Derivatives of Function in Parametric Form - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives of Functions in Parametric Forms”.

1. Find \(\frac{dy}{dx}\), if x=3a² cos²⁡θ and y=4a sin²⁡θ.
a) \(\frac{3}{4a}\)
b) –\(\frac{4}{3a}\)
c) \(\frac{4}{3a}\)
d) –\(\frac{3}{4a}\)
View Answer

Answer: b
Explanation: Given that, x=3a² cos²⁡θ and y=4a sin²⁡θ
Then, \(\frac{dx}{dθ}\)=3a².(2 cos⁡θ)(-sin⁡θ)=-3a² sin⁡2θ
\(\frac{dy}{dθ}\)=4a(2 sin⁡θ)(cos⁡θ)=4a sin⁡2θ
\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}×\frac{dθ}{dx}=-\frac{4a \,sin⁡2θ}{3a^2 \,sin⁡2θ}=-\frac{4}{3a}\)

2. Find \(\frac{dy}{dx}\), if x=9t⁴ and y=t.
a) \(\frac{1}{36t^3}\)
b) \(\frac{1}{36t^2}\)
c) \(\frac{-1}{36t^3}\)
d) \(\frac{1}{32t^3}\)
View Answer

Answer: a
Explanation: Given that, x=9t⁴ and y=t
\(\frac{dx}{dt}\)=36t³
\(\frac{dy}{dt}\)=1
∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{36t^3}\)

3. Find \(\frac{dy}{dx}\), if x=sin⁡3t and y=t² tan⁡2t.
a) \(\frac{3t(tan⁡2t+tsec^2 2t)}{4 cos⁡3t}\)
b) \(\frac{(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)
c) \(\frac{-2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)
d) \(\frac{2t(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)
View Answer

Answer: d
Explanation: Given that, x=sin⁡3t and y=t² tan⁡2t
\(\frac{dx}{dt}\)=3 cos⁡3t
By using u.v rule, we get
\(\frac{dy}{dt}\)=\(\frac{d}{dx} \,(t^2) \,tan⁡2t+\frac{d}{dx} \,(tan⁡2t)\) t²
\(\frac{dy}{dt}\)=2t tan⁡2t+2t² sec²⁡2t
\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{(2t \,tan⁡2t+2t^2 \,sec^2⁡2t)}{3 \,cos⁡3t}\)
∴\(\frac{dy}{dx}=\frac{2t(tan⁡2t+tsec^2 \,2t)}{3 \,cos⁡3t}\)

4. Find \(\frac{dy}{dx}\), if x=log⁡t² and y=\(\frac{1}{t}\).
a) \(\frac{1}{2t}\)
b) –\(\frac{t}{2}\)
c) –\(\frac{1}{2t}\)
d) \(\frac{t}{2}\)
View Answer

Answer: c
Explanation: Given that, x=log⁡t² and y=\(\frac{1}{t}\)
\(\frac{dx}{dt}\)=\(\frac{1}{t^2}.2t=\frac{2}{t}\)
\(\frac{dy}{dt}\)=-\(\frac{1}{t^2}\)
∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=-\frac{1}{t^2}.\frac{t}{2}=-\frac{1}{2t}\)

5. Find \(\frac{dy}{dx}\), if x=6 sin^-1⁡2t and y=\(\frac{1}{\sqrt{1-4t^2}}\).
a) \(\frac{t}{1-4t^2}\)
b) –\(\frac{1}{3(1-4t^2)}\)
c) –\(\frac{t}{3(1-4t^2)}\)
d) \(\frac{1}{3(1-4t^2)}\)
View Answer

Answer: d
Explanation: Given that, x=6 sin^-1⁡2t and y=\(\frac{1}{\sqrt{1-4t^2}}\)
\(\frac{dx}{dt}\)=\(\frac{6}{\sqrt{1-4t^2}}.2=\frac{12}{\sqrt{1-4t^2}}\)
\(\frac{dy}{dt}\)=-\(\frac{1}{2(1-4t^2)^{3/2}}.(-8t)=\frac{4t}{(1-4t^2)^{3/2}}\)
\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t}{(1-4t^2)^{3/2}}.\frac{\sqrt{1-4t^2}}{12}\)
\(\frac{dy}{dx}\)=\(\frac{t}{3(1-4t^2)}\)

