Let

be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function).

Question

Let $[t]$ denote the greatest integer less than or equal to $t$.
If the function
\[ f(x)= \begin{cases} b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x < 0 \\ \dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases} \] is continuous at $x=0$, then $a^2+b^2$ is equal to

Answer 1

Step 1: Evaluate the left hand limit at x = 0

LHL = lim_x→0⁻ (sin x − sin 2x) / x³

Using identity:

sin x − sin 2x = sin x (1 − cos x)

So,

LHL = lim_x→0 [ (sin x / x) · ((1 − cos x) / x²) ]

= (lim_x→0 sin x / x) · (lim_x→0 (1 − cos x) / x²)

= 1 × 1/2

LHL = 1/2

Step 2: Apply continuity at x = 0

Given f(0) = a,

a = 1/2

⇒ a² = 1/4

Step 3: Evaluate the right hand limit at x = 0

As x → 0⁺,

(sin x + cos x) cos x → 1

Hence,

RHL = b² sin(π/2)

= b²

Step 4: Use continuity condition

RHL = LHL

b² = 1/2

Step 5: Required value

a² + b² = 1/4 + 1/2

= 3/4

Final Answer:

The value of (a² + b²) is
3/4

Let

be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function).

Show Hint

Correct Answer: 34

Solution and Explanation

Top Questions on Continuity

Questions Asked in JEE Main exam

Let be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function).

Show Hint

Correct Answer: 34

Solution and Explanation

Top Questions on Continuity

Questions Asked in JEE Main exam

Let

be a continuous function at $x=0$, then the value of $(a^2+b^2)$ is (where $[\ ]$ denotes greatest integer function).