CUET UG 2025 — Mathematics Calculus

For the function to be continuous at $x = 0$x=0:

$$\lim_{x \to 0} f(x) = f(0) = K$$x→0lim​f(x)=f(0)=K

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Finding the limit as $x$x approaches $0$0:

$$\lim_{x \to 0} \left(\frac{\sin 2x}{x} + \cos x\right)$$x→0lim​(xsin2x​+cosx)

$$= \lim_{x \to 0} \frac{\sin 2x}{x} + \lim_{x \to 0} \cos x$$=x→0lim​xsin2x​+x→0lim​cosx

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For $\displaystyle \lim_{x \to 0} \frac{\sin 2x}{x}$x→0lim​xsin2x​:

$$\frac{\sin 2x}{x} = \frac{2 \sin 2x}{2x}$$xsin2x​=2x2sin2x​

$$= 2 \cdot \frac{\sin 2x}{2x}$$=2⋅2xsin2x​

Using $\displaystyle \lim_{u \to 0} \frac{\sin u}{u} = 1$u→0lim​usinu​=1 where $u = 2x$u=2x:

$$\lim_{x \to 0} \frac{\sin 2x}{x} = 2 \times 1$$x→0lim​xsin2x​=2×1

$$= 2$$=2

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For $\displaystyle \lim_{x \to 0} \cos x$x→0lim​cosx:

$$\lim_{x \to 0} \cos x = \cos(0)$$x→0lim​cosx=cos(0)

$$= 1$$=1

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Combining the results:

$$\lim_{x \to 0} f(x) = 2 + 1$$x→0lim​f(x)=2+1

$$= 3$$=3

Since the function is continuous at $x = 0$x=0:

$$K = 3$$K=3