Q.1. If then find the value of a.
Ans. 1
Since, the given limit is in the form of 0/0, we will use L’ Hospital’s Rule
It is satisfied only when a = 1.
Q.2.
Ans. 0
Q.3.
Ans. 1
where α = x^{4 } + x^{2} = 1
Q.4. The value of is
Ans. 3
Q.5. The value of must be
Ans. 8
Q.6. Let f(x) = [x]+ 1  x, 1 ≤ x ≤ 3 and [x] is the largest integer not exceeding x. The number of points in [1, 3] where f is not continuous is
Ans. 4
f(x) = [x] + 1  x, 1 ≤ x ≤ 3
= x if 1 ≤ x < 0
=1  x if 0 ≤ x <1 = x if 1 ≤ x < 2
= x +1 if 2 ≤ x < 3 = 5 if x = 3
Clearly f is not continuous at x = 0, 1, 2 and 3
Q.7. Let f : R → R be a differential function satisfying for all real x and y and f'(2) = 2. If g(x) = f( x ) 3 for all x ∈ R , then for g(x) total nondifferentiable point are
Ans. 3
⇒ f '(2) = f'(0) = 2 (∵ f'(2) = 2) ⇒ f'(x) = 2 ⇒f(x)= 2x + c
Put x = y = 0 in
Now, from equation (i), f(0) = 0 + c = 2
∴ c = 2
From equation (i), f(x) = 2x + 2
So, function g(x) = 2x  1, hence the points of nondifferentiability of g(x) are
x = ±(1/2), 0
Q.8. Let f(x) = g(x) = f (x 1) + f (x + 1). Then the number of points where g is not differentiable is
Ans. 5
⇒ (x + y)2007 = xy ⇒ xy  2007x  2007y = 0
(x  2007)(y  2007) = 2007^{2 }= 3^{4 }x 223^{2}
The number of pairs is equal to the number of divisors of 2007^{2} that is (4 + 1) x (2 + 1) = 15
Since, x < y, so required number of pairs = 7
Q.9. Let f (x) = If m is the number of points in [0, 3]where f is discontinuous and n is the number of points in [0, 3]where f is not differentiable, then the value of m + n is
Ans. 5
f is discontinuous at x = 1 and x = 3 only
Also, f'(0) = 0 and f is not derivable at x = 2
So m = 2 and n = 3.
Q.10. Let f(x) = x tan^{1}(x^{2}) + x^{4}. Let f^{k}(x) denotes kth derivative of f(x) w.r.t. x, k ∈ N. If f ^{2m}(0) ≠ 0 m ∈ N then m equals
Ans. 2
Let g(x) = x tan^{1} (x^{2}). It is an odd function. So, g^{2m} (0) = 0.
Let h(x) = x^{4}.
So, f(x) = g(x) + h(x) ⇒ f^{2m} (0) = g^{2m}(0) + h^{2m}(0) = h^{2m} (0) ≠ 0.
It happens when 2m = 4 ⇒ m = 2.
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