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The correct option is (a) 3√3/4 ab SQ units

Easiest explanation: LET A, B, C be the vertices of the isosceles triangle.

Let B = (a cosθ, b sinθ)

=> C – (a cosθ. -b sinθ)(by SYMMETRY).

Now, area of a triangle is given by

A = ½(BC)(AD)

= ½(2b sinθ)(1 + cosθ)

=> dA/dθ = ab[cosθ(1 + cosθ) – sin^2θ]

= ab(1 + cosθ)(2 cosθ – 1)

Now, putting dA/dθ = 0,

We get, θ = π/3, π

Now, for θ = π, triangle ABC is not possible.

And, dA/dθ]θ = π/3 < 0,

Thus, for θ = π/3, the area of isosceles triangle will be maximum.

So, A = 3√3/4 ab sq units.

Right choice is (b) 0

Best explanation: We have, y = tan^-1 √(X^2 – 1)

Differentiating both sides with respect to x, we get,

y’ = 1/(1 + x^2 – 1)* d/dx(x^2 – 1)^1/2

1/x^2 * 1/2((x^2 – 1))^1/2 * 2X

Squaring both sides,

x^2(x^2 – 1)(y’)^2 = 1

Differentiating again with respect to x we get,

x^2(x^2 – 1)(y’)^2 + (y’)^2d/dx(x^4 – x^2) = d/dx(1)

Or, x^2(x^2 – 1)^2y’y” + (y’)^2(4x^3 – 2x) = 0

Or, x^2(x^2 – 1)y” + (y’)(2x^3 – x) = 0

The correct answer is (C) ⁡\(\frac{-9}{x^3}\)

For EXPLANATION: We use chain rule to FIND \(\frac{d}{dx}\)(ln ⁡\(\frac{3}{x}\)).

\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{1}{\frac{3}{x}} \frac{d}{dx}(\frac{3}{x})\)

\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{3}{x}(-\frac{3}{x^2})\)

\(\frac{d}{dx}(ln \frac{3}{x}\)) = \(\frac{-9}{x^3}\)

The CORRECT answer is (a) x^4 – 8x^3 + 22x^2 – 24x – 55

Explanation: Here it is given that f(x) is a polynomial of degree 4

So, it has critical POINTS at x=1, x=2 and x=3

So, f’(x) = A(x – 1)(x – 2)(x – 3)

 = A(x^3 – 6x^2 + 11x -6)

On INTEGRATING this, gives,

f(x) = A/4(x^4 – 8x^3 + 22x^2 – 24x) + B

Also, f(-1) = 0

So, we get,

B = -554A/4

Also, it is given that -2∫^2 f(x) dx = -1348/15

So, A/4 -2∫^2(x^4 – 8x^3 + 22x^2 – 24x – 55)dx = -1348/15

By changing the RANGE of the integrals, we get,

=> A/2 0∫^2(x^4 + 22x^2 – 55) dx = -1348/15

=> -337A/15 = -1348/15

=> A = 4

Thus, f(x) = x^4 – 8x^3 + 22x^2 – 24x – 55.

Right CHOICE is (d) 2e^x.COS⁡ x

Explanation: We need to use product RULE in both the TERMS to get the answer.

\(\frac{d}{dx}\) (f.g) = g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)

\(\frac{d}{dx}\) (e^x SIN x + e^x cos ⁡x) = (e^x.\(\frac{d}{dx}\) (sin⁡ x) + sin ⁡x.\(\frac{d}{dx}\) (e^x)) + (e^x.\(\frac{d}{dx}\) (cos ⁡x) + cos⁡ x.\(\frac{d}{dx}\) (e^x))

\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) =(e^x.cos⁡ x + sin ⁡x . e^x) + (e^x.(-sin ⁡x) + cos ⁡x.e^x)

\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) = e^x.cos⁡ x + sin⁡ x . e^x – e^x.sin⁡ x + cos ⁡x.e^x

\(\frac{d}{dx}\) (e^x sin x + e^x cos ⁡x) = 2e^x.cos⁡ x

Correct CHOICE is (b) 1

Best explanation: \(\lim\limits_{X \rightarrow 0}\frac{sin^2⁡x}{x^2}\) =

= (\(\lim\limits_{x \rightarrow 0}\frac{sin ⁡x}{x}\) x \(\lim\limits_{x \rightarrow 0}\frac{sinx}{x}\))

We apply L’Hospital’s rule and DIFFERENTIATE numerator and denominator.

