# Important Questions for Class 12 Chapter 5 - Continuity and Differentiability

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**Important questions based on NCERT syllabus for Class 12 Chapter 5 - Continuity and Differentiability:**

*Question 1*: Find the value of k, for which f(x) = (√1+kx - √1-kx)/x if -1≤ x<0

(2x+1)/(x-1) if 0≤ x<1

is continuous at x = 0

*Solution*:

f(x) = (√1+kx - √1-kx)/x if -1≤ x<0

(2x+1)/(x-1) if 0≤ x<1

is continuous at x = 0

Now, f(0) = 2 x 0+ 1/0-1 = 1/-1 = -1

LHL = lim h->0 f(0-h)

= lim h-> 0 (√1 - kh - √1 + kh)/-h

= lim h-> 0 (√1 - kh - √1 + kh)/-h (√1 - kh + √1 + kh/ √1 - kh + √1 + kh)

= lim h-> 0 (1-kh)(1+kh)/-h (√1 - kh + √1 + kh)

Since (a+b)(a-b) = a^2 - b^2

= lim h->0 2kh/ -h (√1 - kh + √1 + kh)

= lim h-> 0 2k/√1 - kh + √1 + kh

= 2k/(1 + 1) = 2k/2 = k

Since f(x) is continuous at x = 0

Therefore, f(0) = LHL => -1 = k

=> k = -1

*Question 2*: Find the value of a for which the function f is defined as f(x) = asin pi/2(x+1), x<=0

(tan x - sin x)/x^3, x>0

is continuous at x = 0

*Solution*: Given

f(x) = asin pi/2(x+1), x<=0

(tan x - sin x)/x^3, x>0

Also given that f(x) is continuous at x = 0

Therefore LHL = RHL = f(0)......(1)

Consider LHL = lim x-> 0 a sin pi/2 (x+1)

LHL = lim h->0 a sin pi/2(-h + 1) [put h = 0-h = -h]

= lim h->0 (a sin[pi(-h+1)/2])/(pi(-h+1)/2) * pi(-h+1)/2

Since multiplying numerator and denominator by pi(-h+1)/2

= lim h->0 a pi(-h+1)/2

Since lim x->0 sin x/x = 1

=> LHL = a pi/2

Now consider RHL = lim x -> 0 (tan x - sin x)/x^3

Put x = 0+ h = h, we get

RHL = lim h-> 0 (tan h - sin h)/h^3 = lim h->0 (sin h/cos h - sin h)/h^3

= lim h->0 (sin h - sin h cos h)/h^3 cos h = lim h->0 (sin h(1 - cos h))/h^3 cos h

= lim sin h/h . lim h-> 0 (1-cos h)/h^2. lim h-> 0 1/cos h

= 1 * lim h->0 (1-cos h)/h^2 * 1

since lim h-> 0 sin h/h = 1 and lim h->0 1/cos h = 1/cos 0 = 1/1 = 1

= lim h->0 (1-cos h)/h^2

= lim h->0 (2sin^2 h/2)/h^2 [since 1-cosx = 2sin^2 x/2]

= lim h->0 (2 * sin^2 h/2)/(h^2/4 * 4) = lim h->0 2/4 * lim h->0 (sin^2 h/2)/h^2/4

[since multiply and divided by 4]

= 1/2 * lim h->0 [sin h/2)/(h/2) = 1/2 * 1 * 1

[since lim x->0 sin x/x = 1]

=> RHL = 1/2

Now from equation (1), we have LHL = RHL

Therefore api/2 = 1/2 => api = 1=> a = 1/pi