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Right OPTION is (d) \(\frac{2T(tan⁡2t+tsec^2 2t)}{3 cos⁡3t}\)

To EXPLAIN I would say: GIVEN that, x=sin⁡3t and y=t^2 tan⁡2t

\(\frac{dx}{dt}\)=3 cos⁡3t

By using u.v rule, we get

\(\frac{DY}{dt}\)=\(\frac{d}{dx} \,(t^2) \,tan⁡2t+\frac{d}{dx} \,(tan⁡2t)\) t^2

\(\frac{dy}{dt}\)=2t tan⁡2t+2t^2 sec^2⁡2t

\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{(2t \,tan⁡2t+2t^2 \,sec^2⁡2t)}{3 \,cos⁡3t}\)

∴\(\frac{dy}{dx}=\frac{2t(tan⁡2t+tsec^2 \,2t)}{3 \,cos⁡3t}\)

The correct option is (B) 9t^4

The EXPLANATION is: Given that, x=2t^2 and y=6t^6

\(\frac{DX}{dt}\)=4t

\(\frac{DY}{dt}\)=36t^5

\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{36t^5}{4t}=9t^4\)

Correct choice is (c) \(\FRAC{2e^{x^4} (4x^4 log⁡x+1)}{x}\)

Easy explanation: CONSIDER y=2e^x^4 log⁡x

\(\frac{dy}{dx}\)=\(\frac{d}{dx}\) (2e^x^4 log⁡x)

Differentiating w.r.t x by using CHAIN rule, we GET

\(\frac{dy}{dx}\)=2(log⁡x \(\frac{d}{dx} (e^{x^4})+e^{x^4}\frac{d}{dx}\) (log⁡x))

\(\frac{dy}{dx}\)=\(2(log⁡x.e^{x^4}\frac{d}{dx} {x^4}+e^{x^4}.\frac{1}{x})\)

\(\frac{dy}{dx}\)=\(2(log⁡x.e^{x^4}.4x^3+e^{x^4}.\frac{1}{x})\)

\(\frac{dy}{dx}\)=\(2e^{x^4} (4x^3 log⁡x+\frac{1}{x})\)

∴\(\frac{dy}{dx}=\frac{(2e^{x^4} (4x^4 log⁡x+1))}{x}\)

Correct option is (b) -sin (x^2+5).2x

For explanation I would SAY: We differentiate the given function with help of CHAIN rule and hence the OUTER function becomes –sin and the inner function is DIFFERENTIATED into 2x, therefore the answer COMES out to be -sin (x^2+5).2x.

Right answer is (a) a.cos (ax + b)

The explanation is: We differentiate the given function with the help of CHAIN rule and hence

the OUTER function is differentiated into cos, and the INNER function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be

=> a.cos (ax + b).

The CORRECT OPTION is (B) \(\frac{tan⁡θ \,sin⁡θ}{2at}\)

To explain: Given that, x=a^2 t^2 cotθ and y=at sin⁡θ

\(\frac{dx}{dt}\)=2ta^2 cot⁡θ

\(\frac{DY}{dt}\)=asin⁡θ

\(\frac{dy}{dx}\)=\(\frac{asin⁡θ}{2ta^2 \,cot⁡θ}=\frac{a sin⁡θ}{2a^2 t cos⁡θ}.sin⁡θ=\frac{tan⁡θ \,sin⁡θ}{2at}\)

The correct ANSWER is (a) \(\FRAC{1}{36t^3}\)

The BEST I can explain: Given that, x=9t^4 and y=t

\(\frac{DX}{dt}\)=36t^3

\(\frac{dy}{dt}\)=1

∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=\frac{1}{36t^3}\)

The correct CHOICE is (d) \(\frac{7(4+3x^3)}{X}\)

The best explanation: CONSIDER y=7 log⁡(x^4.5e^x^3)

y=\(7(log⁡x^4 +log⁡5e^{x^3})\)

y=\(7(4 log⁡x+log⁡5e^{x^3})\)

\(\frac{dy}{dx}=7(4 \frac{d}{dx} (log⁡x)+\frac{d}{dx} (log⁡5e^{x^3}))\)

\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}} \frac{d}{dx} (5e^{x^3}))\)

\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}}.5e^{x^3}.\frac{d}{dx} {x^3})\)

\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}}.5e^{x^3}.3x^2)\)

\(\frac{dy}{dx}=7(\frac{4}{x}+3x^2)\)

∴\(\frac{dy}{dx}=\frac{7(4+3x^3)}{x}\)

The correct choice is (a) \(\frac{6E^2x}{\sqrt{1-E^{4X}}}\)

Explanation: Consider y=3 sin^-1⁡(e^2x)

\(\frac{dy}{dx}=\frac{d}{dx}\)(3 sin^-1⁡(e^2x))

\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\RIGHT )\frac{d}{dx}\)(e^2x)

\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\)2e^2x

∴\(\frac{dy}{dx}\)=\(\frac{6e^{2x}}{\sqrt{1-e^{4x}}}\)

The correct CHOICE is (a) False

The explanation: According to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that

i) f is continuous on [a,b]

ii) f is differentiable on (a,b) then there EXISTS a least POINT C ∈ (a,b) such that f’(c) = \(\FRAC {f(b)-f(a)}{b-a}\).

