Class 12 Maths MCQ – Derivatives Application – Tangents and Normals

This set of Class 12 Maths Chapter 6 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
View Answer

Answer: a
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) –\(\frac{5}{4}\)
View Answer

Answer: c
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) –\(\frac{49}{24}\)
View Answer

Answer: d
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(\(\frac{49}{144})-\frac{49}{12}=49(-\frac{1}{24})=-\frac{49}{24}\)
advertisement

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 \((\frac{1}{2},\frac{15}{16})\)
View Answer

Answer: b
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
View Answer

Answer: b
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
Free 30-Day Java Certification Bootcamp is Live. Join Now!

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) –\(\frac{1}{6}\)
View Answer

Answer: c
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
View Answer

Answer: a
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)
View Answer

Answer: c
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
View Answer

Answer: c
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
advertisement

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
View Answer

Answer: a
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 chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). 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!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
I’m Manish - Founder and CTO at Sanfoundry. I’ve been working in tech for over 25 years, with deep focus on Linux kernel, SAN technologies, Advanced C, Full Stack and Scalable website designs.

You can connect with me on LinkedIn, watch my Youtube Masterclasses, or join my Telegram tech discussions.

If you’re in your 40s–60s and exploring new directions in your career, I also offer mentoring. Learn more here.