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Gujarati JEE Mathematics : નિશ્વાયક

Multiple Choice Questions

1. જો 1, w, w² એ 1 નાં ઘન મુળ હોય, તો $open vertical bar table row bold 1 cell bold w to the power of bold n end cell cell bold w to the power of bold 2 bold n end exponent end cell row cell bold w to the power of bold n end cell cell bold w to the power of bold 2 bold n end exponent end cell bold 1 row cell bold w to the power of bold 2 bold n end exponent end cell bold 1 cell bold w to the power of bold n end cell end table close vertical bar bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.$

જો x + 2ay + az = 0; x + 3by + bz = 0, x + 4cz + cz = 0 સમીકરણો સુસંગત હોય, તો a, b, c એ ........ છે.

સ્વરિત શ્રેણીમાં
સમાંતર શ્રેણીમાં
સમગુણોત્તર શ્રેણીમાં
ત્રણમાંથી એક પણ નહી

3. a ની કઈ કિંમત માટે સમીકરણ સંહિત
ax + y + z = a - 1
x+ ay + z = a - 1
x + y + az = a - 1 ને એક પણ ઉકેલ નથી.

-2
2 અથવા 1
-2 સિવાય
1

-2

Tips: -

ધારો કે $open vertical bar table row bold a bold 1 bold 1 row bold 1 bold a bold 1 row bold 1 bold 1 bold a end table close vertical bar bold space bold equals bold space bold 0 bold space$

$bold therefore bold space bold left parenthesis bold a bold space bold plus bold space bold 2 bold right parenthesis bold space open vertical bar table row bold 1 bold 1 bold 1 row bold 1 bold a bold 1 row bold 1 bold 1 bold a end table close vertical bar bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold C subscript bold 21 bold left parenthesis bold 1 bold right parenthesis bold comma bold space bold C subscript bold 31 bold left parenthesis bold 1 bold right parenthesis bold space bold અન ે bold space bold C subscript bold 1 bold space open parentheses fraction numerator bold 1 over denominator bold a bold plus bold 2 end fraction close parentheses bold therefore bold space bold left parenthesis bold a bold plus bold 2 bold right parenthesis bold space bold left square bracket bold a to the power of bold 2 bold space bold minus bold space bold 1 bold space bold minus bold space bold 1 bold space bold left parenthesis bold a bold space bold minus bold space bold 1 bold right parenthesis bold space bold plus bold space bold 1 bold space bold left parenthesis bold 1 bold space bold minus bold space bold a bold right parenthesis bold right square bracket bold space bold equals bold space bold 0 bold space bold space bold space bold space bold space bold left parenthesis bold a bold space bold plus bold space bold 2 bold right parenthesis bold space bold left parenthesis bold a to the power of bold 2 bold space bold minus bold space bold 1 bold space bold minus bold space bold a bold space bold plus bold space bold 1 bold space bold plus bold space bold 1 bold space bold minus bold space bold a bold right parenthesis bold space bold equals bold space bold 0 bold space bold space bold space bold space bold left parenthesis bold a bold space bold plus bold space bold 2 bold right parenthesis bold space bold left parenthesis bold a to the power of bold 2 bold space bold minus bold space bold 2 bold a bold space bold plus bold space bold 1 bold space bold right parenthesis bold space bold equals bold space bold 0 bold. bold space bold space bold space bold આથ ી bold space bold left parenthesis bold a bold space bold plus bold space bold 2 bold right parenthesis bold space bold left parenthesis bold a bold space bold minus bold space bold 1 bold right parenthesis bold space bold equals bold equals bold space bold 0 bold space$

∴ a = -2 અથવા a = 1

a = 1 લેતાં, ત્રણેય સમીકરણ સમાન બને છે. આથી ઉકેલગણ અનંતગણ મળે.

a = -2 લેતાં $open vertical bar table row bold 1 bold 1 cell bold minus bold 3 end cell row cell bold minus bold 2 end cell bold 1 cell bold minus bold 3 end cell row bold 1 cell bold minus bold 2 end cell cell bold minus bold 3 end cell end table close vertical bar bold space bold not equal to bold space bold 0 bold space$ આથી a = -2 માટે x નો છેદ શુન્ય હોવાથી સમીકરણનો એક પણ ઉકેલ નથી.

