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The correct option is (b) -sin (sinx).cosx

The best I can EXPLAIN: We differentiate the given function with the help of chain rule so we FIRST differentiate the OUTER function which BECOMES –sin and then we differentiate the inner function sinx which is differentiated and comes out to be cosx, HENCE the differentiated function comes out to be -sin (sinx).cosx.

The correct answer is (c) 2 sec^2⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)

EXPLANATION: Given that, y=tan^2⁡⁡x+3 tan⁡x

\(\FRAC{DY}{DX}\)=2 tan⁡x sec^2⁡⁡x+3 sec^2⁡x=sec^2⁡⁡x (2 tan⁡x+3)

By USING the u.v rule, we get

\(\frac{d^2 y}{dx^2}=\frac{d}{dx}\) (sec^2⁡⁡x).(2 tan⁡x+3)+\(\frac{d}{dx}\) (2 tan⁡x+3).sec^2⁡⁡x

\(\frac{d^2 y}{dx^2}\)=2 sec^2⁡⁡x tan⁡x (2 tan⁡x+3)+sec^2⁡⁡x (2 sec⁡x tanx)

=2 sec^2⁡x tan⁡x (2 tan⁡x+sec⁡x+3).

Correct option is (a) Equals to

Easy explanation: According to ROLLE’s 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)

iii) f(a) = f(b) then there exists at least one point C ∈ (a,b) such that f’(c) = 0

Correct OPTION is (d) (3 cos⁡X)^x (LOG⁡(3 cos⁡x)-x tan⁡x)

The best I can explain: Consider y=(3 cos⁡x)^x

Applying log on both sides, we get

log⁡y=log⁡(3 cos⁡x)^x

log⁡y=x log⁡(3 cos⁡x)

log⁡y=x(log⁡3+log⁡(cos⁡x))

Differentiating both sides with respect to x, we get

\(\frac{1}{y} \frac{dy}{dx} =\frac{d}{dx} (x log⁡3)+\frac{d}{dx} (x) log⁡(cos⁡x)+\frac{d}{dx}(log⁡(cos⁡x)).x\)

\(\frac{1}{y} \frac{dy}{dx}=log⁡3+log⁡(cos⁡x)+\frac{1}{cos⁡x}.-sin⁡x.x\)

\(\frac{1}{y} \frac{dy}{dx}\)=log⁡3+log⁡(cos⁡x)-x tan⁡x

\(\frac{dy}{dx}\)=y(log⁡(3 cos⁡x)-x tan⁡x)

\(\frac{dy}{dx}\)=(3 cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)

Correct ANSWER is (c) 15x^2

The BEST I can explain: Consider y=log⁡(e^5x^3)

y=5x^3 (∴log⁡e^x=x)

⇒\(\FRAC{dy}{DX}\)=\(\frac{d}{dx} (5x^3)\)

∴\(\frac{dy}{dx}=5(3x^2)=15x^2\)

Correct choice is (c) First and SECOND kinds

Best explanation: Kinds of DISCONTINUITY are CLASSIFIED as follows.

i. Discontinuity of the first kind: Removable and JUMP discontinuities.

ii. Discontinuity of the second kind: OSCILLATING and infinite discontinuities.

Correct choice is (d) \(12(4x^2-8x-1)\)

For EXPLANATION I would say: Given that, \(y=4x^4+2x\)

\(\FRAC{dy}{DX}\)=16x^3+2

\(\frac{d^2 y}{dx^2}\)=48x^2

\(\frac{d^2 y}{dx^2}\)-6 \(\frac{dy}{dx}=48x^2-96x^3-12\)

=12(4x^2-8x-1)

Correct choice is (a) 0

Best EXPLANATION: Given that, y=2 sin^-1⁡(cos⁡x)

\(\FRAC{dy}{DX}=2.\frac{1}{\SQRT{1-cos^2⁡x}}\).-sin⁡x=-2 (∵\(\sqrt{1-cos^2⁡x}\)=sin⁡x)

