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Let y = tan^-1(sinx/(1 + cosx)) = tan^-1((2sin(x/2) cos(x/2))/2cos²(x/2))

= tan^-1(sin(x/2)/cos(x/2)) = tan^-1(tan(x/2)) = x/2

Differentiating both sides w.r.t. x, we get

dy/dx = 1/2

Option : (b)

Formula : - 

(i) standard limit  \(\lim\limits_{x \to 0}\frac{log(1-x)}{x}\) = 1

(ii) A function f(x) is said to be continuous at a point x=a of its domain, if \(\lim\limits_{x \to a}f(x)\) = f(a)

 \(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a)

(iii) \(\lim\limits_{x \to a}{\{f(x)±g(x)}\}\) = 1 ± m,

Where \(\lim\limits_{x \to a}f(x)\) = 1, \(\lim\limits_{x \to a}g(x)\) = m

Given :-

f(x) =\(\begin{cases} \frac{log(1+ax)-log(1-bx)}{x} &, \quad x ≠{0}\\ k&, \quad \text x =0 \end{cases} \) 

And,

f(x) is continuous at x = 0

\(\lim\limits_{x \to 0}\frac{log(1+ax)-log(1-bx)}{x} \) = k

Using formula (ii),

\(a\lim\limits_{x \to 0}\frac{log(1+ax)}{ax} \) - \(b\lim\limits_{x \to 0}\frac{log(1-bx)}{bx} \) = k

Using formula (i),

a + b = k

Given function is f(x) = x² +2 x − 8, x ∈ [ −4, 2]

Since, a polynomial function is continuous and derivable on R, therefore 

(i) f(x) is continuous on [− 4,2]. 

(ii) f(x) is derivable on (− 4,2). 

Also, f(− 4) = (− 4)² + 2 (− 4) − 8 = 0 

(since f (x) = x² + 2x − 8)

and f(2) = 2² + 2 x 2 – 8 = 0 

⇒ f(− 4) = f(2) 

This means that all the conditions of Rolle's theorem are satisfied by f(x) in [− 4,2]. 

Therefore, it exists at least one real c ∈ [ −4, 2] such that f′(c) = 0. 

Now, f (x) = x² + 2x − 8 ⇒ f'(x) = d/dx(x² + 2x - 8) = 2x + 2 

Putting f′(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = − 1. 

Thus, f′( −1) =0 and − 1 ∈ (− 4,2). 

Hence, Rolle's theorem is verified with c = − 1.

Here, f(x) = x² − 4x − 3, x ∈ [1, 4] which is a polynomial function, so it is continuous and derivable at all x ∈ R, therefore 

(i) f(x) is continuous on [1, 4] 

(ii) f(x) is derivable on (1, 4). 

Therefore, Conditions of Lagrange's theorem are satisfied on [1, 4]. 

Hence, there is atleast one real number. c ∈ (1, 4) such that

Now, f'(x) = d/dx(x² - 4x - 3) = 2x - 4

⇒ f(4) = 4² - 4(4) - 3 = 16 - 16 - 3 = - 3

and f(1) = 1 - 4 - 3 = - 6

∴ f'(c) =(f(4) - f(1))/(4 - 1) = (-3 - (-6))/3 = (-3 + 6)/3 = 1

⇒ 2c - 4 = 1 ⇒ 2c = 1 + 4 = 5 ⇒ c = 5/2 ∈ (1, 4)

We have,

y = e^{sin x}

⇒ \(\frac{dy}{dx}=e^{sinx} .cosx\)

^∴ \(\frac{dy}{dx}|_\frac{\pi}{2}=e^1 .0=0\)

y = \(e^{x^3}\)

∴ \(\frac{dy}{dx} = e^{x^3}\) x 3x² x 1 

= 3x² . \(e^{x^3}\) 

Yes, every differentiable function is continuous. 

No, some continuous functions may not be differentiable.

Let y = \(\sqrt{e^{\sqrt x}}\)

∴ \(\frac{dy}{dx} = \frac{1}{2\sqrt{e^\sqrt x}}\times\; e^\sqrt x \times \;\frac{1}{2\sqrt x}\)