= (\(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\) x \(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\))

= 1

The CORRECT option is (b) (1 – x^2)y” – xy’ = 0

The best I can explain: We have, y = a sin^-1x + b cos^-1x

Differentiating both sides with respect to x we get,

dy/DX = 12∫ e^2xcos3x dx – 5∫ e^2x sin3x dx

y’ = a* 1/√(1 – x^2) + b * (-1/√(1 – x^2))

or, y’√(1 – x^2) = a – b

now, squaring both sides,

or, y’^2(1 – x^2) = (a – b)^2

Differentiating again with respect to x we get,

(1 – x^2) * 2y’y” + y’^2 . (-2x) = 0

As, y’ ≠ 0,

(1 – x^2)y” – xy’ = 0

Right answer is (b) 0

Explanation: We have, y = COS(2sin^-1x)

DIFFERENTIATING both the sides,

y’ = d/dxcos(2sin^-1x)

= -SIN(2sin^-1x) * d/dx(2sin^-1x)

or, y’ = -sin(2sin^-1x)*2*1/√(1 – x^2)

or, y’*√(1 – x^2) = -2sin(2sin^-1x)

SQUARING both sides,

or, (y’)^2 *(1 – x^2) = 4sin^2(2sin^-1x)

or, (y’)^2 *(1 – x^2) = 4(1 – y^2)

Differentiating again with respect to x, we get,

(1 – x^2)*2y’y” + (y’)^2 (-2X) = 4 * (-2yy’)

(1 – x^2)y” – xy’ = – 4y

=> (1 – x^2)y” – xy’ + 4y = 0

Right OPTION is (b) At least one root in (a, b) and at least one root in (b, c)

The best explanation: Here, f (x) being a polynomial is CONTINUOUS and differentiable for all real values of x.

We also have f(a) = f(b) = f(c). If we apply Rolle’s theorem to f (x) in [a, b] and [b, c] we will OBSERVE that f'(x) = 0 will have at least one root in (a, b) and at least one root in (b, c).

But f'(x) is a polynomial of degree two, so that f'(x) = 0 can’t have more than two roots. It implies that exactly one root of f'(x) = 0 will lie in (a, b) and exactly one root off'(x) = 0 will lie in (b, c).

Let y = f(x) be a polynomial function of degree n. If f (x) = 0 has real roots only, then f'(x) = 0, f”(x) = 0, … , f^n-1(x) = 0 will have real roots. It is in FACT the general version of above mentioned APPLICATION, because if f (x) = 0 have all real roots, then between two consecutive roots of f(x) = 0, exactly one root of f'(x) = 0 will lie.

Right ANSWER is (d) 0

For EXPLANATION: \(\lim\limits_{X \RIGHTARROW 0}\frac{x}{sin⁡ x}\)x \(\lim\limits_{x \rightarrow 0}⁡\frac{x}{cos⁡ x}\)

= 1 x 0

= 0

The correct answer is (d) 6

To explain I would SAY: The denominator BECOMES 0, as x approaches 4.

\(\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}\) Here, if we factorize the numerator we get:

\(\lim\limits_{x \rightarrow 4} \frac{(x – 4)(x + 2)}{x – 4}\)

We can now CANCEL out (x – 4) from both the numerator and denominator.

We get, \(\lim\limits_{x \rightarrow 4}\)(x + 2) = 6

Right option is (a) Yes

The explanation: Obviously, F(X) is CONTINUOUS at [1, 2]

And, f(x) DIFFERENTIABLE at [1, 2]

Also, f(1) = f(2) = 2

Now, f’(x) = 0

=> 2x – 3 = 0

=> x = 3/2

Thus, x belongs to [1, 2]

Hence, it is verified.

The CORRECT choice is (b) 3x^2 cos⁡ x^2 cos⁡ x^3 – 2xsin⁡ x^3 sin x^2

Best explanation: We FOLLOW product RULE\(\frac{d}{dx}\) (f.G)= g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)

Here f = sin⁡ x^3 and g = cos⁡ x^2

\(\frac{d}{dx}\) (f) = 3x^2 cos⁡ x^3

\(\frac{d}{dx}\) (g) = -2X sin x^2

We now substitute this in our main equation,

= cos⁡ x^2.3x^2 cos⁡ x^3 + sin⁡ x^3.(-2x sin x^2)

