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As we know that polynomial functions are always continuous and differentiable, we also know that the point of derivability is a sharp corner or a stop on the on going curve.

Hence f(x) = │x│ is a function which is non-derivable at a point which is the sharp corner, at that point, there are more than one slopes possible due to this it is non-derivable but the modulus function is always continuous.

So a function which is non derivable at exactly 5 points and continuous always is,

f(x) = │x - 1│ + │x - 2│ + │x - 3│ + │x - 4│ + │x - 5│

this can be done by drawing the graph of the function or by algebraic method also.

We can choose any value along with x in the modulus function as we only need the points.

To draw the graph, we will solve the function by taking some points or intervals to open the modulus one by one.

Answer is (a) f(a) – af'(a) 

(a) 6

f(x) = x|x| 

f(x) = x(-x) ⇒ f(x) = – x² 

f ‘(x) = -(2x)

f ‘(-3) = -(2) (-3) = 6

(b) 1

f(x) = x + 2 

f'(x) = 1 

f'(x) (at x = 4) = 1

(c) 3 

y = mx + c 

dy/dx = m 

y = x + c (i.e.) f(x) = x + c 

y(at x = 0) = f(0) 0 + c = 1 ⇒ c = 1 

y = x + 1 ⇒ f(x) = x + 1 

f(2) = 2 + 1 = 3

y = e^x sin x 

⇒ y = uv’ + vu’ 

Now u = e^x ⇒ u’ = (du/dx) e^x 

v = sin x ⇒ v’ = (dv/dx) = cos x 

i.e. y’ = e^x (cos x) + sin x (e^x) 

= e^x [sin x + cos x]

(i) dy/dx = -sin x = 2 (sec² x) 

= – sin x – 2 sec² x

(ii) g(t) = t³ cos t (i.e.) u = t³ and v = cos t 

Let u’ = du/dx and v’ = dv/dx = (-sin t)

g'(t) = uv’ + vu’ 

g'(t) = t³ (-sin t) + cos t (3t²)

= -t³ sin t + 3t² cos t

g(t) = 4 sec t + tan t

g'(t) = 4(sec t tan t) + sec² t 

= 4sec t tan t + sec² t

(i) dy/dx = cos x + (-sin x) = cos x – sin x

(ii) f(x) = x sin x

f(x) = uv 

⇒ f'(x) = uv’ + vu’ = u(du/dx) + v(dv/dx) 

Now u = x ⇒ u’ = 1 

v = sin x ⇒ v’ cos x 

f'(x) = x (cos x) + sin x(1) 

= x cos x + sin x

f(x) = x – 3 sin x 

= f'(x) = 1 – 3(cos x) 

= 1 – 3 cos x

y = (3 sec x – 4 cosec x) (2 sin x + 5cos x) 

Let u = 3 sec x-4 cosec x and v = 2 sin x + 5 cos x 

u’ = 3 (sec x tan x) – 4 (-cosec x cot x) ; v’ = 2 (cos x) + 5 (- sin x) 

u’ = 3 sec x tan x + 4 cosec x cot x); v’ = 2 cos x – 5 sin x. 

∴ y’ = uv’ + vu'

So dy/dx = (3 sec x – 4 cosec x) (2 cos x – 5 sin x) + (2 sin x + 5 cos x) (3 sec x tan x + 4 cosec x cot x) 

= 6 sec x cos x – 15 sec x sin x – 8 cosec x cos x + 20 cosec x sin x + 6 sin x sec x tan x + 8 sin x cosec x cot x + 15 cos x sec x tan x + 20 cos x cosec x cot x 

= 6 (1/cos x) cos x – 15 (1/cos x) sin x – 8 (1/sin x) cos x + 20 (1/sin x) sin x + 6 sin x (1/cos x) tan x + 8 sin x (1/sin x) cot x + 15 cos x (1/cos x) tan x + 20 cos x (1/sin x) cot x 

= 6 – 15 tan x – 8 cot x + 20 + 6 tan² x + 8 cot x + 15 tan x + 20 cot² x 

= 26 + 6 tan² x + 20 cot² x

y = sin³ x + cos³ x 

Here u = sin³ x = (sin x)³ 

⇒ du/dx = 3 (sin x)² (cos x) 

= 3sin² x cos x 

v = cos³ x = (cos x)³ 

⇒ dv/dx = 3 (cos x)² (-sin x) = -3 sin x cos² x 

Now y = u + v ⇒ dy/dx = du/dx + dv/dx

= 3 sin² x cos x – 3sin x cos² x 

= 3 sin x cos x (sin x – cos x)

y = e^{tan^-1 x} 

y = e^{tan^-1 x} 1/(1 + x²)

