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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 164 questions

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121 Easy
121 of 164
Application +4 | -1
The points on the curve y=12x−x3y=12 x-x^3 at which the gradient is zero are
View answer and explanation
Correct option: D
(2,16),(−2,−16)(2,16),(-2,-16)
Option C ((2,−16),(−2,16)(2,-16),(-2,16)) is correct because setting the derivative of the curve equal to zero and solving for x gives x=±\pm2, and substituting these values back into the original equation yields the corresponding y-values.
122 Easy
122 of 164
Application +4 | -1
Let slope of the tangent line to a curve at any point P (x, y)P\,(x,\,y) be given by xy2+yx\frac{{x{y^2} + y}}{x}. If the curve intersects the line x+2y=4x + 2y = 4 at x=−2x = - 2, then the value of yy, for which the point (3, y)(3,\,y) lies on the curve, is
View answer and explanation
Correct option: C
−1819- \frac{{18}}{{19}}
Option C (−1819- \frac{{18}}{{19}}) is correct because it is the value of y that satisfies the differential equation and the given condition.
123 Easy
123 of 164
Application +4 | -1
The subtangent, ordinate and subnormal to the parabola y2=4ax{y^2} = 4ax at a point (different from the origin) are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the ordinate squared equals the product of the subtangent and subnormal, indicating a geometric progression.
124 Easy
124 of 164
Application +4 | -1
If at each point on the curve y=x3−ax2+x+1y=x^3-a x^2+x+1 the tangent is inclined at an acute angle with the positive direction of xx-axis, then
View answer and explanation
Correct option: C
∣a∣≤3|a| \leq \sqrt{3}
Option C (∣a∣≤3|a| \leq \sqrt{3}) is correct because the condition for the tangent to be inclined at an acute angle with the positive x-axis leads to the discriminant of the quadratic being non-positive, resulting in this inequality.
125 Easy
125 of 164
Application +4 | -1
If the rate of change of sine of an angle θ\theta is kk, then the rate of change of its tangent is
View answer and explanation
Correct option: B
1k2\frac{1}{{{k^2}}}
Option B (1k2\frac{1}{{{k^2}}}) is correct because it is derived by finding the derivative of tan(θ\theta) with respect to θ\theta and using the given rate of change of sin(θ\theta).
126 Easy
126 of 164
Application +4 | -1
The lengths of tangent, subtangent, normal and subnormal for the curve y=x2+x−1y=x^2+x-1 at (1,1)(1,1) are A,B,CA, B, C and DD respectively, then their increasing order is
View answer and explanation
Correct option: D
B,A,D,CB, A, D, C
Option A (B,D,A,CB, D, A, C) is correct because, after calculating the lengths of the tangent, subtangent, normal, and subnormal, their increasing order is indeed B < A < D < C.
127 Easy
127 of 164
Application +4 | -1
The equation of the normal to the curve y=sin⁡xy=\sin x at (0,0)(0,0), is
View answer and explanation
Correct option: C
x+y=0x+y=0
Option C (x+y=0x+y=0) is correct because the slope of the normal at (0,0) is -1, and the line passes through the origin.
128 Easy
128 of 164
Application +4 | -1
Equation of normal to x = 2et,y=e−t at t=0x{\text{ = }}2{e^t},y = {e^{ - t}}{\text{ }}at{\text{ }}t = 0 is
View answer and explanation
Correct option: C
2x−y−3=02x - y - 3 = 0
Option C (2x−y−3=02x - y - 3 = 0) is correct because it is the equation of the normal line to the curve at the given point, calculated using derivatives.
129 Easy
129 of 164
Application +4 | -1
If the function f(x)=x2+α/xf(x)=x^2+\alpha / x has a local minimum at x=2x=2, then the value of α\alpha is
View answer and explanation
Correct option: C
16
Option C (16) is correct because the value of alpha is 16, which is found by setting the first derivative of the function equal to zero at x=2 and solving for alpha.
130 Easy
130 of 164
Application +4 | -1
The gradient of the curve passing through (4,0)(4,0) is given by dydx−yx+5x(x+2)(x−3)=0\frac{dy}{dx}-\frac{y}{x}+\frac{5x}{\left( x+2 \right)\left( x-3 \right)}=0, if the point (5,a)(5,a) lies on the curve, then the value of aa is
View answer and explanation
Correct option: C
5 log⁡7125\,\log \frac{7}{12}
The correct answer is D (5 log⁡1275\,\log \frac{12}{7}) because it is derived by solving the given differential equation and applying the provided conditions.
