Tangents and Normals - Class 12 Maths MCQ

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

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

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

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
\(-\frac{1}{slope\, of \,the \,tangent \,at \,x=5}=-\frac{1}{26}\).

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