= 3x^2 cos x^2 cos⁡ x^3 – 2x sin⁡ x^3 sin x^2

The correct OPTION is (a) [1 + (d^2y/dx^2)^2]^3/2

For explanation I would say: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)

=> x = 2A cos^2θ*sin2θ and y = 2a sin^2θ*cos2θ

Now DIFFERENTIATING x and y with respect to θ we GET,

dx/dθ = 2a[cos^2θ*2cos2θ + sin2θ*2cos2θ cos2θ]

= 4a cosθ (cosθ cos2θ – sinθ sin2θ)

= 4a cosθ cos(θ + 2θ)

= 4a cosθ cos3θ

dy/dθ = 2a[cos2θ*2cosθ sinθ + sin^2θ (-2sin2θ)]

= 4a sinθ (cosθ cos2θ – sinθ sin2θ )

= 4a sinθ cos(θ + 2θ)

= 4a sinθ cos3θ

Thus, dy/dx = (dy/dθ)/(dx/dθ)

= (= 4a cosθ cos3θ)/( 4a sinθ cos3θ)

= tanθ

So, (d^2y/dx^2) = d/dx(tanθ)

= sec^2θ*dθ/dx

= sec^2θ*(1/(dx/dθ))

= sec^2θ*1/(4a cosθ cos3θ)

Or, 4a cos3θ (d^2y/dx^2) = sec^3θ

= (sec^2θ)^3/2

= (1 + tan2θ)^3/2

As, dy/dx = tanθ

So, 4a cos3θ(d^2y/dx^2) = [1 + (d^2y/dx^2)^2]^3/2

The correct answer is (c) -1

Easiest explanation: Given, y = 1/(1 + x + x^2 + x^3)

Assuming x ≠ 1

Or, y = (x – 1)/(x – 1)( x^3 + x^2 + x + 1)

Differentiating both the SIDES with RESPECT to x, we GET,

dy/dx = [(x^4 – 1)*1 – (x – 1)*4x^3]/ (x^4 – 1)^2

Thus, PUTTING x = 0 in the above equation, we get,

(dy/dx) = -1/(-1)^2

= -1

The correct option is (a) 0

To ELABORATE: \(\lim\limits_{X \rightarrow 0}\frac{x TANX}{cot x}\) = \(\lim\limits_{x \rightarrow 0}\frac{x\frac{sin⁡ x}{cos ⁡x}}{\frac{cos⁡ x}{sin⁡ x}}\)

 = \(\lim\limits_{x \rightarrow 0}\) ⁡x

= 0

The correct CHOICE is (c) \(\frac{a^x (LN\, ⁡a-1)}{e^x}\)

The best EXPLANATION: Using quotient rule, we know that, \(\frac{d}{dx} (\frac{F}{g}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)

Here, f = a^x and g = e^x

\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x.\frac{d}{dx}(a^x)-a^x.\frac{d}{dx}(e^x)}{(e^x)(e^x)}\)

\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{e^x a^x ln⁡\, a-a^x e^x}{(e^x)(e^x)}\)

\(\frac{d}{dx} (\frac{a^x}{e^x}) = \frac{a^x (ln⁡ \,a-1)}{e^x}\)

Right option is (B) 6.4

To explain: Use L’HOSPITAL’s Rule, and differentiate the NUMERATOR and denominator.

\(\lim\limits_{x \rightarrow 5}\frac{⁡32}{2x–5}\)

= \(\frac{32}{5}\)

= 6.4

The correct answer is (d) (b – a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)

The BEST explanation: Let, F(X) = \(\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}\) = f(a)g(x) – f(x)g(a)…..(1)

=> F’(x) = f’(a)g’(x) – f’(x)g(a)

Since, f(x) and g(x) are continuous in [a, b] and differentiable in (a, b),

So, F(x) is continuous in [a, b] and differentiable in (a, b)

Also from (1), F(a) = f(a)g(a) – f(a)g(a) = 0

And F(b) = f(a)g(b) – f(b)g(a)

Now, by the mean value theorem, there exists at least one point c, a < c < b, such that,

F’(c) = (F’(b) – F’(a)) / (b – a)

=> f(a) g’(c) – g(a) f’(c) = (f(a)g(b) – f(b)g(a) – 0)/b – a

Or, f(a)g(b) – f(b)g(a) = (b – a)( f(a) g’(c) – g(a) f’(c))

=>\(\begin{vmatrix}f(a) & f(b) \\g(a) & g(b) \end {vmatrix}\) = (b – a)\(\begin{vmatrix}f(a) & f'(c) \\g(a) & g'(c) \end {vmatrix}\)

The correct CHOICE is (a) 0

To elaborate: \(\lim\limits_{X \rightarrow 4}\frac{x^2-4-3x}{x-3}\)

= \(\frac{4^2-4-3(4)}{4-3}\)

= \(\frac{0}{1}\)

= 0.