\(\frac{d^2 y}{dx^2}\)=\(\frac{d}{dx} (\frac{dy}{dx})=\frac{d}{dx}\) (-2)=0

The correct OPTION is (a) 9^tan⁡3x (3 LOG⁡9 sec^2⁡x)

For explanation: CONSIDER y=9^tan⁡3x

Applying log on both sides, we get

log⁡y=log⁡9^tan⁡3x

Differentiating both sides with respect to x, we get

\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \)(tan⁡3x.log⁡9)

\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \,(tan⁡3x) \,log⁡9+\frac{d}{dx} \,(log⁡9).TAN3X \,(∵ Using \,u.v=u’ \,v+uv’)\)

\(\frac{dy}{dx}\)=y(3 sec^2⁡⁡x.log⁡9+0)

\(\frac{dy}{dx}\)=9^tan⁡3x (3 log⁡9 sec^2⁡x)

Right option is (d) the mean value theorem

The best EXPLANATION: Lagrange’s mean value theorem is also CALLED the mean value theorem and ROLLE’s theorem is just a special CASE of Lagrange’s mean value theorem when f(a) = f(b).

The correct choice is (d) \(\frac{3}{1+x^2}\)

For explanation I would say: Given function is \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2})\),

Now taking RHS and substituting x = tang in it and then we get,

y = \(tan^{-1}(\frac{3tang-tang^3}{1-3tang^2})\), Now it becomes the expansion of the function tan3g,

Hence the given function becomes y= tan^-1(tan3g). Which is equal to 3g, now substituting the VALUE of G= tan^-1x, now after differentiating both sides we get the answer \(\frac{3}{1+x^2}\).

Correct answer is (C) –cosec^2x

The explanation is: The derivative of cotx is –cosec^2x, as this function has a FIXED derivative LIKE SINX has its derivative cosx. Therefore the answer to the above question is –cosec^2x.

The correct option is (B) True

Easiest 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).

Right choice is (c) F is differentiable and continuous on (a,b)

For 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) = \(\frac {f(b)-f(a)}{b-a}\).

Right answer is (b) f’(C) = \(\FRAC {f(b)-f(a)}{b-a}\)

Explanation: ACCORDING to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that f is differentiable on (a,b) then the FORMULA for Lagrange’s theorem is f’(c) = \(\frac {f(b)-f(a)}{b-a}\).

Right CHOICE is (b) –\(\FRAC{4}{3a}\)

Easy explanation: Given that, x=3a^2 cos^2⁡θ and y=4A sin^2⁡θ

Then, \(\frac{DX}{dθ}\)=3a^2.(2 cos⁡θ)(-sin⁡θ)=-3a^2 sin⁡2θ

\(\frac{DY}{dθ}\)=4a(2 sin⁡θ)(cos⁡θ)=4a sin⁡2θ

\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}×\frac{dθ}{dx}=-\frac{4a \,sin⁡2θ}{3a^2 \,sin⁡2θ}=-\frac{4}{3a}\)

The correct choice is (a) continuous

The explanation is: ACCORDING to ROLLE’s 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)

iii) f(a) = f(b) then there EXISTS at least one POINT c ∈ (a,b) such that f’(c) = 0

Correct answer is (C) nx^n-1

To elaborate: It is a standard rule for derivative of a function of this form in which the original POWER comes in FRONT and the value in the power is decreased by one. Therefore the only option of this form is nx^n-1 from the GIVEN OPTIONS.

Right ANSWER is (B) 2xcos(x^2)

Explanation: Differentiation of the function f(x) = sin(x^2) is DONE with chain RULE. First we differentiate sin function which becomes cos and then differentiate the INNER (x^2) which becomes 2x, hence it comes out to be 2xcos(x^2).

