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Bihar Board Class 12 Science Mathematics · Ch 2: Inverse Trigonometric Functions

Bihar Board Class 12 Science Mathematics · Ch 2: Inverse Trigonometric Functions is an online chapter test in the Bihar Board Class 12 Science pattern, covering Inverse Trigonometric Functions. Attempt it on SeatBeat with a timer like the real exam, then see a detailed solution for every question and your score and weak-topic analysis.

30
Questions
20
Minutes
30
Marks
None
Negative marking
Unlock all 19 tests — ₹149Try Hindi · Ch 1: Chandradhar Sharma Guleri free →

Topics in this test

TopicQuestions
Inverse Trigonometric Functions30

Difficulty: 7 medium · 23 hard

Inverse Trigonometric Functions MCQ questions with answers

The first 3 of 30 questions. Unlock the series for the rest, each with a solution.

  1. Q1. If $\tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}\tan^{-1}x$ for $x>0$, then the value of $x$ is:

    यदि $x>0$ के लिए $\tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}\tan^{-1}x$ है, तो $x$ का मान है:

    • A. $1$
    • B. $\sqrt3$
    • C. $\frac{1}{\sqrt2}$
    • D. $\frac{1}{\sqrt3}$
    Solution

    Since $\tan^{-1}\frac{1-x}{1+x}=\frac{\pi}{4}-\tan^{-1}x$, the equation becomes $\frac{\pi}{4}-\tan^{-1}x=\frac{1}{2}\tan^{-1}x$, giving $\tan^{-1}x=\frac{\pi}{6}$, so $x=\frac{1}{\sqrt3}$.

    चूँकि $\tan^{-1}\frac{1-x}{1+x}=\frac{\pi}{4}-\tan^{-1}x$ है, समीकरण $\frac{\pi}{4}-\tan^{-1}x=\frac{1}{2}\tan^{-1}x$ बन जाता है, जिससे $\tan^{-1}x=\frac{\pi}{6}$ प्राप्त होता है, अतः $x=\frac{1}{\sqrt3}$।

  2. Q2. The value of $2\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}$ is equal to:

    $2\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}$ का मान बराबर है:

    • A. $\frac{\pi}{2}$
    • B. $\frac{\pi}{3}$
    • C. $\frac{\pi}{4}$
    • D. $\frac{3\pi}{4}$
    Solution

    Using $2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}$ gives $2\tan^{-1}\frac13=\tan^{-1}\frac34$, and then $\tan^{-1}\frac34+\tan^{-1}\frac17=\tan^{-1}\frac{(3/4+1/7)}{1-(3/28)}=\tan^{-1}1=\frac{\pi}{4}$.

    $2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}$ का प्रयोग करने पर $2\tan^{-1}\frac13=\tan^{-1}\frac34$ प्राप्त होता है, और फिर $\tan^{-1}\frac34+\tan^{-1}\frac17=\tan^{-1}\frac{(3/4+1/7)}{1-(3/28)}=\tan^{-1}1=\frac{\pi}{4}$।

  3. Q3. The principal value of $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ is:

    $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ का मुख्य मान है:

    • A. $-\frac{5\pi}{6}$
    • B. $\frac{5\pi}{6}$
    • C. $\frac{7\pi}{6}$
    • D. $\frac{\pi}{6}$
    Solution

    The range of $\cos^{-1}$ is $[0,\pi]$, so the answer must lie there. Since $\cos\frac{7\pi}{6}=-\frac{\sqrt3}{2}$, the angle in $[0,\pi]$ with this cosine is $\frac{5\pi}{6}$.

    $\cos^{-1}$ का परिसर $[0,\pi]$ है, अतः उत्तर इसी में होना चाहिए। चूँकि $\cos\frac{7\pi}{6}=-\frac{\sqrt3}{2}$ है, $[0,\pi]$ में वह कोण जिसका यह कोज्या (cosine) है, $\frac{5\pi}{6}$ है।

Frequently asked questions

How many questions are in the Bihar Board Class 12 Science Mathematics · Ch 2: Inverse Trigonometric Functions?

30 questions for 30 marks, to be answered in 20 minutes.

Is there negative marking in the Bihar Board Class 12 Science Mathematics · Ch 2: Inverse Trigonometric Functions?

No — wrong answers are not penalised in this test.

Which topics does the Bihar Board Class 12 Science Mathematics · Ch 2: Inverse Trigonometric Functions cover?

Inverse Trigonometric Functions (30 questions).

Is the Bihar Board Class 12 Science Mathematics · Ch 2: Inverse Trigonometric Functions free?

It is part of the BSEB Class 12 Science Test Series 2027 — Chapter-wise + Full Tests (19 tests, ₹149). You can first try Hindi · Ch 1: Chandradhar Sharma Guleri for free.

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