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6. Find \(\frac{dy}{dx}\), if x=2t² and y=6t⁶.
a) -9t⁴
b) 9t⁴
c) t⁴
d) 9t³
View Answer

Answer: b
Explanation: Given that, x=2t² and y=6t⁶
\(\frac{dx}{dt}\)=4t
\(\frac{dy}{dt}\)=36t⁵
\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{36t^5}{4t}=9t^4\)

7. Find \(\frac{dy}{dx}\), if x=2e^t and y=log⁡t
a) \(\frac{1}{2te^t}\)
b) –\(\frac{1}{2te^t}\)
c) \(\frac{1}{te^t}\)
d) \(\frac{1}{e^t}\)
View Answer

Answer: a
Explanation: Given that, x=2e^t and y=log⁡t
\(\frac{dx}{dt}\)=2e^t
\(\frac{dy}{dt}\)=1/t
∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{t}.\frac{1}{2e^t}=\frac{1}{2te^t}\).

8. Find \(\frac{dy}{dx}\), if x=tan⁡2θ and y=cos⁡2θ+sin²⁡θ.
a) –\(\frac{tan^2⁡2θ \,sin⁡2θ}{2}\)
b) \(\frac{3 tan^2⁡2θ sin⁡2θ}{2}\)
c) 0
d) \(\frac{tan^2⁡2θ sin⁡2θ}{2}\)
View Answer

Answer: a
Explanation: Given that, x=tan⁡2θ and y=cos⁡2θ+sin²⁡θ
\(\frac{dx}{dθ}\)=2 sec²⁡2θ
\(\frac{dy}{dθ}\)=-2 sin⁡2θ+2 sin⁡θ cos⁡θ=-2 sin⁡2θ+sin⁡2θ=-sin⁡2θ
∴\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}.\frac{dθ}{dx}=-\frac{sin⁡2θ}{2 sec^2⁡2θ}=-\frac{sin⁡2θ}{2 cos^2⁡2θ}.sin^2⁡2θ=-\frac{tan^2⁡2θ sin⁡2θ}{2}\)

9. Find \(\frac{dy}{dx}\), if x=log⁡(tan⁡t) and y=log⁡(sin⁡t).
a) 2 cos²⁡t
b) cos²⁡2t
c) cos²t
d) -cos²t
View Answer

Answer: c
Explanation: Given that, x=log⁡(tan⁡t) and y=log⁡(sin⁡t)
\(\frac{dx}{dt}\)=\(\frac{1}{tan⁡t}.sec^2⁡t=cot⁡t sec^2⁡t\)
\(\frac{dy}{dt}=\frac{1}{sin⁡ \,t}.cos⁡ \,t=cot⁡ \,t\)
∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{cot⁡\,t}{cot\,⁡t sec^2⁡t}=\frac{1}{sec^2⁡t}=cos^2⁡t\).

10. Find \(\frac{dy}{dx}\), if x=a² t² cotθ and y=at sin⁡θ.
a) \(\frac{tan⁡θ \,sin⁡θ}{at}\)
b) \(\frac{tan⁡θ \,sin⁡θ}{2at}\)
c) \(\frac{tan⁡θ \,sin⁡θ}{2t}\)
d) \(\frac{tan⁡θ \,sin⁡θ}{2a}\)
View Answer

Answer: b
Explanation: Given that, x=a² t² cotθ and y=at sin⁡θ
\(\frac{dx}{dt}\)=2ta² cot⁡θ
\(\frac{dy}{dt}\)=asin⁡θ
\(\frac{dy}{dx}\)=\(\frac{asin⁡θ}{2ta^2 \,cot⁡θ}=\frac{a sin⁡θ}{2a^2 t cos⁡θ}.sin⁡θ=\frac{tan⁡θ \,sin⁡θ}{2at}\)

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