⇒ y’ = y/(1 + x²) ⇒ y'(1 + x²) = y 

Differentiating w.r.to x 

y’ (2x) + (1 + x²) (y”) = y’ 

(i.e.) (1 + x²) y” + y’ (2x) – y’ = 0 

(i.e.) (1 + x²) y” + (2x – 1) y’ = 0

y = 3 sin x + 4 cos x – e^x 

dy/dx = 3 (cos x) + 4 (- sin x) – (e^x) 

= 3 cos x – 4 sin x – e^x

y = sin 5 + log₁₀ x + 2 sec x

dy/dx = 0 + (1/x) log₁₀ e + 2[sec x + tan x] 

= log₁₀ e/x + 2 sec x tan x

(i) y = tan 3x 

Put u = 3x

du/dx = 3

Now y = tan u

du/dx = sec² u

So dy/dx = (dy/du) x (du/dx) = sec² u (3)

= 3 sec² 3x

(ii) y = cos (tan x)

Put u = tan x 

du/dx = sec² x 

Now y = cos u ⇒ du/dx = – sin u 

Now dy/dx = (dy/du) x (du/dx) 

= (-sin u) (sec² x) 

= – sec² (sin (tan x))

y = (3x² + 1)² = (3x² + 1) (3x² + 1) 

Let u = 3x² + 1 and v = 3x² + 1 

∴ u’ = 3(2x) = 6x and v’ = 6x 

y’ = uv’ + vu’ 

(i.e.,) dy/dx = (3x² + 1) (6x) + (3x² + 1) 6x = 12x (3x² + 1)

Let = u = x² + 4x + 6

du/dx = 2x + 4

Now y = u⁵ 

dy/dx = 5u⁴

∴ dy/dx = (dy/du) x (du/dx) = 5u⁴(2x + 4)

= 5(x² + 4x + 6)⁴ (2x + 4) 

= 5(2x + 4) (x² + 4x + 6)⁴

Let u = x⁴ – 6x³ + 7x² + 4x + 2 and v = x³ – 1 

u’ = 4x³ – 6(3x²) + 7(2x) + 4(1) + 0 

= 4x³ – 18x² + 14x + 4 

v’ = 3x³ 

y = uv’ + vu’ 

i.e. dy/dx = (x⁴ – 6x³ + 7x² + 4x + 2) (3x²) + (x³ – 1)

(4x³ – 18x² + 14x + 4) 

= 3x⁶ – 18x⁵ + 21x⁴ + 12x³ + 6x² + 4x⁶ – 18x⁵ + 14x⁴ + 4x³ – 4x³ + 18x² – 14x – 4 

= 7x⁶ – 36x⁵ + 35x⁴ + 12x³ + 24x² – 14x – 4

y = x² e^x sin x 

Let u = x², v = e^x and w = sin x 

u’ = 2x, v’ = e^x and w’ = cos x 

y’ = uvw’ + vwu’ + uwv’ 

= (x²e^x) cos x + (e^x sin x)(2x) + (x² sin x)e^x 

= x² e^x cos x + 2xe^x sin x + x² e^x sin x 

= xe^x {x cos x + 2 sin x + x sin x}

y = e^-x log x = uv (say) 

Here u = e^-x and v = log x 

⇒ u’ = -e^-x and v’ = 1/x

Now y = uv ⇒ y’ = uv’ + vu’

(i.e.) dy/dx = e^-x (1/x) + log(-e^-x)

= e^-x(1/x - log x)

(a) e^x .x⁴ (x + 5) 

y = e^{x + 5 log x} = e^x.e^{5 log x} = e^x.e^{log x^5}  

= x⁵ e^x 

∴ dy/dx = x⁵ (e^x) + e^x (5x⁴) 

= e^x.x⁴ (x + 5)

(d) 2

y = (ax – 5)e^3x 

dy/dx = y’ = (ax – 5) (3e^3x) + e^3x (a) 

= e^3x[3ax – 15 + a] 

Given dy/dx = -13 at x = 0 

⇒ [-15 + a] = -13 

⇒ a = -13 + 15 

a = 2

Answer is (d) 0

y = cos (sin x²) 

dy/dx = – sin (sin x²) [cos (x²)] (2x) 

∴ dy/dx at x = √π/2 = -sin (1) [0] = 0

No, a function may be continuous at a point but not differentiable at that point.

For example,

Function f(x)=|x| is continuous at x = 0 but not differentiable at x =0.