The correct choice is (a) limx→a+⁡f(x)=f(a)

To explain I would SAY: A function is said to be continuous when it is both LEFT continuous and right continuous. MATHEMATICAL expression for a function f is right continuous on (a,B) is limx→a+⁡f(x)=f(a).

Correct answer is (d) 4E^2x+\(\frac{e^x}{(1-e^2x)^{3/2}}\)

The explanation: Given that, y=e^2x+sin^-1⁡e^x

\(\frac{DY}{dx}\)=2e^2x+\(\frac{1}{\sqrt{1-e^{2x}}} e^x\)

\(\frac{d^2 y}{dx^2} = 4e^2x+\bigg(\frac{\frac{d}{dx} (e^x) \sqrt{1-e^{2x}} – \frac{d}{dx} (\sqrt{1-e^{2x}}).e^x}{(\sqrt{1-e^{2x}})^2}\bigg)\)

\(=4e^{2x}+\frac{(e^x \sqrt{1-e^{2x}})-e^x \left(\frac{1}{2\sqrt{1-e^{2x}}}.-2e^{2x}\RIGHT)}{1-e^{2x}}\)

\(=4e^{2x}+\frac{(e^x (1-e^{2x})+e^{3x})}{(1-e^{2x})^{\frac{3}{2}}}\)

\(=4e^{2x}+\frac{e^x (1-e^{2x}+e^{2x})}{(1-e^{2x})^{\frac{3}{2}}}\)

4e^2x+\(\frac{e^x}{(1-e^2x)^{3/2}}\).

Right option is (a) True

The EXPLANATION: ACCORDING to Rolle’s 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)

iii) f(a) = f(b) then there exists at least one point C ∈ (a,b) such that f’(c) = 0

Right choice is (d) 2e^X-8e^-x

To EXPLAIN: To solve:y=(8e^-x+2e^x)

DIFFERENTIATING w.r.t x we get,

\(\FRAC{DY}{dx}\)=8(-e^-x+2e^x)

∴\(\frac{dy}{dx}\)=2e^x-8e^-x.

Right OPTION is (d) 4x^e^x-1 e^x (x log⁡x+1)

Explanation: Consider y=4x^e^x

Applying log on both sides, we GET

log⁡y=log⁡4x^e^x

log⁡y=log⁡4+log⁡x^e^x (∵log⁡ab=log⁡a+log⁡b)

DIFFERENTIATING both sides with respect to x, we get

\(\FRAC{1}{y} \frac{DY}{dx}=0+\frac{d}{dx}(e^x \,log⁡x)(∵log⁡a^b=b \,log⁡a)\)

\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \,(e^x) \,log⁡x+e^{x} \,\frac{d}{dx} \,(log⁡x)\)

\(\frac{dy}{dx}=y(e^x log⁡x+\frac{e^x}{x})\)

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

The correct option is (b) -16 sin⁡2X e^cos2x

Best explanation: Consider y=8e^cos2x

Differentiating w.r.t X by USING CHAIN rule, we get

\(\frac{dy}{dx}=\frac{d}{dx}\)(8e^cos2x)

=8e^cos2x\(\frac{d}{dx}\) (cos⁡2x)

=8e^cos2x(-sin⁡2x)\(\frac{d}{dx}\) (2x)

=8e^cos2x(-sin⁡2x)(2)

∴\(\frac{dy}{dx}\)=-16 sin⁡2x e^cos2x

The CORRECT answer is (a) \(\frac{cosx-2}{3}\)

Easiest EXPLANATION: Differentiating on both SIDES we GET 2 + 3\(\frac{dy}{DX}\) = cosx.

3\(\frac{dy}{dx}\) = cosx-2.

\(\frac{dy}{dx} = \frac{cosx-2}{3}\).