131 Easy
131 of 164
Analyzing +4 | -1
The equation of the tangent to the curve y=(2x−1)e2(1−x)y=(2 x-1) e^{2(1-x)} at the point of its maximum, is
View answer and explanation
Correct option: A
y−1=0y-1=0
Option A (y−1=0y-1=0) is correct because it represents the equation of the tangent to the curve at its maximum point.
132 Easy
132 of 164
Analyzing +4 | -1
The angle between the curves y2=4x+4{y^2} = 4x + 4 and y2=36(9−x){y^2} = 36\left( {9 - x} \right) is
View answer and explanation
Correct option: D
900{90^0}
Option D (900{90^0}) is correct because the product of the derivatives of the two curves at their intersection points is -1, indicating perpendicularity.
133 Easy
133 of 164
Application +4 | -1
If the curve y=ax2+bx+cy=a x^2+b x+c passes through the point (1,2)(1,2) such that the slope of the tangent at the origin is equal to 1, then (a,b,c)=(a, b, c)=
View answer and explanation
Correct option: B
(1,1,0)(1,1,0)
Option B ((1,1,0)) is correct because substituting the point (1,2) and the slope of the tangent at the origin into the curve equation yields these values for a, b, and c.
134 Easy
134 of 164
Analyzing +4 | -1
The angle between the curves y=axy = {a^x} and y=bxy = {b^x} is equal to
View answer and explanation
Correct option: D
tan⁡−1(∣log⁡a−log⁡b1+log⁡alog⁡b∣  )  {\tan ^{ - 1}}\left( {\left| {\frac{{\log a - \log b}}{{1 + \log a\log b}}} \right|\;} \right)\;
Option D (tan⁡−1(∣log⁡a−log⁡b1+log⁡alog⁡b∣;);{\tan ^{ - 1}}\left( {\left| {\frac{{\log a - \log b}}{{1 + \log a\log b}}} \right|\\;} \right)\\;) is correct because it represents the angle between the curves, calculated using the derivatives at the point of intersection.
135 Easy
135 of 164
Application +4 | -1
Let y be the number of people in a village at time tt. Assume that the rate of change of the population is proportional to the number of people in the village at any time and further assume that the population never increase in time. Then, the population of the village at any fixed tt is given by
View answer and explanation
Correct option: B
y=cekt,y = c{e^{kt}}, for some constants c≥0c \ge 0 and k≤0k \le 0
Option B (y=cekt,y = c{e^{kt}}, for some constants c≥0c \ge 0 and k≤0k \le 0) is correct because it represents exponential decay, which aligns with a population that never increases.
136 Easy
136 of 164
Application +4 | -1
The curves x=y2x=y^2 and xy=a3x y=a^3 cut orthogonally at a point, then a=a=
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (1/2) is correct because the value of 'a' for the curves to cut orthogonally is calculated to be 1/2.
137 Easy
137 of 164
Application +4 | -1
The curve y=ax3+bx2+cx+5y=a x^3+b x^2+c x+5 touches the x\mathrm{x}-axis at P(−2,0)\mathrm{P}(-2,0) and cut the y\mathrm{y}-axis at a point Q\mathrm{Q} where its gradient is 3 . Then, a+b+c=a+b+c=
View answer and explanation
Correct option: B
7/47 / 4
Option B (7/47 / 4) is correct because it is obtained by using the given conditions to find the coefficients of the cubic polynomial and then calculating a+b+ca+b+c.
138 Easy
138 of 164
Application +4 | -1
The equation of the tangent to the curve x=tcos⁡t,  y=tsin⁡tx = t\cos t,\;y = t\sin t at the origin is
View answer and explanation
Correct option: B
y=0y = 0
Option B (y=0y = 0) is correct because at the origin (t=0), the slope dydx\frac{{dy}}{{dx}} is 0, indicating a horizontal tangent.
139 Easy
139 of 164
Application +4 | -1
Point of contact of tangents to the curve y2−2x3−4y+8=0y^{2}-2 x^{3}-4 y+8=0 from the point (1,2)(1,2) is/are :
View answer and explanation
Correct option: A
(2,2±3)(2,2 \pm \sqrt{3})
Option A is correct because the points of contact are found by solving the tangent equation and the curve equation simultaneously.
140 Easy
140 of 164
Application +4 | -1
The number of points on the curve y=x3−2x2+x−2y = {x^3} - 2{x^2} + x - 2 where tangents are parallel to xx-axis, is
View answer and explanation
Correct option: C
2
The correct answer is 2 because the derivative of the given curve, when set to zero (indicating a tangent parallel to the x-axis), yields two distinct real roots.
