

Option A ($\left(\frac{{35}}{{9}}, \pm \frac{{16 \sqrt{{2}}}}{{27}}-1\right)$) is correct because it satisfies the condition where the tangent at one point is normal at another point on the curve.

Option C ($m \in R - [-1, 1]$) is correct because the slope of the normal satisfies the condition $|m| \ge 1$.

Option B (${e^{\left( {\frac{{1}}{{e - 1}}} \right)}}$) is correct because it is the value of $c$ that satisfies the condition of the tangent being parallel to the line segment.

Option B (12) is correct because it represents the maximum value of the slope function, found by taking its derivative and setting it to zero.

Option D ((-2t, -8t³)) is correct because it represents the other intersection point of the tangent line with the curve y=x³.

Option D ($\left( { - 1,\;5} \right),\;\left( {1,\; - 1} \right)$) is correct because by finding the equation of the normal and solving it simultaneously with the curve's equation, we find the intersection points.

Option D ($\frac{9}{2}$) is correct because the value of b is found by using the condition that the curves intersect at right angles, implying the product of their slopes at the point of intersection is -1.

Option B ($a > 0,\;b < 0$) is correct because for the line to be normal to the curve, the slope of the normal must be positive, leading to the specified conditions on a and b.

Option B (-$\frac{2}{3}$) is correct because finding the inverse function and its derivative, then evaluating at the y-intercept, gives the slope of the tangent as -2/3.

Option C ($\frac{2}{{\sqrt 5 }}$) is correct because it represents the perpendicular distance from the origin to the normal line of the given curve at x=0.

Option B ($x=e^{-1}$) is correct because the normal to the curve at its maximum point is a vertical line at x=1/e.

Option A $((-\infty, 1 / 2))$ is correct because finding the derivative of the curve and setting it less than 0 gives the interval where the gradient is negative.

Option B ((0, a)) is correct because the normal being parallel to the x-axis implies dx/dy = 0, leading to x=0, and substituting in the original equation gives y=a.

Option D (3) is correct because it is the absolute value of the tangent of the angle between the curves, calculated using their derivatives at the point of intersection.

The number of points is 4. The slope of the tangent is found, set equal to 90, and the resulting quartic equation has 4 real roots, indicating 4 points where the normal lines are parallel to the given line.

Option C (-1/2,0) is correct because it is the x-intercept of the tangent line to the curve at the point (0,1).

Option D ((-2,-7)) is correct because the point (-2, -7) does not satisfy the equation of the tangent to the curve at the given point.

Option C ($\frac{\pi}{2}$) is correct because the minimum distance between a point and a curve is always along the normal to the curve.

Option A ($\pm \tan^{-1} 2 \sqrt{2}$) is correct because it is the angle between the curves calculated using the derivatives at their intersection point.

Option A $\left(0,-2 t^3\right)$ is correct because it represents the coordinates of the point that divides the line segment PQ in the ratio 1:2 internally.

Option A ($-(\frac{{2+\log _{e} 8}}{{3+\log _{e} 4}})$) is correct because it is the result of differentiating the given equation implicitly and then substituting the given point (2,2).

Option C (${\tan ^{ - 1}}t$) is correct because the angle between the tangents is found using the slopes of the tangent to the parabola and the tangent to the circle at point P.

Option D ($2x=\pi$) is correct because the normal to the curve at the given point is a vertical line passing through x = pi/2.

Option C ($(9 / 4,3 / 8)$) is correct because the tangent line equation is found, and then solved simultaneously with the original curve to find the intersection point P.