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CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for
Class 10 Class 12

ગણ, સંબંધ અને વિધેય

જો 2 ≤ i ≤ 4; i ∈ A, 1 ≤ j ≤ 5; j ∈ B તથા 2 ≤ k ≤ 4; k ∈ C હોય તેમજ S₉ { (a_i, b_j, c_k)| i + j + k = 9} હોય તો S₉ ના કુલ ઘટકો ....... હોઈ શકે.

જો U = {x|x⁵ - 6x⁴ + 11x³ - 6x² = 0; x ∈ R} હોય તથા A = {x|x² - 5x + 6 = 0; x ∈ R}અને B {x | x² - 3x + 2 = 0; x ∈ R }હોય, તો નીચેના જોડકાં જોડો :

(i) -(a), (ii) - (b), (iii) - (c), (iv) - (d)
(i) - (d), (ii) - (a), (iii) - (b), (iv) - (c)
(i) - (c), (ii) - (b), (iii) - (a), (iv) - (d)
આપેલ પૈકી એક પણ નહી

10000 કુટુંબની વસતિ ધરાવતા એક શહેરમાં સર્વેક્ષણ પરથી માલૂમ પડ્યું કે 40 % કુટુંબ વર્તમાનપત્ર P વાંચે છે, 20 % કુટુંબ વર્તમાનપત્ર Q વાંચે છે તથા 10 % કુટુંબ વર્તમાનપત્ર R વાંચે છે. તેમજ 5 % કુટુંબ વર્તમાનપત્ર P તથા Q વાંચે છે, 3 % કુટુંબ વર્તમાનપત્ર Q તથા R વાંચે છે તેમજ 4 % કુટુંબ વર્તમાનપત્ર P તથા R વાંચે છે. જ્યારે 2 % કુટુંબ ત્રણેય વર્તમાનપત્ર વાંચે છે તો માત્ર વર્તમાનપત્ર P વાંચતા હોય તેવાં કુટુંબની સંખ્યા ........... હોય.

3200
3300
3100
3000

જો A = {(x, y) | x² + y² = 25; x, y ∈ R } અને B = {(x, y) | 9x² + y² = 144, x, y ∈ R } હોય તો A ∩ B નાં ઘટકોની સંખ્યા ...... મળે.

$bold A bold space bold equals bold space bold left curly bracket bold x bold vertical line bold 2 bold sin bold space bold x bold space bold minus bold space square root of bold 3 bold space bold equals bold space bold 0 bold semicolon bold space bold 0 bold space bold less than bold space bold x bold space bold less than bold space bold pi bold space bold right curly bracket$ તથા $bold B bold space bold equals bold space bold left curly bracket bold x bold vertical line bold 2 bold space bold cos bold space bold x bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold semicolon bold space bold 0 bold space bold less than bold space bold x bold space bold less than bold space bold 2 bold pi bold space bold right curly bracket$ હોય, તો A ∩ B શોધો.

$open curly brackets fraction numerator 2 straight pi over denominator 3 end fraction close curly brackets$
$open curly brackets fraction numerator 5 straight pi over denominator 6 end fraction close curly brackets$
$open curly brackets straight pi over 3 close curly brackets$
$up diagonal strike 0$

open curly brackets fraction numerator 2 straight pi over denominator 3 end fraction close curly brackets

Tips: -

bold A bold space bold equals bold space bold left curly bracket bold x bold vertical line bold 2 bold sin bold space bold x bold space bold minus bold space square root of bold 3 bold space bold equals bold space bold 0 bold space bold right curly bracket bold space bold therefore bold 2 bold space bold sin bold space bold x bold space bold equals bold space square root of bold 3 bold therefore bold space bold sin bold space bold x bold space bold equals bold space fraction numerator square root of bold 3 over denominator bold 2 end fraction bold space bold greater than bold space bold 0

હવે $bold x bold space bold element of bold space bold left parenthesis bold 0 bold comma bold space bold pi bold right parenthesis$ હોવાથી

$bold x bold space bold equals bold space bold pi over bold 3$ અથવા $bold x bold space bold equals bold space fraction numerator bold 2 bold pi over denominator bold 3 end fraction$

$bold therefore bold space bold A bold space bold equals bold space open curly brackets bold pi over bold 3 bold comma bold space fraction numerator bold 2 bold pi over denominator bold 3 end fraction close curly brackets$

$bold B bold space bold equals bold space bold left curly bracket bold x bold space bold vertical line bold space bold 2 bold space bold cos bold space bold x bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold right curly bracket bold space bold therefore bold space bold 2 bold space bold cos bold space bold x bold space bold equals bold space bold minus bold 1 bold therefore bold space bold cos bold space bold x bold space bold equals bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold space bold space bold less than bold space bold 0$

વળી, $bold x bold space bold element of bold space bold left parenthesis bold 0 bold comma bold space bold 2 bold pi bold right parenthesis$ આપેલ છે.

પરંતુ cos x < 0 હોવાથી, $bold x bold space bold element of bold space open parentheses bold pi over bold 2 bold comma fraction numerator bold 3 bold pi over denominator bold 2 end fraction close parentheses$ મળે.

$bold therefore bold space bold x bold space bold equals bold space bold pi bold space bold minus bold space bold pi over bold 3$ અથવા, $bold pi bold space bold plus bold space bold pi over bold 3$