Right choice is (d) x^3e^3x.3e^3x (3 log⁡x+\(\FRAC{1}{x}\))

For EXPLANATION: CONSIDER y=x^3e^3x

Applying log on both sides, we get

log⁡y=3e^3x log⁡x

Differentiating both sides with respect to x, we get

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

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

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

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

The CORRECT option is (d) 8e^2x+\(\FRAC{12}{(2x-3)^2}\)

To ELABORATE: Given that, y=2e^2x-3 log⁡(2x-3)

\(\frac{dy}{dx}\)=4e^2x-3.\(\frac{1}{(2x-3)}\).2=4e^2x–\(\frac{6}{(2x-3)}\)

\(\frac{d^2 y}{dx^2}=\frac{d}{dx} (\frac{dy}{dx})\)=8e^2x+\(\frac{12}{(2x-3)^2}\)

Correct option is (C) (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

The best explanation: CONSIDER y=(cos⁡3x)^3x

Applying log on both sides, we get

log⁡y=log⁡(cos⁡3x)^3x

log⁡y=3x log⁡(cos⁡3x)

Differentiating both sides with respect to x, we get

\(\frac{1}{y} \frac{DY}{dx}=\frac{d}{dx} (3x \,log⁡(cos⁡3x))\)

By USING u.v=u’ v+uv’, we get

\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{d}{dx} \,(3x) \,log⁡(cos⁡3x)+\frac{d}{dx} \,(log⁡(cos⁡3x)).3x\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \frac{d}{dx} \,(cos⁡3x).3x)\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).\frac{d}{dx}(3x).3x)\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).3.3x)\)

\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) – 9x tan⁡3x)

\(\frac{dy}{dx}\)=(cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

Correct choice is (B) \(LOG⁡2x^{sin⁡3x} \,(3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{X \,log⁡2x})\)

The explanation is: Consider y=\((log⁡2x)^{sin⁡3x}\)

Applying log on both sides, we get

log⁡y=\(log⁡(log⁡2x)^{sin⁡3x}\)

log⁡y=sin⁡3x log⁡(log⁡2x)

Differentiating with RESPECT to x, we get

\(\frac{1}{y} \,\frac{dy}{dx}=log⁡(log⁡2x)\frac{d}{dx} (sin⁡3x)+sin⁡3x \frac{d}{dx} \,(log⁡(log⁡2x))\)

By using chain rule, we get

\(\frac{1}{y} \,\frac{dy}{dx}=log⁡(log⁡2x).3 \,cos⁡3x+sin⁡3x.\frac{1}{log⁡2x}.\frac{1}{2x}.2 \,(∵u.v=u’ \,v+uv’)\)

\(\frac{dy}{dx}\)=y(3 cos⁡3x log⁡(log⁡2x)+\(\frac{sin⁡3x}{x \,log⁡2x}\))

∴\(\frac{dy}{dx}\)=log⁡2x^sin⁡3x \(\left (3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x \,log⁡2x} \right )\)

The CORRECT choice is (b) limx→C⁡f(x) = f(c) ∀ c ∈ (a,b)

The best explanation: A function f defined on (a,b) is said to be CONTINUOUS on (a,b) if it is continuous at EVERY point of (a,b) i.e., if limx→c⁡f(x)=f(c) ∀ c ∈ (a,b).

Right choice is (a) True

The explanation: According to Lagrange’s mean VALUE THEOREM, if f : [a,b] → R is a FUNCTION such that f is DIFFERENTIABLE on (a,b) then there exists a least point c ∈ (a,b) such that f’(c) = \(\frac {f(b)-f(a)}{b-a}\). This shows Function f is differentiable on [a,b].

Right choice is (d) f is DIFFERENTIABLE on (a,a+h)

BEST 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 answer is (C) 4e^2x^2 (4x^2+1)

To EXPLAIN: GIVEN that, y=E^2x^2

\(\frac{DY}{dx}\)=e^2x^2.4x

By using u.v rule, we get

\(\frac{d^2 y}{dx^2}=\frac{d}{dx} (e^{{2x}^2}).4x+\frac{d}{dx} (4x).e^{{2x}^2}\)

16x^2 e^2x^2+4e^2x^2=4e^2x^2 (4x^2+1)

Right choice is (B) False

The best explanation: The GIVEN statement is false. Given that, y=6x^2+3

\(\FRAC{dy}{dx}\)=12x

⇒\(\left (\frac{dy}{dx}\right )^2=(12x)^2=144x^2\)

\(\frac{d^2 y}{dx^2}=\frac{d}{dx}\) (12x)=12

∴\(\left (\frac{dy}{dx}\right )^2≠\frac{d^2 y}{dx^2}\)

The CORRECT OPTION is (a) –\(\FRAC{tan^2⁡2θ \,sin⁡2θ}{2}\)

The explanation is: GIVEN that, x=tan⁡2θ and y=cos⁡2θ+sin^2⁡θ

\(\frac{DX}{dθ}\)=2 sec^2⁡2θ

\(\frac{dy}{dθ}\)=-2 sin⁡2θ+2 sin⁡θ cos⁡θ=-2 sin⁡2θ+sin⁡2θ=-sin⁡2θ

∴\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}.\frac{dθ}{dx}=-\frac{sin⁡2θ}{2 sec^2⁡2θ}=-\frac{sin⁡2θ}{2 cos^2⁡2θ}.sin^2⁡2θ=-\frac{tan^2⁡2θ sin⁡2θ}{2}\)

Correct option is (b) True

The BEST EXPLANATION: If a ∈ R then f(a) = c

limx→a⁡f(X)=limx→a⁡c=c=f(a)

HENCE, f(x) is continuous at any point a ∀ R.

Correct option is (a) 14e^2x x^(2e^2x) (2 log⁡x+\(\frac{1}{x}\))

The explanation is: Consider y=7x^(2e^2x)

log⁡y=log⁡7x^(2e^2x)

log⁡y=log⁡7+log⁡x^(2e^2x)

log⁡y=log⁡7+2e^2x log⁡x

Differentiating with RESPECT to x on both sides, we GET

\(\frac{1}{y} \frac{DY}{dx}\)=\(\frac{d}{dx}\) (log⁡7+2e^2x log⁡x)

\(\frac{1}{y} \frac{dy}{dx}\)=0+\(\frac{d}{dx}\) (2e^2x) log⁡x+\(\frac{d}{dx}\) (log⁡x)2e^2x (using u.v=u’ v+uv’)

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

\(\frac{dy}{dx}\)=y\( \left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)\)

\(\frac{dy}{dx}\)=7x^(2e^2x) \( \left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)\)

\(\frac{dy}{dx}\)=14e^2x x^(2e^2x) (2 log⁡x+\(\frac{1}{x}\))

Right answer is (b) No

Best explanation: Given function is f(x) = tan x on [ \(\frac {\PI }{4}, \frac {5\pi }{4}\) ]

F(x) = tan x is not DEFINED at x on [ \(\frac {\pi }{4}, \frac {5\pi }{4}\) ]

So, f(x) is not CONTINUOUS on [ \(\frac {\pi }{4}, \frac {5\pi }{4}\) ].

Hence, ROLLE’s theorem is not applicable.

The correct choice is (a) \(\frac{1}{2te^t}\)

EXPLANATION: GIVEN that, x=2e^t and y=log⁡t

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

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

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

Right ANSWER is (a) sec^2(x+4)

Explanation: We know that derivative of TANX is sec^2(x), now in the above question we GET TAN(x+4), hence its derivative COMES out to be sec^2(x+4), as the inside expression (x+4) is differentiated into 1.

Therefore the answer is sec^2(x+4).

The correct option is (c) Lagrange’s mean value theorem

Easy EXPLANATION: ROLLE’s theorem is just a special CASE of Lagrange’s mean value theorem when f(a) = f(b) and Lagrange’s mean value theorem is also called the mean value theorem.