Correct choice is (C) 6-\(\frac{1}{X^2}\)

The BEST I can explain: GIVEN that, y=3x^2+log⁡(4x)

\(\frac{dy}{dx}=6x+\frac{1}{4x}.4=6x+\frac{1}{x}=\frac{6x^2+1}{x}\)

\(\frac{d^2 y}{dx^2}=\frac{\frac{d}{dx} (6x^2+1).(x)-\frac{d}{dx} (x).(6x^2+1)}{x^2} \Big(using\, \frac{d}{dx} (\frac{u}{v})=\frac{(\frac{d}{dx} (u).v-\frac{d}{dx} (v).u)}{v^2}\Big)\)

\(\frac{d^2 y}{dx^2}=\frac{(12x.x-6x^2-1)}{x^2} \)

\(\frac{d^2 y}{dx^2}=\frac{6x^2-1}{x^2} = 6-\frac{1}{x^2}\).

Right option is (c) –\(\frac{1}{2t}\)

Best explanation: GIVEN that, x=log⁡t^2 and y=\(\frac{1}{t}\)

\(\frac{DX}{DT}\)=\(\frac{1}{t^2}.2t=\frac{2}{t}\)

\(\frac{dy}{dt}\)=-\(\frac{1}{t^2}\)

∴\(\frac{dy}{dx}\)=\(\frac{dy}{dt}.\frac{dt}{dx}=-\frac{1}{t^2}.\frac{t}{2}=-\frac{1}{2t}\)

The CORRECT ANSWER is (a) True

Easy explanation: A FUNCTION is said to be continuous when it is both LEFT continuous and right continuous. Mathematical expression for a function f is left continuous on (a,b) is limx→a⁡+f(x)=f(a).

The correct answer is (a) 5e^x^2 (1+tan⁡x)^2

To explain I would SAY: Consider y=5e^x^2 tan⁡x

Differentiating w.r.t x by using chain RULE, we GET

\(\frac{dy}{dx}\)=tan⁡x \(\frac{d}{dx}\) (5e^x^2)+5e^x^2 \(\frac{d}{dx}\) (tan⁡x)

\(\frac{dy}{dx}\)=tan⁡x (5e^x^2.2x)+5e^x^2 (sec^2⁡x)

\(\frac{dy}{dx}\)=5e^x^2 (2x tan⁡x+sec^2⁡x)

\(\frac{dy}{dx}\)=5e^x^2 (1+tan^2⁡x+2x tan⁡x)

\(\frac{dy}{dx}\)=5e^x^2 (1+tan⁡x)^2

Right choice is (d) log\(_e^{e-1}\)

EASY EXPLANATION: Given f(x) = e^x, a = 0, B = 1

f’(c) = \(\frac {f(b)-f(a)}{b-a}\)

e^c = \(\frac {e-1}{1-0}\)

e^c = e – 1

C = log\(_e^{e-1}\)

The CORRECT choice is (b) f(a) = f(a+h)

Easy explanation: According to Rolle’s THEOREM, if f : [a,a+h] → R is a function such that

i) f is continuous on [a,a+h]

ii) f is differentiable on (a,a+h)

iii) f(a) = f(a+h) then there exists at LEAST ONE θ c ∈ (0,1) such that f’(a+θh) = 0

Correct choice is (a) \(\frac{-2}{(3x-1)\SQRT{(3x-1)(x+1)}}\)

The explanation: Consider y=\(\sqrt{\frac{x+1}{3x-1}}\)

Applying log to both SIDES, we get

log⁡y=log⁡\(\sqrt{\frac{x+1}{3x-1}}\)

log⁡y=\(\frac{1}{2} log⁡\left (\frac{x+1}{3x-1}\RIGHT )\)

log⁡y=\(\frac{1}{2}\) (log⁡(x+1)-log⁡(3x-1))

Differentiating with respect to x, we get

\(\frac{1}{y} \frac{DY}{DX}\)=\(\frac{1}{2}\left (\frac{d}{dx} (log⁡(x+1))-\frac{d}{dx} (log⁡(3x-1))\right )\)

\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{1}{x+1}-\frac{3}{3x-1}\right )\)

\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{3x-1-3x-3}{(x+1)(3x-1)})\right )\)

\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{-4}{(x+1)(3x-1)}\right )\)

\(\frac{dy}{dx}\)=\(\sqrt{\frac{x+1}{3x-1}} \left (\frac{-2}{(x+1)(3x-1)}\right )\)

\(\frac{dy}{dx}\)=\(\frac{-2}{(3x-1) \sqrt{(3x-1)(x+1)}}\)

Correct answer is (B) 27x^2 e^3x^3

To explain: CONSIDER y=3e^3x^3

\(\FRAC{dy}{dx}\)=\(\frac{d}{dx}\)(3e^3x^3)

\(\frac{dy}{dx}\)=3e^3x^3 \(\frac{d}{dx}\)(3x^3)

\(\frac{dy}{dx}\)=3e^3x^3 (3(3x^2))

\(\frac{dy}{dx}\)=27x^2 e^3x^3.

Right option is (B) limx→b-⁡F(X)=f(b)

The explanation is: A function is said to be continuous when it is both left continuous and right continuous. Mathematical expression for a function f is left continuous on (a,b) is limx→a+⁡f(x)=f(a).