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NCERT Exemplar · Q54

Q.State True or False: The minimum value of nn for which tan⁡−1nπ>π4\tan^{-1}\frac{n}{\pi}>\frac{\pi}{4}, n∈Nn\in N, is valid is 5.

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The inequality tan⁡−1nπ>π4\tan^{-1}\frac{n}{\pi} > \frac{\pi}{4} simplifies to n>πn > \pi. Since nn is a natural number, the smallest such nn is 4, not 5. Hence the statement is False.

Concept and Intuition

The core idea here is simple: the inverse tangent function tan⁡−1x\tan^{-1}x is strictly increasing for all real xx. That means if you want tan⁡−1(something)\tan^{-1}(\text{something}) to exceed π4\frac{\pi}{4}, that "something" must be greater than the value whose tangent is exactly π4\frac{\pi}{4}.

What value gives tan⁡−1x=π4\tan^{-1}x = \frac{\pi}{4}? It's x=1x = 1, because tan⁡π4=1\tan\frac{\pi}{4} = 1. So the inequality tan⁡−1nπ>π4\tan^{-1}\frac{n}{\pi} > \frac{\pi}{4} is equivalent to nπ>1\frac{n}{\pi} > 1, provided we are careful about the domain (which is all real numbers here, so no issues).

Once we have n>πn > \pi, the smallest natural number nn satisfying this is n=4n = 4, since π≈3.14\pi \approx 3.14. The statement claims it's 5 — that's off by one.

Step-by-Step Solution

  1. Set up the inequality We are given:

tan⁡−1nπ>π4,n∈N\tan^{-1}\frac{n}{\pi} > \frac{\pi}{4}, \quad n \in \mathbb{N}

  1. Apply the monotonicity of tan⁡−1\tan^{-1} The function tan⁡−1x\tan^{-1}x is strictly increasing on R\mathbb{R}. Therefore, for any a,ba, b:

tan⁡−1a>tan⁡−1b  ⟺  a>b\tan^{-1}a > \tan^{-1}b \iff a > b

Here, take a=nπa = \frac{n}{\pi} and b=1b = 1 (since tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}).

So the inequality becomes:

nπ>1\frac{n}{\pi} > 1

  1. Solve for nn Multiply both sides by π\pi (positive, so inequality direction stays):

n>πn > \pi

  1. Find the smallest natural number π≈3.14159…\pi \approx 3.14159\ldots …

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