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ISC Class 12 Linear Programming — practice questions

Explore 50 free Linear Programming questions in ISC Class 12 Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Linear Programming questions and exercises with answers

Showing questions 1–30 of 50.

The principal value of $\sin^{-1}(x)$ lies in the interval:Read the answer and worked solution →The value of $\sin^{-1}\left(\dfrac{1}{2}\right)$ is:Read the answer and worked solution →$\cos^{-1}(-x) = $Read the answer and worked solution →If $\sin^{-1}(x) + \sin^{-1}(y) = \dfrac{\pi}{2}$, then $x^2 + y^2$ equals:Read the answer and worked solution →If $\tan^{-1}\left(\dfrac{1}{2}\right) + \tan^{-1}\left(\dfrac{1}{3}\right) = ?$Read the answer and worked solution →Range of arcsin(x) (principal value):Read the answer and worked solution →Range of arccos(x):Read the answer and worked solution →Range of arctan(x):Read the answer and worked solution →Value of arcsin(1):Read the answer and worked solution →arcsin(x) + arccos(x) =Read the answer and worked solution →arctan(x) + arctan(1/x) (for x > 0) =Read the answer and worked solution →Domain of f(x) = arccos(x):Read the answer and worked solution →d/dx arcsin(x) =Read the answer and worked solution →d/dx arctan(x) =Read the answer and worked solution →Value of arcsin(-1/2):Read the answer and worked solution →Value of arccos(-1):Read the answer and worked solution →sin(arccos(x)) =Read the answer and worked solution →tan(arcsin(x)) =Read the answer and worked solution →Value of cos(arcsin(3/5)):Read the answer and worked solution →cos⁻¹(cos(5π/4)) =Read the answer and worked solution →tan⁻¹(tan(3π/4)) =Read the answer and worked solution →Which function is an antiderivative of $\cos x$?Read the answer and worked solution →The integral $\int x^n\,dx$ (for $n\neq -1$) equalsRead the answer and worked solution →A feasible region is bounded if it can be:Read the answer and worked solution →The region other than the feasible region is called:Read the answer and worked solution →Theorem 1 describes the feasible region as a:Read the answer and worked solution →Corner points of a feasible region are also called its:Read the answer and worked solution →Linear programming is a class of which problems?Read the answer and worked solution →The region satisfying all constraints is called:Read the answer and worked solution →Some problems minimise a linear function subject to given:Read the answer and worked solution →

Worked practice questions

The principal value of $\sin^{-1}(x)$ lies in the interval:

  1. $[0, 2\pi]$
  2. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
  3. $[-\pi, \pi]$
  4. $[0, \pi]$
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Correct answer: B. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

The principal-value range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. This is the unique interval on which $\sin$ is monotonic and covers its full range $[-1, 1]$. Quick reference: - $\sin^{-1}, \tan^{-1}, \csc^{-1}$: range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (with closed/open variations). - $\cos^{-1}, \sec^{-1}, \cot^{-1}$: range $[0, \pi]$ (option A is for these).

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The value of $\sin^{-1}\left(\dfrac{1}{2}\right)$ is:

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{2}$
  3. $\dfrac{\pi}{4}$
  4. $\dfrac{\pi}{3}$
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Correct answer: A. $\dfrac{\pi}{6}$

$\sin\theta = \dfrac{1}{2}$ has the principal solution $\theta = \dfrac{\pi}{6}$ (30°), which lies in the standard range $[-\pi/2, \pi/2]$. Memorise the three classic values: $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$, $\sin^{-1}\dfrac{1}{\sqrt 2} = \dfrac{\pi}{4}$, $\sin^{-1}\dfrac{\sqrt 3}{2} = \dfrac{\pi}{3}$.

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$\cos^{-1}(-x) = $

  1. $-\cos^{-1}(x)$
  2. $2\pi - \cos^{-1}(x)$
  3. $\pi - \cos^{-1}(x)$
  4. $\cos^{-1}(x)$
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Correct answer: C. $\pi - \cos^{-1}(x)$

For $x \in [-1, 1]$: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$. Reason: $\cos(\pi - y) = -\cos y$. Setting $y = \cos^{-1}(x)$ gives $\cos(\pi - \cos^{-1} x) = -x$ so applying $\cos^{-1}$ to both sides (both LHS and target lie in $[0, \pi]$) gives the result. Compare $\sin^{-1}(-x) = -\sin^{-1}(x)$ (option A would apply here). The odd-function form works for $\sin^{-1}$ but not $\cos^{-1}$ because their principal ranges differ.

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