The CORRECT choice is (a) 1/e

Explanation: We have, y = x^(1/(1 – x))

So, log y = (log x)/(1 – x)

So, \(\lim\limits_{x \rightarrow 1}\)log y = \(\lim\limits_{x \rightarrow 1}\)(log x)/(1 – x)

Now, using L’HOSPITAL’s rule,

\(\lim\limits_{x \rightarrow 1}\)log y = \(\lim\limits_{x \rightarrow 1}\)((\(\FRAC{1}{x})\) / – 1)

=>, \(\lim\limits_{x \rightarrow 1}\)log y = -1

So, \(\lim\limits_{x \rightarrow 1} x^{\frac{1}{1-x}}\) = 1/e.

Correct answer is (b) 1

Easy explanation: We NEED to use product rule in both the TERMS to GET the answer.

\(\frac{d}{DX}\) (f.g) = g.\(\frac{d}{dx}\) (f)+ f.\(\frac{dy}{dx}\) (g)

Here f = e^x and g = tan ⁡x

\(\frac{d}{dx}\) (e^x tan x) = tan⁡ x.\(\frac{d}{dx}\) (e^x) + e^x.\(\frac{d}{dx}\) (tan ⁡x)

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

At x = 0 we get,

= tan ⁡0.e^0 + e^0.sec^2⁡0

= 0.(1) + 1.(1)

= 1

Correct answer is (c) \(\frac{-1}{2\sqrt 2}\)

The BEST I can explain: This is of the form\(\frac{0}{0}\), therefore we USE L’Hospital’s rule and differentiate the NUMERATOR and DENOMINATOR.

= \(\lim\limits_{x \rightarrow 0}\frac{2sin⁡ \,x cos \,⁡x + cos \,⁡x \sqrt 2}{2x – 4}\)

= \(\frac{0+\sqrt 2}{-4}\)

= \(\frac{-1}{2\sqrt 2}\)

Correct CHOICE is (c) 2

For explanation I would say: The limit can be WRITTEN as, \(\lim\limits_{X \rightarrow 0}\FRAC{32x^2}{sin^2⁡4X}\)

= 2 x \(\lim\limits_{x \rightarrow 0}\frac{4x}{sin 4x}\) x \(\lim\limits_{x \rightarrow 0}\frac{4x}{sin 4x}\)

= 2 x 1 x 1

= 2

Correct choice is (d) 6

For explanation I would say: When x tends to 3, both the NUMERATOR and the denominator become 0 and it becomes of the form, \(\frac{0}{0}\).

Therefore, we use L’Hospital’s rule, which states the we differentiate the numerator and the denominator, until a definite answer is reached.

On differentiating once we GET,

\(\lim\limits_{x \rightarrow 3}\frac{2X}{1}\)

Since, this not an indeterminate form now, we can substitute the value of x.

= 2 x 3

= 6

Correct CHOICE is (c) 6

To explain: We have, y = (sin^-1x)^2 ………..(1)

Differentiating with respect to x, we get,

y’ = 2(sin^-1x)*1/(1 – x^2)^1/2

or, y’(1 – x^2)^1/2 = 2(sin^-1x)

Squaring both SIDES,

(1 – x^2)(y’)^2 = 4(sin^-1x)^2

From (1),

(1 – x^2)( y’)^2 = 4y

Differentiating with respect to x, we get

(1 – x^2)y” + (y’)^2 d/dx(1 – x^2) = 4 y’

=>(1 – x^2)2y’y” + (y’)^2 (-2X) = 4y’

Or, (1 – x^2)y” – XY’ = 2

Or, (1 – x^2)y” – xy’ + 4 = 2 + 4 = 6

Right option is (c) x = 3 is the GLOBAL maximum

To explain: Clearly, x = 1, 3 are the points of global minimum (as the values being EQUAL).

And, x = 0, 4 are the points of global maximum (as the values being equal).

And, x = 2, is the POINT of local maximum (as the values being equal).

Thus, x = 3 is the global maximum.

