Mathematics Questions and Answers – Derivatives Application – Tangents and Normals

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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
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.
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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(\(\frac{49}{144})-\frac{49}{12}=49(-\frac{1}{24})=-\frac{49}{24}\)
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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
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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
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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
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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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter