Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen Exemplar Solution Maths Mathematical Reasoning

Which of the following sentences are statements? Justify.

i. A triangle has three sides.

ii. 0 is a complex number.

iii. Sky is red.

iv. Every set is an infinite set.

v. 15 + 8 > 23.

vi. y + 9 = 7.

vii. Where is your bag?

viii. Every square is a rectangle.

ix. Sum of opposite angles of a cyclic quadrilateral is 180°.

x. sin²x + cos² x = 0.

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Payal Gupta | Contributor-Level 10

7 months ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} A s w e k n o w, a s t a t e m e n t i s a s e n t e n c e w h i c h i s e i t h e r t r u e o r f a l s e b u t n o t b o t h simultaneously . \\ (i) I t i s t r u e s t a t e m e n t . A s t r i a n g l e i s a c l o s e d p o l y g o n o f t h r e e s i d e s . \\ (i i) I t i s t r u e s t a t e m e n t . A s 0 + 0 i \\ (i i i) I t i s f a l s e s t a t e m e n t . A s s k y i s b l u e . \\ (i v) I t i s f a l s e s t a t e m e n t . N o, e v e r y s e t i s n o t a n infinite s e t . \\ (v) I t i s f a l s e s t a t e m e n t . 15 + 8 = 23 > 23 \\ (v i) I t i s n o t a s t a t e m e n t . B e c a u s e y + 9 = 7, t h e v a l u e o f y i s n o t g i v e n . \\ (v i i) I t i s n o t a s t a t e m e n t . I t i s a q u e s t i o n . \\ (v i i i) I t i s t r u e s t a t e m e n t . A s s q u a r e i s a rectangle w h o s e a l l s i d e s a r e e q u a l . \\ (i x) I t i s t r u e s t a t e m e n t . \\ (x) I t i s f a l s e s t a t e m e n t . A s s i n^{2} x + c o s^{2} x = 1 \end{array}$

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Let $Δ, \nabla \in {\land, \lor}$ be such that $p \nabla q \Rightarrow ((p Δ q) \nabla r)$ is a tautology. Then $(p \nabla q) Δ r$ is logically equivalent to:

Case – I $Δ \equiv \nabla \equiv \land$

$(p \land q) \to ((p \land q) \land r)$

it can be false if r is false,

so not a tautology

&n

Negation of the Boolean statement $(p \lor q) \Rightarrow ((\sim r) \lor p)$ is equivalent to

pva $(\sim r v p)$

$\sim (p \lor q) \lor (\sim r v p)$

its negation as asked in question

$(p \lor q) \land (\sim p \land r)$

= $(p \land \sim p \land r) \lor (q \land r \land \sim p)$

= $(p \land r \land \sim p) [a s p \land \sim p i s f a l s e]$

The remainder when (2021)²⁰²³ is divided by 7 is:

$2021 \equiv - 2$ mod (7)

$\Rightarrow {(2021)}^{2023} \equiv {(- 2)}^{2023} m o d (7)$ …… (i)

Now, ${(- 2)}^{3} \equiv - 1 m o d (7)$

$\Rightarrow {(- 2)}^{2023} \equiv (- 2) m o d (7) \equiv 5 m o d (7)$ ……. (ii)

(i) & (ii)

$\Rightarrow {(2021)}^{2023} \equiv 5 m o d (7)$

$∴$ Remainder = 5

Let p, q, r be three logical statements. Consider the compound statements

$S_{1} : ((\sim p) \lor q) \lor ((\sim p) \lor r) a n d$

$S_{2} : p \to (q \lor r)$

Then, which of the following is NOT true?

kindly consider the following Image

The statement $(\sim (p \Leftrightarrow \sim q)) \land q i s :$

$\sim (p \Leftrightarrow \sim q) \land q = (p \Leftrightarrow \sim q) \land q i s :$

$∴ (\sim (P \Leftrightarrow \sim Q)) \land$ q is equivalent to $(p \Rightarrow q) \land$

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Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen 2025

Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen 2025

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