250+ TOP MCQs on Derivatives Application – Tangents and Normals | Class 12 Maths

Mathematics Multiple Choice Questions 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
Answer: a
Clarification: 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})
Answer: c
Clarification: 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} tanfrac{⁡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})
Answer: d
Clarification: 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}))
Answer: b
Clarification: 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
Answer: b
Clarification: 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})
Answer: c
Clarification: 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{cotpi}{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
Answer: a
Clarification: 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)
Answer: c
Clarification: Given that, y=(sqrt{4x^2+1})-2
(frac{dy}{dx})=(frac{1}{2sqrt{4x^2+1}} (16x)=frac{16x}{2sqrt{4x^2+1}})
Given that the slope is 2
∴2=(frac{16x}{2sqrt{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}{2sqrt{12} (sqrt{(1/3+1)})}=frac{16}{2sqrt{12} (2/sqrt{3})}=2)
When x=-(frac{1}{sqrt{12}})
y=(frac{-16}{2sqrt{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
Answer: c
Clarification: 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
Answer: a
Clarification: 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}).