Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 451-460 of 600 questions.
If (f(x)=x-2-1) and (g(x)=2x+3) then what is the correct form of ((f+g)(x))?
A. \(x^2+2x+2\)
((f+g)(x)=f(x)+g(x)=x-2-1+2x+3=x-2+2x+2)। परीक्षा में समान पदों को ध्यान से जोड़ें। / ((f+g)(x)=f(x)+g(x)=x-2-1+2x+3=x-2+2x+2). In exams combine like terms carefully.
Login to save your score, XP, coins and progress.
If (f(x)=3x-5) and (g(x)=x-2+4) then what is ((g-f)(x))?
A. \(x^2-3x+9\)
((g-f)(x)=g(x)-f(x)=x-2+4-(3x-5)=x-2-3x+9)। घटाते समय पूरे दूसरे फलन पर ऋण लगाएँ। / ((g-f)(x)=g(x)-f(x)=x-2+4-(3x-5)=x-2-3x+9). Put a minus before the whole second function.
Login to save your score, XP, coins and progress.
If (f(x)=x+2) and (g(x)=x-3) then ((fg)(x)) is equal to which expression?
A. \(x^2-x-6\)
((fg)(x)=f(x)g(x)=(x+2)(x-3)=x-2-x-6)। फलनों के गुणन में बीजगणितीय विस्तार सही रखें। / ((fg)(x)=f(x)g(x)=(x+2)(x-3)=x-2-x-6). In product of functions expand algebraically with care.
Login to save your score, XP, coins and progress.
A. (x+2), \(x\ne 2\)
(\left\(\frac{f}{g}\right\)(x)=\frac{x-2-4}{x-2}=x+2) पर \(x\ne 2\)। सरल करने के बाद भी मूल हर का प्रतिबंध न भूलें। / (\left\(\frac{f}{g}\right\)(x)=\frac{x-2-4}{x-2}=x+2) but \(x\ne 2\). Even after simplifying do not forget the original denominator restriction.
Login to save your score, XP, coins and progress.
If (f(x)=\frac{1}{x-1}) and (g(x)=\frac{1}{x+1}) then what is the domain of ((f+g)(x))?
A. \(\mathbb{R}-{-1,1})
दोनों हर शून्य नहीं हो सकते इसलिए \(x\ne 1\) और \(x\ne -1\)। जोड़ में सामान्य प्रांत दोनों प्रांतों का प्रतिच्छेद होता है। / Both denominators cannot be zero so \(x\ne 1\) and \(x\ne -1\). For addition the common domain is the intersection of both domains.
Login to save your score, XP, coins and progress.
If (f(x)=\sqrt{x-2}) and (g(x)=\sqrt{5-x}) then what is the domain of ((f+g)(x))?
A. \([2,5])
वर्गमूल के लिए \(x-2\ge 0\) और \(5-x\ge 0\) चाहिए इसलिए \(2\le x\le 5\)। ऐसे प्रश्नों में दोनों शर्तें साथ लगाएँ। / For square roots we need \(x-2\ge 0\) and \(5-x\ge 0\) so \(2\le x\le 5\). In such questions apply both conditions together.
Login to save your score, XP, coins and progress.
If (f(x)=x-2+1) and (g(x)=x-4) then what is (\(f\circ g\)(x))?
A. \(x^2-8x+17\)
(\(f\circ g\)(x)=f(g(x))=(x-4)2+1=x-2-8x+17)। संयोजन में पहले अंदर वाला फलन लगाएँ। / (\(f\circ g\)(x)=f(g(x))=(x-4)2+1=x-2-8x+17). In composition apply the inner function first.
Login to save your score, XP, coins and progress.
If (f(x)=2x+1) and (g(x)=x-2) then what is (\(g\circ f\)(x))?
A. \(4x^2+4x+1\)
(\(g\circ f\)(x)=g(f(x))=(2x+1)2=4x-2+4x+1)। ((a+b)2) का विस्तार सही करें। / (\(g\circ f\)(x)=g(f(x))=(2x+1)2=4x-2+4x+1). Expand ((a+b)2) correctly.
Login to save your score, XP, coins and progress.
A. (1), \(x\ne 1,-1\)
गुणन से ((fg)(x)=1) मिलता है लेकिन मूल हरों से \(x\ne 1,-1\)। सरलीकरण प्रतिबंधों को मिटाता नहीं है। / Multiplication gives ((fg)(x)=1) but original denominators require \(x\ne 1,-1\). Simplification does not remove restrictions.
Login to save your score, XP, coins and progress.
If (f(x)=|x|) and (g(x)=x) then which statement is correct for ((f-g)(x))?
A. (0) जब \(x\ge 0\) और (-2x) जब (x<0)
यदि \(x\ge 0\) तो (|x|-x=0) और यदि (x<0) तो (|x|-x=-x-x=-2x)। मापांक में चिन्ह के अनुसार केस बनाएं। / If \(x\ge 0\) then (|x|-x=0) and if (x<0) then (|x|-x=-x-x=-2x). For modulus use cases according to sign.
Login to save your score, XP, coins and progress.
Use Google login or mobile OTP. Admin can block users and control login providers.