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f(b) = pb+3

\(\lim\limits_{x\to b^+} f (x)\) = 5b²-q

∴ pb + 3= 5b²-q

∴ p = 5b²-q-3/b is the required condition.

f(x) = x² + 5, x ∈ R 

∴ f(1) = 1 + 5 = 6 

If f(x) = ax + b is continuous at x = 1, then 

f(1) = \(\lim\limits_{x\to1}(ax+b) =a +b\) 

∴ 6 = a + b where, a, b ∈ R 

∴ There are infinitely many values of a and b.

Let f(x) = x³ – 3x 

f(x) is a polynomial function and hence it is continuous for all x ∈ R. 

A root of f(x) exists, if f(x) = 0 for at least one value of x. 

f(1) = (1)³ – 3(1) = -2 < 0 

f(2) = (2)³ – 3(2) = 2 > 0 

∴ f(1) < 0 and f(2) > 0 

∴ By intermediate value theorem, there has to be point ‘c’ between 1 and 2 such that f(c) = 0. 

There is a root of the given equation between 1 and 2

Let f(x) = 2x³ – x – 16 

f(x) is a polynomial function and hence it is continuous for all x ∈ R. 

A root of f(x) exists, if f(x) = 0 for at least one value of x. 

f(2) = 2(2)³ – 2 – 16 = -2 < 0 

f(3) = 2(3)³ – 3 – 16 = 35 > 0 

∴ f(2) < 0 and f(3) > 0 

∴ By intermediate value theorem, there has to be point ‘c’ between 2 and 3 such that f(c) = 0. 

∴ There is a root of the given equation between 2 and 3.

Left Hand Limit: = \(\lim\limits_{x \to2^-} \)f(x) =  \(\lim\limits_{x \to2^-} \) x² 

= 4

Right Hand Limit: = \(\lim\limits_{x \to2^+} \)f(x) =  \(\lim\limits_{x \to2^+} \) x² 

= 4 

ƒ(2) = 4

Since,   \(\lim\limits_{x \to2} \) f(x) = f(2) 

f is continuous at x = 2.

sin (A + B) = sin A cos B + cos A sin B 

Diff : w:r.to B

cos(A + B) x \(\frac{dA}{dB}\)= (sinA x (- sinB) + cosA x cosB) x \(\frac{dA}{dB}\)

cos(A +b) = cosA cosB – sinA sinB 

Hence cosine formulae

Correct answer is

(A) f is continuous at x = -1.

Let y = sin (ax + b) 

\(\frac{dy}{dx}\) = cos (ax + b)x a = a cos (ax + b)

\(y = \frac{sin(ax + b)}{cos(cx + d)}\)

\(\frac{dy}{dx} = \frac{cos(cx + d)\times cos(ax + b)\times a - sin(ax + b) \times - sin(cx + d) \times c}{cos^2(cx + d)}\)

\(= \frac{acos(cx + d)cos(ax + b) + csin(ax + b)sin(cx + d)}{cos^2(cx + d)}\)

Let y = cos (sin x) 

\(\frac{dy}{dx}\)= – sin (sin x)x cos x – cos x sin(sin x)

(i) Given; g(x) = 1 – x + |x| ⇒ g(x) (1 – x) + |x|

Here g(x) is the sum of two functions continuous functions hence continuous.

(ii) We have;

fog(x) = f(gx))

= f(1-x+|x|) = |1-x-|x|| = h(x)

The composition of two continuous functions is again continuous. Therefore h(x) continuous.

(i) f(x) = |x] being greatest integer function it is not differentiable at integral points. 

Hence not continuous on integral points 

∴ f (x) is not continuous [5, 9] 

Rolle’s theorem is not applicable to f (x) in [5,9] 

How ever f’ (x) = 0 ∀ non integral values in [5, 9] thus converse of Rolle’s theorem is not true.

(ii) f (x) = |x] being greatest integer function it is not differentiable at integral points. 

Hence not continuous on integral points 

∴ f (x) is not continuous [-2, 2] 

Rolle’s theorem is not applicable to f (x) in [-2,2] 

How ever f’ (x) = 0 ∀ non integral values in [-2, 2] thus converse of Rolle’s theorem is not true.

(iii) f (x) = x² – 1 being polynomial function f(x) is differentiable ∀ x ∈ [1, 2] 

∴ f (x) is continuous is [1, 2] 

Also f’ (x) exist, f (x) is derivable on (1, 2) 

f(1) = (1)² – 1 = 0, f(2) = (2)² – 1 =3 

f(1) ≠ f(2) 

Since all the conditions are not satisfied, the Rolle’s theorem is not applicable in the given intervals. 

Thus the converse of Rolle’s theorem does not hold. That is if the condition of Rolle’s theorem are not satisfied by a function f (x) on some interval [a, b] then f’ (x) may or may not be zero in some point in (a, b).

The domain is R Let g(x) = |x|, and h(x) = |x +1| 

∴ g(x)-h(x)= |x| – |x + 1| = f(x) 

g(x) and h(x) are continuous on R 

∴ g(x) – h(x) are also continuous on R 

f(x) is continuous on R. 

∴ So no points of discontinuity present on R.

Let g (x) = sin x and h (x) = |x|. 

Both g (x) and h (x) are continuous on R 

goh (x) = g |x| = sin |x| = f (x) is continuous on R.