જો a < 0 અને ax² + 2bx + c નો વિવેચક ઋણ હોય, તો $open vertical bar table row bold a bold b cell bold ax bold plus bold b end cell row bold b bold c cell bold bx bold plus bold c end cell row cell bold ax bold plus bold b end cell cell bold bx bold plus bold c end cell bold 0 end table close vertical bar$ નું મૂલ્ય ......... છે.

ઋણ
(ac-b²) (ax²+2bx+c)
0
ધન

5. જો $bold D bold space bold equals bold space open vertical bar table row bold 1 bold 1 bold 1 row bold 1 cell bold 1 bold plus bold x end cell bold 1 row bold 1 bold 1 cell bold 1 bold plus bold y end cell end table close vertical bar$ જ્યાં x ≠ 0, y ≠ 0 તો D એ ....... x, y ∈ N

x અને y વડે વિભાજ્ય છે.
x વડે વિભાજ્ય છે પરંતુ y વડે વિભાજ્ય નથી.
x વડે વિભાજ્ય નથી પરંતુ y વડે વિભાજ્ય છે.
x અને y પૈકી એક પણ વડે વિભાજ્ય નથી.

6. જો $open vertical bar table row bold a bold b cell bold ax bold plus bold by end cell row bold b bold c cell bold ax bold plus bold cy end cell row cell bold ax bold plus bold by end cell cell bold bx bold plus bold cy end cell bold 0 end table close vertical bar bold space bold equals bold space bold 0$ અને ax² + 2bxy + cy² ≠ 0 તો.........

a, b, cસમગુણોત્તર શ્રેણીમાં છે.
a, b, c સમાંતર શ્રેણીમાં છે.
a, b, c સ્વરિત શ્રેણીમાં છે.
a, b, c એ સમગુણોત્તર, સમાંતર કે સ્વરિત શ્રેણીમાં નથી.

જો $open vertical bar table row bold a cell bold a to the power of bold 2 end cell cell bold 1 bold plus bold a to the power of bold 3 end cell row bold b cell bold b to the power of bold 2 end cell cell bold 1 bold plus bold b to the power of bold 3 end cell row bold c cell bold c to the power of bold 2 end cell cell bold 1 bold plus bold c to the power of bold 3 end cell end table close vertical bar bold space bold equals bold space bold 0$ અને સદિશો (1, a, a²) ; (1, b, b²) અને (1, c, c²) એ સમતલીય હોય, તો abc = ......

1
2
0
-1

8. જો, w એ 1 નું ઘન મૂળ હોય, અને w ≠ 1 $open vertical bar table row bold 1 cell bold 1 bold plus bold i bold plus bold w to the power of bold 2 end cell cell bold w to the power of bold 2 end cell row cell bold 1 bold minus bold i end cell cell bold minus bold 1 end cell cell bold w to the power of bold 2 bold minus bold 1 end cell row cell bold minus bold i end cell cell bold minus bold 1 bold plus bold w bold minus bold i end cell cell bold minus bold 1 end cell end table close vertical bar bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space$

9. જો s = p + q + r, તો $open vertical bar table row cell bold s bold plus bold r end cell bold p bold q row bold r cell bold s bold plus bold p end cell bold q row bold r bold p cell bold s bold plus bold q end cell end table close vertical bar$ ની કિંમત ....... છે.

s³
2s³
2s²
3s³

10.

જો a²+ b² + c² + 2 = 0 અને $bold f bold left parenthesis bold x bold right parenthesis bold space bold equals bold space open vertical bar table row cell bold 1 bold plus bold a to the power of bold 2 bold x end cell cell bold left parenthesis bold 1 bold plus bold b to the power of bold 2 bold right parenthesis bold x end cell cell bold left parenthesis bold 1 bold plus bold c to the power of bold 2 bold right parenthesis bold x end cell row cell bold left parenthesis bold 1 bold plus bold a to the power of bold 2 bold right parenthesis bold x end cell cell bold 1 bold plus bold b to the power of bold 2 bold x end cell cell bold left parenthesis bold 1 bold plus bold c to the power of bold 2 bold right parenthesis bold x end cell row cell bold left parenthesis bold 1 bold plus bold a to the power of bold 2 bold right parenthesis bold x end cell cell bold left parenthesis bold 1 bold plus bold b to the power of bold 2 bold right parenthesis bold x end cell cell bold 1 bold plus bold c to the power of bold 2 bold x end cell end table close vertical bar$ તો, f(x) એ ..........ઘાતવાળી બહુપદી થાય.

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Gujarati JEE Mathematics : નિશ્વાયક

Multiple Choice Questions

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