141 Easy
141 of 164
Analyzing +4 | -1
The curve y=x15y=x^{\frac{1}{5}} at (0,0)(0,0) has
View answer and explanation
Correct option: B
a horizontal tangent (parallel to xx aria)
Option A (a vertical tangent (parallel to y-axis)) is correct because the derivative approaches infinity at (0,0), indicating a vertical tangent.
142 Easy
142 of 164
Application +4 | -1
The subtangent at any point on the curve xmyn=am+nx^m y^n=a^{m+n} varies as
View answer and explanation
Correct option: C
Abscissa
Option C (Abscissa) is correct because the subtangent varies directly with the abscissa (x-coordinate).
143 Easy
143 of 164
Application +4 | -1
If the tangent to the curve, y=x3+ax−b y = {x^3} + ax - b\, at the point (1, -5) is perpendicular to the line, −x+y+4=0, - x + y + 4 = 0,\, then which one of the following points lies on the curve?
View answer and explanation
Correct option: B
(2, -2)
Option B ((2, -2)) is correct because, after finding the values of a and b using the tangent condition and the point (1, -5), the point (2, -2) satisfies the curve equation.
144 Easy
144 of 164
Application +4 | -1
The point at which the tangent to the curve y=x2−4xy=x^2-4 x is parallel to xx-axis, is
View answer and explanation
Correct option: D
(2,−4)(2,-4)
Option D ((2,-4)) is correct because at this point the derivative of the curve is zero, indicating a horizontal tangent parallel to the x-axis.
145 Easy
145 of 164
Application +4 | -1
The slope of the normal to the curve y=2x2+3sin⁡xy=2 x^2+3 \sin x at x=0x=0 is
View answer and explanation
Correct option: D
−13-\frac{1}{3}
Option D (−13-\frac{1}{3}) is correct because it is the negative reciprocal of the slope of the tangent to the curve at x=0, which represents the slope of the normal.
146 Easy
146 of 164
Application +4 | -1
The equation of tangent to the curve x23−y22=1\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1, which is parallel to y=xy = x, is
View answer and explanation
Correct option: A
y=x±1y = x \pm 1
Option A (y=x±1y = x \pm 1) is correct because it represents the equations of the tangents to the hyperbola that are parallel to the line y=x.
147 Easy
147 of 164
Application +4 | -1
The slope of normal at any point (x,y),x>0,y>0(x,y),x > 0,y > 0 on the curve y=y(x)y = y(x) is given by x2xy−x2y2−1.\frac{{{x^2}}}{{xy - {x^2}{y^2} - 1}}. If the curve passes through the point (1, 1) then e.y(e)e.y(e) is equal to
View answer and explanation
Correct option: D
1+tan⁡(1)1−tan⁡(1)\frac{{1 + \tan (1)}}{{1 - \tan (1)}}
Option D (1+tan⁡(1)1−tan⁡(1)\frac{{1 + \tan (1)}}{{1 - \tan (1)}}) is correct because by solving the differential equation and applying the initial condition, we can find the value of e.y(e)e.y(e).
148 Easy
148 of 164
Application +4 | -1
The equation of the tangent to the curve (1+x2)y=2−x\left( {1 + {x^2}} \right)y = 2 - x, where it crosses the xx-axis, is
View answer and explanation
Correct option: A
x+5y=2x + 5y = 2
Option A (x+5y=2x + 5y = 2) is correct because it represents the equation of the tangent line to the curve at the point where it crosses the x-axis, calculated using derivatives.
149 Easy
149 of 164
Application +4 | -1
If y=4x−5y = 4x - 5 is a tangent to the curve y2=px3+q{y^2} = p{x^3} + q at (2, 3) then
View answer and explanation
Correct option: A
p=2,  q=−7p = 2,\;q = - 7
Option A (p=2,  q=−7p = 2,\;q = - 7) is correct because these values of p and q satisfy both the equation of the curve at the given point and the slope of the tangent at that point.
150 Easy
150 of 164
Application +4 | -1
The tangent to the curve y=xex2y=x e^{x^2} passing through the point (1,e)(1, e) also passes through the point
View answer and explanation
Correct option: A
(43,2e)\left(\frac{4}{3}, 2 e\right)
Option A (43,2e)\left(\frac{4}{3}, 2 e\right) is correct because it satisfies the equation of the tangent line to the curve at the point (1, e).
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