Consider f(x) = |x| and g(x) = cos x

We know that

(gof)(x) = g[f(x)] = g[|x|] = cos |x|

Here, f and g is continuous and their composite (gof) is continuous.

Therefore, cos |x| is a continuous function.

Answer is (d)

Left hand limit

f (3 – 0) = lim_h→0 f (3 – h)

= lim_h→0 (3 – h) + λ

= (3 – 0) + λ

= 3 + λ

∵ At x = 3, function is continuous then

f (3 – 0) = f (3)

3 + λ = 4

⇒ λ = 4 – 3

λ = 1

The domain of f (x) is R 

Let h (x) = cos x, g (x) = |x| 

goh (x) = g (cos x) = |cos x| = f (x) 

cos x and |x| are continuous on R as they are trigonometric function and modulus function. g(x) and h (x) are continuous and hence 

goh (x) is also continuous, hence |cosx| is continuous on R.

Let h (x) = cos x, g (x) x² , x ∈ R 

hog (x) = h [x²] = cos x² = f (x). 

Being polynomial function x² is continuous on R. 

Being trigonometric function cos x is continuous on R in the domain. 

Since h (x) is continuous and g (x) is continuous hog (x) is also continuous on R. 

∴ f (x) is also continuous on R.

  \(\lim\limits_{x \to0} \)  sin\(\frac{1}{x}\) = 0

 sin\(\frac{1}{x}\) is bounded function between -1 and +1. 

Also, f(0)=0

Since,   \(\lim\limits_{x \to0} \) f(x) = f(0) 

Hence, f is a continuous function.

Here, we will test the continuity of function at x = 0 and 0 ∈ [-1,2].

Left hand limit

f(0 – 0) = lim_h→0 f(0 – h)

= lim_h→0 (0 – h)²

= lim_h→0 h²

= 0

Right hand limit

f (0 + 0) = lim_h→0 f(0 + h)

= lim_h→0 4(0 + h) – 3

= lim_h→0 Ah – 3

= 0 – 3

= – 3

∴ f (0 – 0) ≠ f(0 + 0)

LHL ≠ RHL

So, function is not continuous at x = 0 and x ∈ [- 1, 2]

At x = 1

Left hand limit

f(1 – h) = limh→0 4(1 – h) – 3

= lim_h→0 4 – 3 – Ah

=4 – 3 – 0 = 1

Right hand limit

f(1 + h) = lim_h→0 5(1 + h)² – 4 (1 + h)

= lim_h→0 5(1 + h² + 2h) – (4 + Ah)

= lim_h→0 5h² +10h + 5 – 4 – 4h

= 5 × 0 + 10 × 0 + 1 – 4(0)

= 1

Value of function at x = 1

f(1) = 4 × 1 – 3 = 1

∵ lim_h→0 f(1 – A) = f(1) = lim_h→0 f(1 + h)

∴Function is continuous at x = 1.

Hence, function is not continuous in interval [-1, 2],

Let a be any constant number. 

Then, f(x) = a

f'(x) = \(\lim\limits_{h \to 0}\) \(\frac{f(x+h)-f(x)}{h}\) 

We know that coefficient of a linear function is

 a = \(\frac{y_2-y_1}{x_2-x_1}\)

Since our function is constant, y₁ = y₂ 

Therefore, a = 0 

Now,

f'(x) = \(\lim\limits_{h \to 0}\) \(\frac{a-a}{h}\)  =   \(\lim\limits_{h \to 0}\) \(\frac{0}{h}\)  =   \(\lim\limits_{h \to 0}\)  0 = 0 

Thus, the derivative of a constant function is always 0.

LHL: \(\lim\limits_{x \to3^-} \)f(x) =  \(\lim\limits_{x \to3^-} \)  \(\frac{​​​​x^2-x-6}{x-3}\) 

 = \(\lim\limits_{x \to3^-} \)   \(\frac{​​​(​x+2)(x-3)}{x-3}\)   [By middle term splitting]

  = \(\lim\limits_{x \to3^-} \) x + 2

= 5 

RHL: \(\lim\limits_{x \to3^-} \)f(x) =  \(\lim\limits_{x \to3^-} \)  \(\frac{​​​​x^2-x-6}{x-3}\) 

 = \(\lim\limits_{x \to3^-} \)   \(\frac{​​​(​x+2)(x-3)}{x-3}\)   [By middle term splitting]

 = \(\lim\limits_{x \to3^-} \) x + 2

 = 5

f(3) = 5 

Since, \(\lim\limits_{x \to3} \) f(x) = f(3)

f is continuous at x=3.

LHL: = \(\lim\limits_{x \to5^-} \)f(x) =  \(\lim\limits_{x \to5^-} \) \(\frac{x^2 - 25}{x-5} \)

 = \(\lim\limits_{x \to5^-} \)  \(\frac{(x+5)(x-5)}{x-5} \)[By middle term splitting]

 = \(\lim\limits_{x \to5^-} \)  x + 5

= 10

RHL: = \(\lim\limits_{x \to5^+} \)f(x) =  \(\lim\limits_{x \to5^+} \) \(\frac{x^2 - 25}{x-5} \)

 = \(\lim\limits_{x \to5^+} \)  \(\frac{(x+5)(x-5)}{x-5} \)[By middle term splitting]

 = \(\lim\limits_{x \to5^+} \)  x + 5

= 10 

f(5)= 10 

Since, = \(\lim\limits_{x \to5} \) f(x) = f(5)

f is continuous at x=5.

Correct answer is

(B). 4