Right answer is (a) 0

The explanation is: \(\frac{d}{DX}(\frac{sec⁡x}{COSEC\, x\, tan⁡x})\) = \(\frac{d}{dx}(\frac{{1} {cos⁡x}}{\frac{1}{sin⁡\, x}.\frac{sin⁡x}{cos⁡x}}\))

= \(\frac{d}{dx}\) (1)

= 0

The correct choice is (a) 0

To explain I would SAY: This is of the FORM \(\frac{\infty}{\infty}\), therefore we use L’Hospital’s rule and differentiate the numerator and DENOMINATOR.

= \(\lim\limits_{x \rightarrow \infty}\frac{2x+\sqrt{2/x}}{2x–4}\)

= \(\lim\limits_{x \rightarrow \infty}⁡\)√2 x^-3/2

= 0

Correct CHOICE is (a) 1

Explanation: Since it is of the form \(\FRAC{\infty}{\infty}\), we use L’Hospital’s rule and differentiate the numerator and denominator

L = \(\lim\limits_{X \rightarrow \infty}\frac{x^2-9}{x^2–3x+2}\)

On differentiating once, we get \(\lim\limits_{x \rightarrow \infty}\frac{2x}{2x}\)

Which is EQUAL to, \(\lim\limits_{x \rightarrow \infty}\) ⁡1 = 1.

Right choice is (a) 1

For explanation: Let, y = x^x

Thus, LOG y = x log x

Differentiating both SIDES with respect to x, we get,

1/y dy/dx = (x*1/x) + log x

=>dy/dx = y(1 + log x)

Or, dx^x/dx = x^x(1 + log x)

Using, L’Hospital’s rule,

\(\lim\limits_{x \rightarrow a}\frac{(a^x-x^a)}{x^x-a^a}\) = \(\lim\limits_{x \rightarrow a}\frac{(a^x*log⁡a-x^{a-1})}{x^x(1+logx)}\)

= \(\lim\limits_{x \rightarrow a}\frac{(a^a*log⁡a-a^{a})}{a^a(1+loga)}\)

= (log a – 1)/(log a + 1)

As per the QUESTION,

(log a – 1)/(log a + 1) = -1

Or, (log a – 1) = -log a – 1

Or, 2 log a = 0

Or, log a = 0

So, a = 1

Correct option is (a) sin⁡ x + tan⁡ x sec⁡ x

Easy explanation: We follow product rule \(\frac{d}{dx}\) (F.G) = g.\(\frac{d}{dx}\) (f) + f.\(\frac{d}{dx}\) (g)

Here, f = sin⁡ x and g = tan⁡ x

\(\frac{d}{dx}\) (sin⁡ x tan⁡ x) = cos⁡ x tan⁡ x + sec^2⁡ x SINX

\(\frac{d}{dx}\) (sin⁡ x tan⁡ x) = sin⁡ x + tan⁡ x sec⁡ x

Correct option is (c) They are indeterminate form

The BEST I can explain: If there are forms like ∞-∞, 0*∞, 0^0, (∞)^0 and 1^∞ then they are said to be indeterminate form.

Then to evaluate the limiting VALUE of the function at the POINT, L’Hospital’s RULE is used.

Correct answer is (d) 2xe^x^2

To ELABORATE: We apply chain RULE. First we differentiate x^2.

\(\frac{d}{DX}\) (x^2) = 2x

Now, we KNOW that \(\frac{d}{dx}\) (e^x) = e^x

We differentiate e^x^2 in the same manner and then multiply with the derivative of x^2

\(\frac{d}{dx}\) (e^x^2) = 2xe^x^2

The CORRECT answer is (a) 40

Explanation: \(\lim\limits_{y \RIGHTARROW 4}\)f(y) = y^2 + 6y

f(4) = 4^2 + 6(4)

f(4) = 16 + 24

f(4) = 40

The CORRECT answer is (a) 0

The explanation is: Any number divided by infinity GIVES us 0.

