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# Fill in the Blanks: Limits, Continuity and Differentiability | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Fill in the Blanks

Q. 1.

be a real-valued function. Then the set of points where f(x) is not differentiable is .......................

Ans. 0

Solution.

We know that | x | is not differentiable at x = 0

is not differentiable at x = 0.

At all other values of x, f(x) is differentiable.
∴ The req. set of points is {0}.

Q. 2.

If f(x) is continuous for all x, then k = ..........

Ans. k = 7

Solution. It will be continuous at x = 2 if

Q. 3. A discontinuous function y =f(x) satisfying x2 +y= 4 is given by f(x) = .......................

Ans.

Solution.

By choosing any arcs of circle x2 – y2 = 4, we can define a discontinuous function, one of which is

Q. 4.

Ans. 2/π

Solution. KEY CONCEPT

(L' Hospital rule)

Q. 5. If f(x) = sin x, x ≠ np,n = 0, ± 1, ± 2,± 3, ..............   = 2, otherwise
and g(x) = x2 + 1,x ≠ 0,2
= 4, x = 0
= 5, x = 2,

Ans. 1

Solution. Given that,
f (x) = sin x, x ≠ p,n = 0, ±1,± 2,... = 2, otherwise
And g(x) = x2 + 1, x ≠ 0,2
= 4, x = 0 = 5, x = 2

Q. 6.

Ans. -1

Solution.

Q. 7.

Ans. 4

Solution.

Q. 8. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC then the triangle ABC has perimeter

Ans.

Solution. In Δ ABC, AB = AC
AD ⊥ BC (D is mid pt of BC)
Let r = radius of circumcircle
∴ OA = OB = OC = r

Q. 9.

Ans. e5

Solution.

Q. 10. Let f(x) = x | x |. The set of poin ts wh ere f(x) is twice differentiable is .....................

Ans. R – {0}

Solution. We have,

Clearly f " (x) exists at every pt. except at x = 0
Thus f (x) is twice differentiable on R – {0}.

Q. 11. denotes the greatest integer function. The domain of f is... and the points of discontinuity of f in the domain are.....

Ans. (-∞, - 1) ∪ [0, ∞), I – {0} where I is the set of ingeger except n = –1

Solution. Thus function is not defined for those values of x for which [x + 1] = 0.
In other words it means that
0 < x + 1 < 1 or – 1 < x < 0 .......(1)
Hence the function is defined outside the region given by (1).
Required domain is ] - ∞, -1[∪[0,∞] Now, consider integral values of x say x = n

Clearly RHL ≠ LHL. Hence the given function is not continuous for integral values of n (n ≠ 0, –1).

At x = 0, f (0) = 0,

The function is not defined for x < 0. Hence we cannot find lim f (0 – h). Thus f (x) is continuous at x = 0. Hence the points of discontinuity are given by I – {0} where I is set of integers n except n = –1

Q. 12.

Ans. e2

Solution. KEY CONCEPT

Q. 13. Let f(x) be a continuous function defined for 1 < x < 3. If f(x) takes rational values for all x and f (2) = 10, then f(1.5) =.............

Ans. 10

Solution. Since f (x) is given continuous on the closed bounded interval [1, 3], f (x) is bounded and assumes all the values lying in the interval [m, M] where

m = min f (x) and M = max f (x)

1 < x < 3 ⇒ f (1) < f (x) < (3)

must assume all the irrational values lying in the [m, M]. But since f (x) takes only rational values, we must have m = M i.e., f (x) must be a constant function.
As f (2) = 10, we get

True / False

Q. 1.  exists then both    and

Ans. F

Solution.

then   exists but neither

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