# Mathematics Questions and Answers – Derivatives Application – Tangents and Normals

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives Application – Tangents and Normals”.

1. Find the tangent to the curve y=3x2+x+4 at x=3.
a) 19
b) 1.9
c) 18
d) 16

Explanation: The slope of the tangent at x=3 is given by
$$\frac{dy}{dx}$$]x=3= 6x+1]x=3=18+1=19.

2. Find the slope of the tangent to the curve x=4 cos3⁡3θ and y=5 sin3⁡⁡3θ at θ=π/4.
a) –$$\frac{3}{4}$$
b) –$$\frac{1}{4}$$
c) $$\frac{5}{4}$$
d) –$$\frac{5}{4}$$

Explanation: Given that, x=4 cos3⁡3θ and y=5 sin3⁡3θ
$$\frac{dx}{dθ}$$=4(3)(3 cos2⁡3θ)(-sin⁡3θ)
$$\frac{dy}{dθ}$$=5(3)(3 sin2⁡3θ)(cos⁡3θ)
$$\frac{dy}{dx}$$=$$\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{5(3)(3 sin^2⁡3θ)(cos⁡3θ)}{4(3)(3 cos^23θ)(-sin⁡3θ)}$$
$$\frac{dy}{dx}$$=-$$\frac{5 tan⁡3θ}{4}$$
$$\frac{dy}{dx}$$]θ=π/4=-$$\frac{5}{4} tan\frac{⁡3 \pi}{4}$$=-$$\frac{5}{4} (-1)=\frac{5}{4}$$.

3. Find the equation of all the lines having slope 0 which are tangent to the curve y=6x2-7x.
a) $$\frac{24}{49}$$
b) –$$\frac{24}{49}$$
c) $$\frac{49}{24}$$
d) –$$\frac{49}{24}$$

Explanation: Given that, y=6x2-7x
$$\frac{dy}{dx}$$=12x-7
It is given that the slope of tangent is 0
∴12x-7=0
⇒x=$$\frac{7}{12}$$
∴ y]x=$$\frac{7}{12}$$=6($$\frac{7}{12}$$)2-7($$\frac{7}{12}$$)
6($$\frac{49}{144})-\frac{49}{12}=49(-\frac{1}{24})=-\frac{49}{24}$$

4. Find the points on the curve y=3x4+2x3-1 at which the tangents is parallel to the x-axis.
a) (0,1) and $$(-\frac{1}{2},-\frac{15}{16})$$
b) (0,-1) and $$(-\frac{1}{2},-\frac{15}{16})$$
c) (0,-1) and $$(\frac{1}{2},-\frac{15}{16})$$
d) (0,1) and $$(\frac{1}{2},\frac{15}{16})$$

Explanation: Given that, y=3x4+2x3-1
Differentiating w.r.t x, we get
$$\frac{dy}{dx}$$=12x3+6x2
The tangent is parallel to x-axis, which implies that the slope $$\frac{dy}{dx}$$ is 0.
∴12x3+6x2=0
6x2 (2x+1)=0
⇒x=0,-$$\frac{1}{2}$$
If x=0
y=3(0)+2(0)-1=-1
If x=-$$\frac{1}{2}$$
y=3$$(-\frac{1}{2})^4+2(-\frac{1}{2})^3-1$$
y=$$\frac{3}{16}-\frac{2}{8}-1$$
y=-$$\frac{15}{16}$$
Hence, the points at which the tangent is parallel to the x-axis is (0,-1) and $$(-\frac{1}{2},-\frac{15}{16})$$.

5. Find the tangent to the curve y=7x3-2x2 at the point x=2.
a) 67
b) 76
c) 46
d) 64

Explanation: The slope of the tangent at x=2 is given by
$$\frac{dy}{dx}$$]x=2 = 21x2-4x]x=2=21(2)2-4(2)=84-8=76

6. Find the equation of the normal to the curve x=12 cosec⁡θ and y=2 sec⁡θ at x=π/4 .
a) $$\frac{1}{6}$$
b) -6
c) 6
d) –$$\frac{1}{6}$$

