CUET Mathematics 2024 - If f(x), defined by f(x)=\arraylllk x+1 & if & x ≤ \\ x & if & x>. is continuous at x=, then the value of k is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Continuity & Differentiability

Easy

If $f(x)$ , defined by $f(x)=\left\{\begin{array}{lll}k x+1 & \text { if } & x \leq \pi \\ \cos x & \text { if } & x>\pi\end{array}\right.$ is continuous at $x=\pi$ , then the value of $k$ is :

0

\pi

\frac{2}{\pi}

-\frac{2}{\pi}

✅ Correct Option: 4

Left-hand limit as

x

approaches

\pi

\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-}(kx + 1) = k\pi + 1

Right-hand limit as

x

approaches

\pi

\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} \cos x = \cos \pi = -1

For continuity at

x = \pi

, these limits must be equal:

k\pi + 1 = -1

k\pi = -2

k = \dfrac{-2}{\pi}

Verification: When

x = \pi

Left side:

f(\pi) = k\pi + 1 = \dfrac{-2}{\pi} \cdot \pi + 1 = -2 + 1 = -1

Right side:

\cos(\pi) = -1

Since both sides equal

-1

, the value of

k

\dfrac{-2}{\pi}

Related questions:

CUET Mathematics 2024

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List-I	List-II
(A) $\|x - 1\| + \|x - 2\|$	(I) is differentiable everywhere except at $x = 0$
(B) $x - \|x\|$	(II) is continuous everywhere
(C) $x - [x]$	(III) is not differentiable at $x = 1$
(D) $x \, \|x\|$	(IV) is differentiable at $x = 1$

CUET Mathematics 2024

5^{\mathrm{x}}

CUET Mathematics 2024

t=e^{2 x}

y=\log _{e} t^{2}

\frac{d^{2} y}{d x^{2}}