Here, SINCE the number 2 is divided by y, as y approaches infinity, we GET 0

Correct choice is (d) Sec^2θ

Explanation: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)

=> x = 2a cos^2θ*sin2θ and y = 2a sin^2θ*cos2θ

Now differentiating x and y with RESPECT to θ we get,

dx/dθ = 2a[cos^2θ*2cos2θ + sin2θ*2cos2θ cos2θ]

= 4a cosθ (cosθ cos2θ – sinθ sin2θ )

= 4a cosθ cos(θ + 2θ)

= 4a cosθ cos3θ

dy/dθ = 2a[cos2θ*2cosθ sinθ + sin^2θ (-2sin2θ)]

= 4a sinθ (cosθ cos2θ – sinθ sin2θ )

= 4a sinθ cos(θ + 2θ)

= 4a sinθ cos3θ

Thus, dy/dx = (dy/dθ)/(dx/dθ)

= (= 4a cosθ cos3θ)/( 4a sinθ cos3θ)

= tanθ

So, (dy/dx)^2 + 1 = 1 + tan^2θ = sec^2θ

The CORRECT option is (d) At least one point C(x3, y3) where the TANGENT will be parallel to the chord AB

The BEST I can explain: Here, y = f(x) = ax^2 + bx + c

As f(x) is a POLYNOMIAL function, it is continuous and differentiable for all x.

So, according to geometrical interpretation of mean value theorem there will be at least one point C (x3, y3) between A (x1, Y1) and B (x2, y2) where tangent will be parallel chord AB.

The correct CHOICE is (c) 1

The explanation: \(\lim\limits_{x \rightarrow 0}\frac{sin^2⁡ x}{x^2}\) x \(\lim\limits_{x \rightarrow 0}\frac{e^x}{x}\)

We apply L’HOSPITAL’s rule and differentiate NUMERATOR and denominator.

= 1 x \(\lim\limits_{x \rightarrow 0}\frac{e^x}{1}\)

= 1

Right choice is (c) -cot^2 x(3 sin^2⁡ x + cos^2 x)

To explain: \(\frac{d}{dx}(\frac{cos⁡x}{sec⁡x \,tan⁡x}) = \frac{d}{dx}(\frac{cos⁡x}{\frac{1}{cos⁡x}\frac{sin⁡x}{cos⁡x}})\)

=\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\)

Using quotient rule, we KNOW that, \(\frac{d}{dx} (\frac{F}{G}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)

Here, f = cos^3⁡ x and g = sin ⁡x

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{sin⁡x.\frac{d}{dx} (cos^3⁡x)-cos^3⁡x.\frac{d}{dx}(sinx)}{sin^2⁡x}\)

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{sin⁡x (-3 cos^2⁡ x \,sin⁡ x) – cos^3⁡x(cos⁡x)}{sin^2⁡x}\)

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{-3 sin^2⁡x \,cos^2⁡x – cos^4⁡x}{sin^2⁡x}\)

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = -cot^2⁡ x(3 sin^2⁡ x + cos^2 x)

Right CHOICE is (a) 1

To explain: This is of the form \(\frac{0}{0}\), so we USE L’Hospital’s rule.

= \(\lim\limits_{x \RIGHTARROW \INFTY}\frac{\frac{-2}{x^2}cos⁡\frac{2}{x}}{\frac{-2}{x^2}}\)

= \(\lim\limits_{x \rightarrow \infty}\)cos\(\frac{2}{x}\)

= 1

Right option is (b) (1/2)^e

To EXPLAIN I would say: Let, f(x) = sinx^(sinx)

So, f(x) = e^sinx log sinx

So, on differentiating it we get,

f’(x) = sinx^(sinx) [cosx log sinx + cos x]

So, f’(x) = 0 when, x = sin^-1(1/e) or x = π/2

Also, f”(x) = sinx^(sinx)[cos^2x(1 + log sinx)^2 – sinx log sinx + (cos^2x/sinx) – sinx]

So that, f”(sin^-1(1/e)) = (e – 1/e)e^-1/e

And, f”(π/2) = -1

Hence, f(x) has local as well as global MINIMUM at x = sin^-1(1/e) also have local and global maximum at x = π/2

Global minimum value of f(x) is (1/e)^1/e

It therefore FOLLOWS that ,

sin π/6 ^(sinπ/6) > (1/e)^1/e

=>(1/2)^1/2 > (1/e)^1/e

=>(1/2)^e > 1/e^2

Correct answer is (b) -1

The EXPLANATION: Here, f(x) = 0∫^x (t + 1)(e^t – 1)(t – 2)(t + 4) dt

So, f’(x) = (x + 1)(e^x – 1)(x – 2)(x + 4)

So, ────^+───-4|─────^–──────-1|──────^+──────0|───────^–──────2|────^+────

Clearly, x = -1 and x = 2 are POINTS of LOCAL minima.

As, x = 2 is not present so it is -1.