Explanation: Given that, x=12 cosecθ and y=2 sec⁡θ
$$\frac{dx}{dθ}$$=-12 cosec θ cot⁡θ
$$\frac{dy}{dθ}$$=2 tan⁡θ sec⁡θ
$$\frac{dy}{dx}$$=$$\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{2 \,tan⁡θ \,sec⁡θ}{-12 \,cosec θ \,cot⁡θ}$$=-$$\frac{1}{6} \frac{sin⁡θ}{cos^2⁡θ} × \frac{cos⁡θ}{sin^2⁡⁡θ}$$ = –$$\frac{cot⁡θ}{6}$$
$$\frac{dy}{dx}]_{x=\frac{\pi}{4}}$$=$$-\frac{\frac{cot\pi}{4}}{6}=-\frac{1}{6}$$
Hence, the slope of normal at θ=π/4 is
–$$\frac{1}{slope \,of \,tangent \,at \,θ=\pi/4}=-\frac{-1}{\frac{1}{6}}$$=6

7. Find the equation of the tangent of the tangent to the curve 2x2+3y2=3 at the point(3,4).
a) x+2y=11
b) x-2y=11
c) -x+2y=11
d) x-2y=-11

Explanation: Differentiating 2x2+3y2=3 w.r.t x, we get
4x+6y $$\frac{dy}{dx}$$=0
$$\frac{dy}{dx} = -\frac{2x}{3y}$$
$$\frac{dy}{dx}$$](3,4)=-$$\frac{2(3)}{3(4)}=-\frac{1}{2}$$
Therefore, the equation of the tangent at (3,4) is
y-y0=m(x-x0)
y-4=-$$\frac{1}{2}$$ (x-3)
2(y-4)=-x+3
2y-8=-x+3
x+2y=11

8. Find the point at which the tangent to the curve y=$$\sqrt{4x^2+1}$$-2 has its slope2.
a) ($$\frac{1}{\sqrt{12}}$$,-1) and ($$\frac{1}{\sqrt{12}}$$,-1)
b) (-$$\frac{1}{\sqrt{12}}$$,3) and (-$$\frac{1}{\sqrt{12}}$$,-1)
c) ($$\frac{1}{\sqrt{12}}$$,2) and (-$$\frac{1}{\sqrt{12}}$$,-2)
d) ($$\frac{1}{\sqrt{12}}$$,3) and ($$\frac{1}{\sqrt{12}}$$,-2)

Explanation: Given that, y=$$\sqrt{4x^2+1}$$-2
$$\frac{dy}{dx}$$=$$\frac{1}{2\sqrt{4x^2+1}} (16x)=\frac{16x}{2\sqrt{4x^2+1}}$$
Given that the slope is 2
∴2=$$\frac{16x}{2\sqrt{4x^2+1}}$$
$$\sqrt{4x^2+1}$$=4x
4x2+1=16x2
12x2=1
x=±$$\frac{1}{\sqrt{12}}$$
When x=+$$\frac{1}{\sqrt{12}}$$
y=$$\frac{16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{16}{2\sqrt{12} (2/\sqrt{3})}=2$$
When x=-$$\frac{1}{\sqrt{12}}$$ y=$$\frac{-16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{-16}{2√12 (2/√3))}=-2$$
Hence, the point on the curve is ($$\frac{1}{\sqrt{12}}$$,2) and (-$$\frac{1}{\sqrt{12}}$$,-2)

9. Find the tangent to the curve y=5x4-3x2+2x-1 at x=1.
a) 15
b) 14
c) 16
d) 17

Explanation: The slope of the tangent at x=1 is given by
$$\frac{dy}{dx}$$]x=1= 20x3-6x+2]x=1=20(1)3-6(1)+2=20-6+2=16

10. Find the slope of the normal to the curve y=4x2-14x+5 at x=5.
a) –$$\frac{1}{26}$$
b) $$\frac{1}{26}$$
c) 26
d) -26

Explanation: The slope of the tangent at x=5 is given by:
$$\frac{dy}{dx}$$=8x-14
$$\frac{dy}{dx}$$]x=5=8(5)-14=40-14=26
∴ The slope of the normal to the curve is
$$-\frac{1}{slope\, of \,the \,tangent \,at \,x=5}=-\frac{1}{26}$$.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! 